ROUTERA


Electromagnetic Induction

Class 12th Concepts Of Physics Part 2 HC Verma Solution



Short Answer
Question 1.

A metallic loop is placed in a non-uniform magnetic field. Will an emf be induced in the loop?


Answer:


Given that the magnetic field is non-uniform. The magnetic field will not be uniform throughout the area of the loop. Due to non-uniformity, the magnetic flux lines will be random. In random motion, the filed does not vary with time. If there is no change in the magnetic field, it can’t induce emf in the loop. So if the metallic loop is placed in a non-uniform magnetic field, the filed will not induce emf in it. Thus, Non-uniform magnetic field does not induce the emf.



Question 2.

An inductor is connected to a battery through a switch. Explain why the emf induced in the inductor is much larger when the switch is opened as compared to the emf induced when the switch is closed.


Answer:

In the above circuit, the inductor is connected to a battery through a switch. In that when the switch is closed, the current will induce in the circuit. The magnetic flux will be increased. Change in flux will induce emf. When the switch has opened the drop in the current occur is more than the increase in the current when the switch is closed.


Due to this large amount of emf is induced in the inductor when the switch is closed as compared to when the switch is opened.



Question 3.

The coil of a moving-coil galvanometer keeps on oscillating for a long time if it is deflected and released. If the ends of the coil are connected together, the oscillation stops at once. Explain.


Answer:

If the two ends of the coil of the moving-coil galvanometer is connected together, then it will act as a closed loop. thus, the coil no longer acts as an inductor. Therefore, all oscillations stop at once.


While If the ends of the coil are not connected, then the coil will act as an indicator, it will oscillate up to the current in it decays slowly.



Question 4.

A short magnet is moved along the axis of conducting loops. Show that the loop repels the magnet if the magnet is approaching the loop and attracts the magnet if it is going away from the loop.


Answer:

Lenz's law: The direction of the induced current is such that it opposes the magnetic field that has induced it.



See fig.(a). In that, the north pole is facing towards the loop. According to Lenz’s law, the direction of induced current will be anticlockwise. And the magnet is coming towards the loop. Then the flux through the loop will increase; it will create a magnetic field. This newly generated magnetic field will cancel the original magnetic field. Therefore the loops get repelled.


See fig.(b) in which the magnetic is moving away from the loop. It will lead to a decrease in the magnetic field intensity. Therefore, flux through the loop decreases. Therefore, flux through the loop decreases. According to Lenz's law in induced current produces a magnetic field in the opposite direction of the original field. Hence the loop attracts the magnet.



Question 5.

Two circular loops are placed coaxially but separated by a distance. A battery is suddenly connected to one of the loops establishing a current in it. Will there be a current induced in the other loop? If yes, when does the current start and when does it end? Do the loops attract each other or do they repel?




Answer:

Let us consider the loops as A and B. When the battery is connected to loop A; the current will flow in a clockwise direction. So the direction of the magnetic field due to the current will be towards left as seen from the loop B. Due to a sudden change in flux through loop A, a current will induce in loop B. But it will only be induced for a moment when the current suddenly jumps from zero to a constant value. After it attained a constant value, there will be no induced current in loop B.


According to the Lenz's law, the direction of the induced current is such that it opposes the magnetic field that has induced it. So the induced current in loop B will be in the opposite direction to the magnetic field of loop A. So the direction of induced current in loop B will be in anticlockwise direction. And the current through loop B will end when the current through loop A becomes zero. Because the directions of the currents in the loops are opposite, they will repel each other.



Question 6.

The battery discussed in the previous question is suddenly disconnected. It a current induced in the A pivoted aluminium bar falls much more slowly through a small region containing a magnetic field than a similar bar of insulating material. loop? If yes, when does it start and when does it end?


Answer:

When the battery is suddenly disconnected, a current induced in B due to a sudden change in the flux through it. But it is only induced for a moment when the current suddenly falls to zero. According to the Lenz's law, the direction of the induced current is such that it opposes the magnetic field that has induced it. So that the induced current is such that it increases the decrease magnetic field. Therefore, if the current in loop A is in a clockwise direction, then the induced current in loop B will also be the clockwise direction. Hence two loops will attract each other.


Conclusion: If two circular loops are placed coaxially. A battery is connected to one of the loops. After some time, the battery is disconnected then the loops will attract each other.



Question 7.

If the magnetic fields outside a copper box are suddenly changed, what happens to the magnetic field inside the box? Such low-resistivity metals are used to form enclosures which shield objects inside them against varying magnetic fields.


Answer:

If the magnetic field is suddenly changing, it will induce eddy currents on the walls of the copper box. And because of these eddy currents there will be a magnetic field and that will be in the opposite direction. Copper is a good conductor of electricity, so a magnetic field due to eddy currents will have strong strength. This newly generated magnetic field will cancel the original magnetic field. So the magnetic field inside the box will become zero. In this way, the copper box will become a shield and protect the objects inside in it form varying magnetic fields.



Question 8.

Metallic (non-ferromagnetic) and nonmetallic particles in solid waste may be separated as follows. The waste is allowed to slide down an incline over permanent magnets. The metallic particles slow down as compared to the nonmetallic ones and hence are separated. Discuss the role of eddy currents in the process.


Answer:

When solid waste (metallic and non-metallic particles) allowed to slide over a permanent magnet, an emf will be induced in metallic particles. According to the Lenz's law, the direction of the induced current is such that it opposes the magnetic field that has induced it. So the induced emf in the metallic particles will oppose the downward notion along the inclined plane of the permanent magnet. And non-metallic particles are free from these effects. In this way, metallic particles slow down and get separated from the non-metallic particles.



Question 9.

A pivoted aluminium bar falls much more slowly through a small region containing a magnetic field than a similar bar of insulating material. Explain.


Answer:

Non-metallic or insulating materials are free of the effects of induced eddy currents or induced emf. According to the Lenz's law, the direction of the induced current is such that it opposes the magnetic field that has induced it. Means the induced eddy current opposes its cause. That’s why an aluminium bar will falls slowly through a small region containing a magnetic field.



Question 10.

A metallic bob A oscillates through the space between the poles of an electromagnet (figure). The oscillations are more quickly damped when the circuit is on, as compared to the case when the circuit is off. Explain.




Answer:

When the circuit is on, eddy currents will produce on the surface of the metallic Bob. Eddy currents will generate thermal energy. Thermal energy comes as the loss of the kinetic energy of the Bob. Therefore, oscillations are more quickly damped when the circuit is on compared to when the circuit is of the bob.


Hence, oscillation is more quickly damped when the circuit is on compared to when the circuit is off.



Question 11.

Two circular loops are placed with their centres separated by a fixed distance. How would you orient the loops to have

(a) the largest mutual inductance

(b) the smallest mutual inductance?


Answer:

(a) The largest mutual induction


Mutual induction will be larger when the two loops are placed coaxially. Then the flux through a loop due to another loop is large. Hence, we will get the largest mutual inductance.


(b) The mutual inductance will be small when the two loops are placed such that their axis are perpendicular to each other. The flux through the loop due to another loop will be small. Therefore, we will get the smallest mutual inductance.


Thus, we can conclude that to get the largest mutual inductance the loop has to place coaxially and to get smallest mutual inductance, the axis of the loops has to place perpendicularly.



Question 12.

Consider the self-inductance per unit length of solenoids at its centre and that near its ends. Which of the two is greater?


Answer:

Self-inductance is given by



Self-inductance per unit length is given by



Where is the permeability of the free space


n is the number of turns per unit length


A is Area of a cross section of the solenoid


L is the self-inductance


l is the length the wire


Therefore, self-inductance at the centre and that near its ends will be the same since the self -induction is independent of the distance of the point from the centre of the solenoid.



Question 13.

Consider the energy density in a solenoid at its centre and that near its ends. Which of the two is greater?


Answer:

Energy density is given by


Energy density,


Where


U is the energy density of the solenoid


B is the magnetic field intensity of the solenoid


is the permeability of the free space.


In the solenoid, energy is stored in the form of magnetic field. For the constant flow of current, the magnetic field inside the solenoid is uniform. So, the energy density in a solenoid is constant. Nothing is greater. The energy density at all the points inside a solenoid is the same.




Objective I
Question 1.

A rod of length ℓ rotates with a small but uniform angular velocity ω about its perpendicular bisector. A uniform magnetic field B exists parallel to the axis of rotation. The potential difference between the centre of the rod and an end is
A. zero

B. 1/8 ωB ℓ2

C. 1/2 ωB ℓ2

D. Bωℓ2


Answer:

Let us consider a small element at a distance from the centre of the rod rotating with angular velocity about its axis perpendicular bisector.


Formula used: The emf induced is given by



The emf is induced in the rod because of the small element is



Where is the emf induced in that small element


B is the magnetic field


is the angular velocity of the small element


is the length of the small element


The emf induced across the centre and the end of the rod is found by the integration of the above relation with respect to x from limit o to





The potential difference between the centre of the rod and an end is .


Thus, option B is correct.


Question 2.

A rod of length ℓ rotates with a uniform angular velocity ω about its perpendicular bisector. A uniform magnetic field B exists parallel to the axis of rotation. The potential difference between the two ends of the rod is
A. zero

B. 1/2 Bℓ ω2

C. Bℓ ω2

D. 2Bℓ ω2


Answer:

Let us consider a small element at a distance from the center of the rod rotating with angular velocity about its axis perpendicular bisector.


Formula used: The emf induced is given by



The emf is induced in the rod because of the small element is



Where is the emf induced in that small element


B is the magnetic field


is the angular velocity of the small element


→ length of the small element


The emf induced across the centre and the end of the rod is





The potential difference between the two ends is



The potential difference between the two ends of the rod is zero.


Question 3.

Consider the situation shown in the figure. If the switch is closed and after some time it is opened again, the closed loop will show.



A. an anticlockwise current-pulse

B. a clockwise current-pulse

C. an anticlockwise current-pulse and then a clockwise current-pulse

D. a clockwise current-pulse and then an anticlockwise current-pulse


Answer:

Let us consider tow points on the circuits A and B.

When the switch S is closed, the current will flow through A&B. Due to this magnetic field introduce in the loop.



According to Lenz’s law the induced current is such that it opposes the increase in the magnetic field that induces it.


So, the induced current will be in clockwise direction, and it opposes the increase in the magnetic field in the upward direction in the loop. When the switch S is opened, the current will fall due to this magnetic field in the loop will decrease. According to Lenz's law, the induced current will be in the anti-clockwise direction opposing the direction in the magnetic field in an upward direction in the loop.


If the switch is closed and after some time it is opened again, the closed loop will show a clockwise current-pulse and then an anticlockwise current-pulse


Question 4.

Solve the previous question if the closed loop is completely enclosed in the circuit containing the switch.



A. an anticlockwise current-pulse

B. a clockwise current-pulse

C. an anticlockwise current-pulse and then a clockwise current-pulse

D. a clockwise current-pulse and then an anticlockwise current-pulse


Answer:


According to Lenz’s law, the induced current is such that it opposes the increase in the magnetic field that induces it. So the induced current will be anticlockwise when the switch S is closed. When the switch S is open, the current will suddenly fall. Then the magnetic field at the centre of the loop will decrease. Then the induced emf will be in the clockwise direction.


If the closed loop is completely enclosed in the circuit containing the switch.


Question 5.

A bar magnet is released from rest along the axis of a very long, vertical copper tube. After some time, the magnet.
A. will stop in tube

B. will move with almost content speed

C. will move with an acceleration g

D. will oscillate.


Answer:

The flux linked with the copper tube will change because of the motion of the magnet. This will produce an eddy current in the body of the copper tube. According to the Lenz's law, the direction of the induced current is such that it opposes the magnetic field that has induced it. So the induced Current will be in the opposite direction to the magnet field of the bar magnet. And it will slow the fall of the magnet. Then negative acceleration nothing retarding force will act on the bar magnet. If the velocity of the magnet is increased then the retarding force will also increase. And it will happen up to this retarding force is equal to the force of gravity. Then the total force acting on the bar magnet will become zero. Then the magnet will move with almost a constant speed. So option B is correct.


Question 6.

The figure shows a horizontal solenoid connected to a battery and a switch. A copper ring is placed on a frictionless track, the axis of the ring is along the axis of the solenoid. As the switch is closed, the ring will



A. remain stationary

B. move towards the solenoid

C. move away from the solenoid

D. move towards the solenoid or away from it depending on which terminal (Positive or negative) of the battery is connected to the left end of the solenoid.


Answer:


For the circuit


Emf introduced in the solenoid,


Where


L is the self-inductance of the solenoid


I is the current in the solenoid.


The direction of the current in the loop is clockwise. When the switch is closed the current will flow in the circuit. Therefore, the current on the solenoid will be increase. Then it will induce a current in the copper ring which is placed along a axis of solenoid. According to Lenz’s law the induced current is such that it opposes the increase in the magnetic field that induces it. So the induce current in the copper ring will be anticlockwise. Because of the opposite direction of the currents the ring will repel. So it will move away from the solenoid.


Horizontal solenoid connected to a battery and a switch. A copper ring is placed on a frictionless track, the axis of the ring being along the axis of the solenoid. As the switch is closed, the ring will move away fro the solenoid.


Question 7.

Consider the following statements:

(A) An emf can be induced by moving a conductor in a magnetic field.

(B) An emf can be induced by changing the magnetic field.

A. Both A and B are true

B. A is true but B is false.

C. B is true but A is false

D. Both A and B are false


Answer:

Formula used: emf is given by



Where v is the speed


B is the magnetic field


l is the length


That above formula states that an emf induced by moving a conductor of length l with some velocity v in a magnetic field B.



Where is the electric flux through the conductor


emf induced in the conductor


Formula (2) states that an emf can be induced by changing the magnetic field that causes the change in flux through a conductor in a loop.


So option A is the answer.


Question 8.

Consider the situation shown in the figure. The wire AB is slid on the fixed rails with a constant velocity. If the wire AB is replaced by a semicircular wire, the magnitude of the induced current will



A. increase

B. remain the same

C. decrease

D. increase or decrease depending on whether the semi-circle bulges towards the resistance or away from it.


Answer:

Draw the equivalent circuit


The formula used: The emf induced across A and B is



Where v is the speed


B is the magnetic field


l is the length


The induced emf will serve as a voltage source. The direction of the current is anticlockwise, according to Lenz’s law.


According to Lenz's law, the direction of the induced current is such that it opposes the magnetic field that has induced it.



Putting the value in the above equation, we get



The induced emf depends on the length of the wire but not on the shape of the wire.


If the wire is replaced by a semicircular wire, the induced emf will be same.


Question 9.

The figure shows a conducting loop being pulled out of a magnetic field with speed v. Which of the four plots shown in the figure may represent the power delivered by the pulling agent as a function of the speed v?



A. (a)

B. (b)


Answer:

Formula used: The emf developed across the ends of the loop is given by


Where v is the speed


B is the magnetic field


L is the length


The power delivered to the loop is



Where P is the Power delivered to the loop


R is the resistance


is the emf induced




From the above equation, P is proportional to v2. Curve b showing this relation.



Question 10.

Two circular loops of equal radii are placed coaxially at some separation. The first is cut and a battery is inserted in between to drive a current in it. The current changes slightly because of the variation in resistance with temperature. During this period, the two loops
A. attract each other

B. repel each other

C. do not exert any force on each other

D. attract or repel each other depending on the sense of the current.


Answer:


By the right-hand screw rule, the magnetic field will be towards the left side. Due to the increase in resistance with temperature, the current through the loop will decrease with time. This change in current will induce the current in loop B. According to Lenz’s law, the induced current is such that it opposes the increase in the magnetic field that induces it. So the direction of induced current in loop B will be in a clockwise direction. Because of the directions of currents, they will attract each other.


Thus, Option A is correct.


Question 11.

A small, conducting circular loop is placed inside a long solenoid carrying a current. The plane of the loop contains the axis of the solenoid. If the current in the solenoid is varied, the current induced in the loop is
A. clockwise

B. anticlockwise

C. zero

D. clockwise or anticlockwise depending on whether the resistance is increased or decreased.


Answer:

The magnetic field inside the solenoid is parallel to its axis. If the plane of the loop contains the axis of the solenoid, the angle between the area vector of a circular loop and the magnetic field will be zero.

The formula used: The flux through the circular loop is given by



Where is the electric flux


B is the magnetic field due to the solenoid


A is the Area of the circular loop


is the Angle between magnetic field and area vector



=constant


Then the induced emf is given by




Because BA is constant


The induced emf does not depend on the varying current through the solenoid. The induced emf will be zero for a constant flux through the loop. So no current will be induced in the loop.


Question 12.

A conducting square loop of side ℓ and resistance R moves in its plane with a uniform velocity v perpendicular to one of its sides. A uniform and constant magnetic field B exists along the perpendicular to the plane of the loop, as shown in the figure. The current induced in the loop is



A. Bℓv/R clockwise

B. Bℓv/R anticlockwise

C. 2Bℓv/R anticlockwise

D. zero


Answer:

if we draw the equivalent circuit, it will be look like


The formula used: The emf induces across the ends AB and CD is given by



Where E is the emf


V is the velocity


B is the magnetic field


l is the length


Apply KVL in the loop in fig(b).




Here the current I become zero means that there is no current induced in the loop. So the current induced in the loop will be zero.



Objective Ii
Question 1.

A bar magnet is moved along the axis of a copper ring placed far away from the magnet. Looking from the side of the magnet, an anticlockwise current is found to be induced in the ring. Which of the following may be true?
A. The south pole faces the ring and the magnet moves towards it.

B. The north pole faces the ring and the magnet moves towards it.

C. The south pole faces the ring, and the magnet moves away from it.

D. The north pole faces the ring, and the magnet moves away from it.


Answer:

A bar magnet is moving along the axis of a copper ring. The movement of a magnet there will be a magnetic field. Then the current will be induced in the copper ring. It is observed that the current is in an anticlockwise direction looking from the magnet. Then the magnetic field induced in the copper ring will be towards the observer (We are observing the current direction form the magnet).

According to Lenz’s law, the induced current is such that it opposes the increase in the magnetic field that induces it. If the south pole faces the ring and the magnet is moving towards the then the current induced in the ring will be the clockwise direction. In the same case if the magnet is moving away from the ring, then the current will be in the anticlockwise direction. If the magnet is moving towards the ring and north pole is facing the ring, then the current induced in the ring will be in the anticlockwise direction. If the magnet is moving away from the ring, in this case, the induced current will be in a clockwise direction. So option B and C are correct.


Question 2.

A conducting rod is moved with a constant velocity v in a magnetic field. A potential difference appears across the two ends.
A. If

B. If

C. If

D. None of the above


Answer:

The potential difference across the two ends is given by


Where is the emf or potential difference


v→ velocity of conducting rod


B is the magnetic field


L is the length of the rod


Consider some conditions,


i. The magnetic field is in the perpendicular direction to the velocity of the rod


ii. The magnetic field is in the perpendicular direction to the length of the rod


iii. The velocity of the rod is perpendicular to the direction to the length of the rod


In the above-mentioned conditions, only the potential difference across the two ends will be non zero.


In the given conditions in the A, B and C, the potential difference across the two ends is zero. So option D is correct.


Question 3.

A conducting loop is placed in a uniform magnetic field with its plane perpendicular to the field. An emf is induced in the loop if
A. it is translated

B. it is rotated about its axis

C. it is rotated about a diameter

D. it is deformed


Answer:

Conducting loop is placed in a uniform magnetic field with its plane perpendicular to the field. An emf will be induced only when the magnetic flux changes. If there is no change in magnetic flux, then induced emf will be zero. An emf will be induced in the loop only when

i. The loop is rotated about a diameter


ii. Deforming the loop.


If the loop is deformed, the area of the loop inside the magnetic field changes. So magnetic flux will change leads to induce an emf. On rotating about its axis, the magnetic flux does not change. So an emf induced in the loop is zero.


Thus, Option C and D are correct.


Question 4.

A metal sheet is placed in front of a strong magnetic pole. A force is needed to
A. hold the sheet there if the metal is magnetic

B. hold the sheet there if the metal is non-magnetic

C. move the sheet away from the pole with uniform velocity if the metal is magnetic.

D. move the sheet away from the pole with uniform velocity if the metal is non-magnetic.

Neglect any effect of paramagnetism, diamagnetism and gravity.


Answer:

A metal sheet is placed in front of the strong magnetic pole. This pole has q strong magnetic field. A strong magnetic pole will attract the magnet if the metal is magnetic then a force is needed to hold the metal sheet. If the metal is magnetic or not if we want to move the metal sheet away from that strong magnetic pole we need some force do it. Because of the movement, eddy currents will be induced in the sheet. These eddy currents produce thermal energy. Thermal energy comes at the cost of kinetic energy. Thus, the plates slow down. So we need a force to hold the sheet if the metal is magnetic. And also the sheet needs a force to move away from that strong magnetic filed even it is nonmagnetic.

So option A, C and D are correct.


Question 5.

A constant current i is maintained in a solenoid. Which of the following quantities will increase if an iron rod is inserted in the solenoid along its axis?
A. The magnetic field at the centre

B. magnetic flux linked with the solenoid

C. self-inductance of the solenoid

D. rate of Joule heating.


Answer:

The iron rod has high permeability. So if we insert the iron rod in the solenoid along its axis, the magnetic flux inside the solenoid will increases. The self-inductance of the coil is proportional to the permeability of the material inside the solenoid. Because of an increase in the permeability inside the coil, the self-inductance will also increase. Thus, a constant current I will be maintained in the solenoid. Then

1. magnetic field at the centre


2. Magnetic flux linked with the solenoid


3. self-inductance of the solenoid is increased if an iron rod is inserted in the solenoid along its axis.


Thus, option A, B and C is the correct option.


Question 6.

Two solenoids have identical geometrical construction, but one is made of thick wire and the other of thin wire.

Which of the following quantities are different for the two solenoids?

A. self-inductance

B. rate of Joule heating if the same current goes through them

C. magnetic field energy if the same current goes through them.

D. time constant if one solenoid is connected to one battery and the other is connected to another battery.


Answer:

The two solenoids are identical. Therefore, the self-inductance of the two solenoids is the same.

The formula used: energy stored in the inductor is given by



Where U is the Magnetic energy


L is the self-Inductance


I is the Current


The current in both solenoids is the same. That implies magnetic field energy is also the same.


The power dissipated as heat is given by



Where P is the power


I is the current


R is the Resistance


The time constant is given by



Where t is the time constant


L is the self-Inductance


R is resistance is given by



Where ρ is the resistivity


l is the length of the wire


A is the Area of cross section


The two solenoids are differed by the Area of the cross section A.


=>


=>


=>


Thus, the time constant and Joule heat is different for the two solenoids. And self-inductance and magnetic field are the same for the both.


Question 7.

An LR circuit with a battery is connected at t = 0. Which of the following quantities is not zero just after the connection?
A. Current in the circuit

B. Magnetic field energy in the inductor.

C. Power delivered by the battery

D. Emf induced in the inductor


Answer:

The formula used: At time t, the current in the LR circuit is given by


Where I is the Current


is the emf


R is the resistance


L is the self-inductance


At time the current is given by



At t=0, the current in the circuit is zero.


The magnetic field energy in the inductor is given by



Current is zero at t=0, So magnetic field energy is also zero.


Therefore, power delivers by the battery is also zero at t=0.


At t=0, the current is on the verge to start growing in the circuit. So there will an induced emf at that time to oppose the growing current.


In the LR circuit, at time t=0 there will be an induced emf in the inductor. Exactly at t=0, the current, magnetic field and power delivered by the battery are zero.


Question 8.

A rod AB moves with a uniform velocity v in a uniform magnetic field, as shown in the figure.



A. The rod becomes electrically charged.

B. The end A becomes positively charged.

C. The end B becomes positively charged.

D. The rod becomes hot because of Joule heating.


Answer:

Due to electromagnetic induction, the emf will be induced in the rod. Induced emf is given by the formula end A becomes more positive because the direction of the induced emf is from A to B. This is because of the magnetic field exerts a same force equal to on each of the electrons, where C. According to Fleming’s left hand rule (If a conductor is placed in a magnetic field then the force acting on the conductor is in the perpendicular direction to the magnetic field and current direction) the, end A becomes positive and, end B becomes negatively charged.

Conclusion: A rod AB moves with a uniform velocity v in a uniform magnetic field then the end A becomes positively charged.


Question 9.

L, C and R represent the physical quantities inductance, capacitance and resistance, respectively. Which of the following combinations have dimensions of frequency?
A.

B.

C.

C. C/L


Answer:

The formula used: For RC circuit the time constant is


Where t is the time constant


R is the resistance


C is the capacitance


Frequency


Then the frequency of the RC circuit is given by



For LR circuit the time constant is given by



Where is the time constant


L is the inductance


R is the resistance


Then the frequency is given by



Let’s take a multiplication of f1 and f2





This combination has a dimension of frequency.


The above combination has a dimension of frequency


i. RC has a frequency component


ii. LR has a frequency component


iii. f1f2 has a frequency component


Thus, option A, B and C are correct.


Question 10.

The switches in figure (a) and (b) are closed at t = 0 and reopened after a long time at t = t0.



A. The charge on C just after t = 0 is ϵ C.

B. The charge on C long after t = 0 is ϵ C.

C. The current in L just before t = t0 in ϵ/R

D. The current in L long after t = t0 is ϵ/R.


Answer:

Formula used:

The charge on capacitor at time t, when the switch is closed is



Where Q is the Charge


is the emf


T is the Time period


R is the Resistance


C is the Capacitance


At t=0, the charge on the capacitor is



After a long time , the charge on the capacitor is




The current in inductor L is given by



Where I is the current


is the voltage or potential difference


R is the Resistance


L is the inductance


T is the time period


At initial we take, t=0, the current in the inductor is given by




After a long time = is





The charge on C long after t=0 is


The current in L long after t = t0 is



Exercises
Question 1.

Calculate the dimension of

(a) (b) vBℓ (c)

The symbols have their usual meanings.


Answer:

(a)using faraday’s law of induction



Where ϵ is the emf or voltage


has thus the dimensions of voltage


Voltage is given by formula



Where


W=work done


Q=charge


Dimensions of W=


Dimensions of Q=


The dimensions of voltage can be given as



……(i)


(b) vBl is the motional emf developed due to motion of conductor of length l with velocity v in a magnetic field (B)


Therefore, vBl has same dimensions as of voltage


Therefore, by eqn.(i)



(c) by faraday’s law of electromagnetic induction



Where


ϵ =emf produced


ϕ =flux of magnetic field


thus has same dimensions as of emf or voltage


therefore, by eqn.(i)



Therefore,andall have same dimensional formula equal to



Question 2.

The flux of magnetic field through a closed conducing ϕ = at2 + bt + c.

(a) Writer the S.I. units of a, b and c.

(b) If the magnitudes of a, b, and c are 0.20, 0.40 and 0.60 respectively, find the induced emf at t = 2s.


Answer:

Given:


flux of magnetic field ϕ = at2 + bt + c


magnitudes of a, b and c= 0.20,0.40 and 0.60


(a) We know that,


Dimensions on both sides of an equation are equal and only terms with same dimensions can be added


Therefore


Dimensions of ϕ, at2, bt and c are same


Thus units of a are given by



Now,


by faraday’s law of electromagnetic induction


…(i)


Where


ϵ =emf produced


ϕ =flux of magnetic field


therefore, units of is same as voltage i.e. Volt



Unit of b is given by



Using eqn.(i)



SI unit of c is given by




(b)


We know that


by faraday’s law of electromagnetic induction



Where


ϵ =emf produced


ϕ =flux of magnetic field


here ϵ is given by



Putting the values of a, b and t=2sec we have



Therefore, the induced emf at t=2sec is given by 1.2Volts



Question 3.

(a) The magnetic field in a region varies as shown in figure. Calculate the average induce emf in a conducting loop of area 2.0 × 10–3 m2 placed perpendicular to the field in each of the 10 ms intervals shown.

(b) In which intervals is the emf not constant? Neglect the behavior near the ends of 10 ms intervals.




Answer:

Given:


Area of loop=


We know that


by faraday’s law of electromagnetic induction



Where


ϵ =emf produced


ϕ =flux of magnetic field


so, the average induced emf in the conducting loop between time intervals t1 and t2 is given by


…(ii)


Also


Magnetic flux through the circular ring of area A is given by



Since loop is placed perpendicular to the field,



(i)


Using eqn.(ii)


. (iii)


Putting the values of we get



(ii)


Using eqn.(ii)


. (iii)


Putting the values of we get



(iii)


Using eqn.(ii)


. (iii)


Putting the values of we get



(iv)


Using eqn.(ii)


. (iii)


Putting the values of we get



emf across the intervals


0-10ms =-2mV


10-20ms=-4mV


20-30ms=4mV


30-40ms=2mV


From the graph we can see that flux varies in a nonlinear fashion between time Intervals 10-20 ms and 20-30 ms and hence the derivative of flux wrt time is not constant in the given interval


emf is not constant in time intervals 10-20 ms and 20-30ms



Question 4.

A conducting circular loop having a radius of 5.0 cm, is placed perpendicular to a magnetic field of 0.50 T. It is removed from the field in 0.50 s. Find the average emf produced in the loop during this time.


Answer:

Given:


Radius if circular loop=5.0cm


Magnetic field intensity=0.50T


Area of circular loop


Initial magnetic flux through the loop is given by



(As loop is placed perpendicular to magnetic field)


After loop is removed from the field after time Δt=0.50s the magnetic flux



(As no magnetic field passes through the loop)


We know that,


Average induced emf in time interval Δt is given by



Where


are flux across the cross section at time intervals respectively.


Putting the values of we get




Therefore, average induced emf produced in the loop during this time interval isVolts.



Question 5.

A conducting circular loop of area 1 mm2 is placed co-planarly with a long, straight wire at a distance of 20 cm from it. The straight wire carries as electric current which changes from 10 A to zero in 0.1 s. Find the average emf induced in the loop in 0.1 s.


Answer:

Given:


Area of circular loop


Separation between wire and loop


Current through the wire



We know that magnetic field (B) due to an infinitely long wire carrying a current i at a perpendicular distance d from the wire is given by



Also,


Average induced emf in time interval Δt is given by



Where


are flux across the cross section at time intervals respectively.


Magnetic flux due to magnetic field B through cross section area A is given by



Since magnetic field due to long wire is perpendicular to the circular loop, initial magnetic flux through the loop is given by



After time interval of 0.1s current in the wire becomes zero and hence magnetic field



Average induced emf is given by




Therefore, average emf induced in the loop in time interval of 0.1s is given by 10-10V



Question 6.

A square-shaped copper coil has edges of length 50 cm and contains 50 turns. It is placed perpendicular to a 1.0T magnetic field. It is removed from the magnetic field in 0.25 s and restored in its original place in the next 0.25 s. Find the magnitude of the average emf induced in the loop during

(a) its removal, (b) its restoration and (c) its motion.


Answer:

Given:


Length of side of square=


No. of turns=50


Intensity of magnetic field =1.0T


Average induced emf in time interval Δt is given by


…(i)


Where


are flux across the cross section at time intervals respectively.


Magnetic flux due to magnetic field B through cross section area A is given by



(a) During removal


Initial magnetic flux through the loop



Final magnetic flux through the loop



Time interval Δt=0.25s


Using eqn.(i) we get



Magnitude of average emf during removal is 50V


(b) During restoration


Initial magnetic flux through the loop



Final magnetic flux through the loop



Time interval Δt=0.25s


Using eqn.(i) we get



Magnitude of average emf during restoration is 50V


(c) During the motion


Initial magnetic flux through the loop



Final magnetic flux through the loop



Time interval Δt=0.25+0.25=0.50s


Using eqm. (i) we get



Therefore, magnitude of average emf during its motion is zero



Question 7.

Suppose the resistance of the coil in the previous problem is 25 Ω. Assume that the coil moves with uniform velocity during its removal and restoration. Find the thermal energy developed in the coil during

(a) its removal

(b) its restoration and

(c) its motion.


Answer:

Given


Resistance of the coil


(a) during its removal


the emf induced in the


so the current flowing through the loop is given by



Thermal energy developed in the coil (H) during its removal in time interval Δt=0.25s is given by



Therefore, energy developed in the coil during removal is 25J


(b) during its restoration


The magnitude of emf induced in the loop ϵ =50V


so the current flowing through the loop is given by



Thermal energy developed in the coil (H) during its removal in time interval Δt=0.25s is given by



Therefore, energy developed in the coil during restoration is 25J


(c) during motion


total thermal energy developed =energy developed during removal+ energy developed during restoration



As heat energy is scalar quantity, it is algebraically added


Therefore, total energy developed in coil during its motion is 50J



Question 8.

A conducting loop of area 5.0 cm2 is placed in a magnetic field which varies sinusoidally with time as B = B0 sin ωt where B0 = 0.20 T and ω = 300 s–1. The normal to the coil makes an angle of 60° with the field. Find

(a) the maximum emf induced in the coil,

(b) the emf induced at τ = (π/900) s and

(c) the emf induced at t = (π/600) s.


Answer:

Given:


Area of conducting loop=


Variation of magnetic field with time


Angle of field with normal to coil θ =60°


Magnetic flux due to magnetic field B through cross section area A is given by



Here flux through the loop is given by


. (i)


Also,


by faraday’s law of electromagnetic induction



Where


ϵ =emf produced


ϕ =flux of magnetic field


using eqn.(i)


we get


…(ii)


Since maximum value of cosωt =1


Therefore, maximum value of magnitude of emf induced in the loop is given by



Putting the values of B0 and ω we get,



Therefore, maximum value of emf induced in the coil is 0.015V


(b) from eqn.(ii), we have



At t=π/900 s magnitude of induced emf is given by



Therefore magnitude of induced emf at t=π/900 is


(c) from eqn.(ii), we have



At t=π/600 s magnitude of induced emf is given by



Therefore magnitude of induced emf at t=π/600 is



Question 9.

Figure shows a conducting square loop placed parallel to the pole-faces of a ring magnet. The pole-faces have an area of 1 cm2 each and the field between the poles is 0.10 T. The wires making the loop are all outside the magnetic field. If the magnet is removed in 1.0 s, what is the average emf induced in the loop?




Answer:

Given:


Area of pole faces A=


Magnetic field intensity =0.1T


Time taken to remove the magnet completely


Initial magnetic flux through the loop is given by formula



Since magnetic field through the square loop is perpendicular to the loop above eqn. reduces to


…(i)


When the magnet is removed after 1s the magnetic flux passing through the square loop becomes zero


…(ii)


Average induced emf in time interval Δt is given by


…(iii)


Where


are flux across the cross section at time intervals respectively.


Using eqns.(i), (ii) and (iii) we get,



Putting the values of B, A and Δt in the above eqn.



Therefore average emf induced in the square loop is



Question 10.

A conducting square loop having edges of length 2.0 cm is rotated through 180° about a diagonal in 0.20 s. A magnetic field B exists in the region which is perpendicular to the loop in its initial position. If the average induced emf during the rotation is 20 mV, find the magnitude of the magnetic field


Answer:

Given:


Average induced emf in the loop


Time taken to rotate the loo


Edge length of square loop


Area of square loop


We know that,


Average induced emf in time interval Δt is given by


…(i)


Where


are flux across the cross section at time intervals respectively.


Magnetic flux(ϕ) through the loop is given by the formula




Where B=magnetic field intensity


A=area of cross section


θ =angle between area vector and magnetic field


Initially, angle between area vector and magnetic field is 0°


Therefore, initial flux through the coil is



When it is rotated by 180° flux passing through the coil is given by



Putting this values in eqn.(i) we get,



Putting the values of ϵ, B and Δt in the above eqn.




Therefore, magnitude of magnetic field intensity is 5T



Question 11.

A conducting loop of face-area A and resistance R is placed perpendicular to a magnetic field B. The loop is withdrawn completely from the field. Find the charge which flows through any cross-section of the wire in the process. Note that it is independent of the shape of the loop as well as the way it is withdrawn.


Answer:

Given:


Face area of loop =A


Resistance of loop=R


Magnetic field intensity =B


Magnetic flux(ϕ) through the loop is given by the formula




Where B=magnetic field intensity


A=area of cross section


θ =angle between area vector and magnetic field


initially loop is perpendicular to the applied magnetic field hence initial flux is



Finally, when the loop is withdrawn from the field flux is given by



Now,


Average induced emf in time interval Δt is given by


…(i)


Where


are flux across the cross section at time intervals respectively


Using eqn.(i) we get



Current flowing in the loop is calculated by using formula



Hence the charge (Q) flowing through the loop is



Therefore, charge flowing through any cross-section of the wire is BA/R



Question 12.

A long solenoid of radius 2 cm has 100 turns/cm and carries a current of 5 A. A coil of radius 1 cm having 100 turns and a total resistance of 20 Ω is placed inside the solenoid coaxially. The coil is connected to a galvanometer. If the current in the solenoid is reversed in direction, find the charge flown through the galvanometer.


Answer:

Given:


Radius of solenoid


No. of turns in the solenoid


Current in the solenoid


Radius of second coil


No. of turns in the coil


Resistance of the coil


We know that,


Magnetic field inside solenoid (B) is given by formula



Where,


n=no. of turns per unit length


i=current through solenoid


Magnetic flux(ϕ) through the coil is given by the formula




Where B=magnetic field intensity


A=area of cross section of the coil


θ =angle between area vector and magnetic field


magnetic field inside solenoid is perpendicular to the coil


initially flux through the coil is given by



When the current in the solenoid is reversed in direction of magnetic field gets reversed and flux through the coil now m=becomes




Now,


Average induced emf in time interval Δt is given by


…(i)


Where


are flux across the cross section at time intervals respectively


Putting these values in eqn.(i) we get



Current (i) through the coil of resistance R can be calculated as



Hence the charge (Q) passing through the coil in time Δt is



Putting the values of μ0, I, N, n π r’ and R in above eqn.



Therefore flowing through the galvanometer is



Question 13.

Figure shown a metallic square frame of edge an in a vertical plane. A uniform magnetic field B exists in the space in a direction perpendicular to the plane of the figure. Two boys pull the opposite corners of the square to deform it into a rhombus. They start pulling the corners at t = 0 and displace the corners at a uniform speed u.

(a) Find the induced emf in the frame at the instant when the angles at these corners reduced to 60°.

(b) Find the induced current in the frame at this instant if the total resistance of the frame is R.

(c) Find the total charge which flows through a side of the frame by the time the square is deformed into a straight line.




Answer:

Given:


Edge length of square frame =a


Magnetic field intensity B


Speed of corners of rhombus =u


(a) when the angles at the corner reduce to 60°



We know that,


motional emf produced due to a conductor of length l moving with velocity v in a magnetic field B is given by



Motional emf is produced in each side of rhombus and the effective length of each side is the length perpendicular to velocity of corners.


From the fig. the effective length of each side is



Since velocity is perpendicular to magnetic field the equation of emf induced in each side given by



Total emf induced in all four side is



Therefore, induced emf in the frame when the angles at the corner reduces to 60° is 2uBa


(b) total resistance of the frame =R


hence current flowing in the frame is



Therefore, current flowing in the frame at this instant is 2uBa/R


(c) initially the frame is in form of square of side a and area


at this time flux through this frame is given by



Finally, when the frame reduces to straight line flux passing through the frame reduces to zero



Average emf induced in the frame in time t is given by



The current flowing through the frame is then given by



Where R is the resistance of frame


Hence the charge (Q) flowing through the side of frame in time t is



Therefore total charge flowing through side of frame by the time the frame reduces to straight line is



Question 14.

The north pole of a magnet is brought down along the axis of a horizontal circular coil (figure). As a result, the flux through the coil changes from 0.35 weber to 0.85 weber in an interval of half a second. Find the average emf induced during this period. Is the induced current clockwise or anticlockwise as you look into the coil from the side of the magnet?




Answer:

Given:


Initial flux


Final flux


Time interval


Average induced emf in time interval Δt is given by


…(i)


Where


are flux across the cross section at time intervals respectively.


Putting the values in eqn.(i) we get,



Magnitude of induced emf =1V


From Lenz’s law,


The direction of induced current is such that it opposes the change that has induced it.


According to fig.



The flux in the downward direction increases, so the current is induced such that flux in upward direction increases. Therefore, current is induced in anticlockwise direction


Therefore, average emf induced in the coil is 1V in anti-clockwise direction



Question 15.

A wire-loop confined in a plane is rotated in its own plane with some angular velocity. A uniform magnetic field exists in the region. Find the emf induced in the loop.


Answer:

Let the area of the wire-loop be A and magnetic field intensity be B.


When the wire rotates in its own plate the area through which flux passes remains same and B is also constant.


Hence, the flux passing through the loop remains constant and is given by



Now,


by faraday’s law of electromagnetic induction



Where


ϵ =emf produced


ϕ =flux of magnetic field


as flux is constant (independent of time), its derivative with respect to time is zero. Therefore, emf induced in the loop is zero


therefore, zero emf is induced in the wire-loop



Question 16.

Figure shows a square loop of side 5 cm being moved towards right at a constant speed of 1 cm/s. The front edge enters the 20 cm wide magnetic field at t = 0. Find the emf induced in the loop at

(a) t = 2s, (b) t = 10 s

(c) t = 22 s, (d) t = 30s




Answer:

Given:


Side length of square loop=5cm


Speed of square loop


Width of magnetic field


Magnetic field intensity =0.6T


(a) t=2s


distance moved by the loop


area of the loop under magnetic field =


area of rectangle of length 0.05m and width 0.02m



Now,


Initial magnetic flux through the loop (at t=0)


Final magnetic flux through the loop is given by




Average induced emf in time interval Δt is given by


…(i)


Where


are flux across the cross section at time intervals respectively.


Putting the values of in eqn.(i),



Therefore magnitude of induced emf at t=2s is


(b) t=10s


distance moved by the square loop


at this moment, square loop is completely inside the magnetic field and area of loop through which flux pass


so the flux linkage does not changes with time


and thus from eqn.(i)



Therefore, magnitude of induced emf in the coil at t=10s is zero


(c) t=22s


distance moved by the loop


the loop is moving out of the field, the area of loop under the field is


the magnetic flux acting on the loop is



(- sign as the flux has decreased)


The induced emf is



Therefore magnitude of induced emf at t=22s is


(d) t=30s


distance moved by the square loop


at this time, square loop is completely outside the magnetic field and the area of loo through which flux passes =0


hence the flux linkage through the loop remains zero


and thus from eqn.(i)




Question 17.

Find the total heat produced in the loop of the previous problem during the interval 0 to 30 s if the resistance of the loop is 4.5 mΩ.


Answer:

Given:


Resistance of the loop R=4.5mΩ


Time interval=30s


As heat produced is a scalar quantity total heat is found by algebraically adding heat produced in different time intervals



Where


heat produced during time interval 0-5s


heat produced during time interval 5-20s


heat produced during time interval 20-25s


heat produced during time interval 25-30s


Now,


(i) during time interval 0-5s emf produced in the loop is given by



Current in the coil is



Heat produced in the coil is given by the formula


…. (i)


where


i=current


R=resistance


t=time interval


putting the values of i, R and t in above eqn. we get,



Emf induced in the time interval 5-20s and 25-30s =0


So current in the coil during this time



Heat produced in the coil is given by eqn.(i)



Emf induced in the time interval 20-25s is same as that induced at 5s



Hence the current and heat produced during this interval is same and given by



Total heat produced is given by





Therefore total heat produced in the loop during interval 0-30s is



Question 18.

A uniform magnetic field B exists in a cylindrical region of radius 10 cm as shown in figure. A uniform wire of length 80 cm and resistance 4.0Ω is bent into a square frame and is placed with one side along a diameter of the cylindrical region. If the magnetic field increases at a constant rate of 0.010 T/s, find the current induced in the frame.




Answer:

Given:


Radius of cylindrical region


Length of wire =80cm


Resistance of the wire


Rate of increase of magnetic field =0.01T/s


Area of loop inside magnetic field =area of semicircle of radius 0.1m



We know that,


Flux (ϕ) of magnetic field (B) through the loop of cross section area A in the magnetic field is given by




Since magnetic field is perpendicular to the loop the flux becomes



Rate of change of magnetic field wrt. time is given by



(since area of cross section in magnetic field does not change with time, A remains constant)


Now,


by faraday’s law of electromagnetic induction


…(i)


Where


ϵ =emf produced


ϕ =flux of magnetic field


using eqn.(i) the emf induced in the loop is given by



Hence the current through the loop (i) of resistance R is



Putting the values of A, R and in the above eqn. the magnitude of current is



Therefore current induced in the frame is



Question 19.

The magnetic field in the cylindrical region shown in figure increases at a constant rate of 20.0 mT/s. Each side of the square loop abcd and defa has a length of 1.00 cm and a resistance of 4.00 Ω. Find the current (magnitude and since) in the wire ad if

(a) the switch S1 is closed but S2 is open,

(b) S1 is open but S2 is closed,

(c) both S1 and S2 are open and

(d) both S1 and S2 are closed.




Answer:

Given:


Rate of increase of magnetic field =


Side length of square loop =


Resistance of each side =4Ω


Area of the coil adef = area of coil abcd =


We know that,


Flux (ϕ) of magnetic field (B) through the loop of cross section area A in the magnetic field is given by




Since magnetic field is perpendicular to the loop the flux becomes



Rate of change of magnetic field wrt. time is given by



(since area of cross section in magnetic field does not change with time, A remains constant)


Now,


by faraday’s law of electromagnetic induction


…(i)


Where


ϵ =emf produced


ϕ =flux of magnetic field


using eqn.(i) the emf induced in the loop is given by



Hence the current through the loop (i) of resistance R is


…(ii)


(a) when the switch S1 is closed but S2 is open


no current flows through loop abcd


net resistance of the loop adef R =4× 4 =16Ω


area of loop adef =


using eqn.(ii) current can be given by



As the magnetic field increases, the flux of magnetic field increases in downward direction so by Lenz’s law


The direction of induced current is such that it opposes the change that has induced it


Therefore, current flows in anticlockwise direction (along ad) to increase the magnetic flux in upward direction


(b) S1 is open but S2 is closed


No current flows in loop adef


Net resistance of loop abcd=4× 4=16Ω


Area of loop abcd =


using eqn.(ii) current can be given by



As the magnetic field increases, the flux of magnetic field increases in downward direction so by Lenz’s law


Therefore, current flows in anticlockwise direction (along da) to increase the magnetic flux in upward direction.


(c) When both S1 and S2 is open


No current flows in both the loop adef and abcd


And hence current in wire ad is zero



(d) When both S1 and S2 is closed


The circuit forms a balanced Wheatstone Bridge and the current flowing through the wire ad is zero



Concept of wheat stone bridge:


When the circuit forms a Wheatstone bridge in balanced condition then the current through galvanometer becomes zero




Question 20.

Figure shows a circular coil of N turns and radius a, connected to a battery of emf ϵ through a rheostat. The rheostat has a total length L and resistance R. The resistance of the coil is r. A small circular loop of radius a’ and resistance r’ is placed coaxially with the coil. The center of the loop is at a distance x from the center of the coil. In the beginning, the sliding contact of the rheostat is at the left end and then onwards it is moved towards right at a constant speed v. Find the emf induced in the small circular loop at the instant

(a) the contact begins to slide and

(b) it has slid through half the length of the rheostat.




Answer:

Given:



Area of coil (2) of radius a’ =


We know that magnetic field due to coil (1) at the center of coil (2) is



Where


N=no. of turns in coil (1)


i= current in coil (1)


a=radius of coil (1)


x=distance of center of coil (2) from center of coil (1)


We know that,


Flux (ϕ) of magnetic field (B) through the loop of cross section area A in the magnetic field is given by




Since the magnetic field due to coil (1) is parallel to axis of coil (2) θ =0° and flux through the coil (2) is given by



Now,


by faraday’s law of electromagnetic induction


…(i)


Where


e =emf produced


ϕ =flux of magnetic field


using eqn.(i) emf induced in the coil (2) is given by


. (ii)


Let y be the distance of sliding contact from its right end


Given,


Total length of rheostat =L


Total resistance of rheostat=R


When the sliding contact is at a distance y from its right end then the resistance (R’) of the rheostat is given by



So the current i flowing through the circuit is given by



Where r is the resistance of the coil and ϵ is the emf of battery


Putting value of i in eqn.(ii) we get,




Since



(a) When the contact begins to slide


Therefore, magnitude of emf induced is




(b) When the contact has slid through half the length of rheostat


Therefore, magnitude of emf induced is





Question 21.

A circular coil of radius 2.00 cm has 50 turns. A uniform magnetic field B = 0.200 T exists in the space in a direction parallel to the axis of the loop. The coil is now rotated about a diameter through an angle of 60.0°. The operation takes 0.100s.

(a) Find the average emf induced in the coil.

(b) If the coil is a closed one (with the two ends joined together) and has a resistance of 4.00 Ω, calculate the net charge crossing a cross-section of the wire of the coil.


Answer:

Given:


Radius of coil


No. of turns in the coil


Magnetic field intensity


We know that,


Flux (ϕ) of magnetic field (B) through the loop of cross section area A in the magnetic field is given by




Where N=no. of turns in the coil


Since magnetic field is perpendicular to the loop the flux becomes



Initial flux through the coil is given by



After 0.1 s the coil is rotated through an angle of 60° =θ


Finally, the flux through the coil becomes



Average induced emf in time interval Δt is given by


…(i)


Where


are flux across the cross section at time intervals respectively.


Using eqn.(i) emf induced in the coil is given by



Putting the values of N, B, A and Δt in above eqn. we get



Therefore average emf induced in the coil is


(b) the current through the coil (i) is calculated using formula



Hence the charge(Q) crossing the cross-section of the wire in time interval Δt is



Putting the values of ϵ, R and Δt we get,



Therefore charge crossing cross-section of the wire in the coil is



Question 22.

A closed coil having 100turns is rotated in a uniform magnetic field B = 4.0 × 10–4 T about a diameter which is perpendicular to the field. The angular velocity of rotation is 300 revolutions per minute. The area of the coil is 25 cm2 and its resistance is 4.0 Ω. Find

(a) the average emf developed in half a turn from a position where the coil is perpendicular to the magnetic field,

(b) the average emf in a full turn and

(c) the net charge displaced in part (a).


Answer:

Given:


No. of turns in the coil


Magnetic field intensity


Angular velocity of rotation


Area of the coil


Resistance of the coil


Magnetic flux through the circular coil ϕ can be given by formula



…. (i)


Where B=magnetic field intensity


A=area of cross section


N=no. of turns in the coil


θ =angle between area vector and magnetic field


(a) Initially, angle between area vector and magnetic field is 0°


Therefore, initial flux through the coil is



When it is rotated by 180° flux passing through the coil is given by



Average induced emf in time interval Δt is given by


…(i)


Where


are flux across the cross section at time intervals respectively.


Average induced emf is then given by


……(ii)


Now,


Angular velocity of coil


Time taken to complete half revolution i.e. rotate by π radian



Putting the values of N, B, A and Δt in eqn.(ii)



Therefore average emf induced in the coil in half a turn is


(b) In a full term coil returns to its original position



And hence emf induce in the coil using eqn.(ii) is



Therefore, average emf induced in full turn in the coil is zero


(c) Emf induced in the coil in part (a) is



Hence the current i flowing through the coil of resistance R is



So the charge displaced in time interval Δt =0.1s is



Therefore net charge displaced in part (a) is



Question 23.

A coil of radius 10 cm and resistance 40 Ω has 1000 turns. It is placed with its plane vertical and its axis parallel to the magnetic meridian. The coil is connected to a galvanometer and is rotated about the vertical diameter through an angle of 180°. Find the charge which flows through the galvanometer if the horizontal component of the earth’s magnetic field is BH = 3.0 × 10–5 T.


Answer:

Given:


Radius of coil r=10cm


Resistance of the coil R=40Ω


No. of turns in the coil N=1000


Horizontal component of earth’s magnetic field=


Angle of rotation θ =180°


Magnetic flux through the circular coil ϕ can be given by formula



…. (i)


Where B=magnetic field intensity


A=area of cross section


N=no. of turns in the coil


θ =angle between area vector and magnetic field


Initially, angle between area vector and magnetic field is 0°


Therefore, initial flux through the coil is



When it is rotated by 180° flux passing through the coil is given by



Now,


by faraday’s law of electromagnetic induction



Where


ϵ =emf produced


ϕ =flux of magnetic field


therefore, emf produced in the coil is given by


………(ii)


Current passing through the loop (i) of resistance R is


using eq.(ii)


Charge flowing through the galvanometer (Q) in time dot is given by formula



Putting the values of N, B, A and R we get,



Therefore charge which flows through the galvanometer is



Question 24.

A circular coil of one turn of radius 5.0 cm is rotated about a diameter with a constant angular speed B = 0.010 T exists in a direction perpendicular to the axis or rotation. Find

(a) the maximum emf induced,

(b) the average emf induced in the coil over a long period and

(c) the average of the squares of emf induced over a long period.


Answer:

Given:


Radius of circular coil


Magnetic field intensity


Angular speed of coil ω =


Magnetic flux through the circular coil ϕ can be given by formula




…(i)


As


Where ω =angular velocity of loop


Where θ is the angle between magnetic field and area vector of loop.


by faraday’s law of electromagnetic induction



Where


ϵ =emf produced


ϕ =flux of magnetic field


putting the value of eqn.(i) in above eqn. we get,


…(ii)


Since maximum value of is equal to 1



Putting the values of B, A and ω we get,


………. (iii)


Therefore maximum emf induced in the circular coil is


(b) from eqn.(ii) emf induced in the coil is given by



Average value of induced emf is given by formula



Where is the time taken by the coil to complete one revolution



Therefore, average induced emf is zero


(c). the average of squares of induced emf is given by the formula







Putting the values of B, A and ω we get


using eqn.(iii)


Therefore average of squares of emf produced is



Question 25.

Suppose the ends of the coil in the previous problem are connected to a resistance of 100 Ω. Neglecting the resistance of the coil, find the heat produced in the circuit in one minute.


Answer:

Given:


Resistance of the coil R=100Ω


Time period T=1min =60s


From previous question induced emf is given by



Current in the coil i is given by


…. (i)


Heat produced in the circuit is calculated by the following formula



Using eqn.(i) we get,






Putting the values of B, A, ω, T and R we have




Therefore heat produced in the circuit in one minute is



Question 26.

Figure shows a circular wheel of radius 10.0 cm whose upper half, shown dark in the figure, is made of iron and the lower half of wood. The two junctions are joined by an iron rod. A uniform magnetic field B of magnitude 2.00 × 10–4 T exists in the space above the central line as suggested by the figure. The wheel is set into pure rolling on the horizontal surface. If it takes 2.00 seconds for the iron part to come down and the wooden part to go up, find the average emf induced during this period.




Answer:

Given:


Radius of circular wheel r =10cm=0.1m


Magnetic field intensity


Area of semicircular part


Initially



Magnetic flux through the wheel ϕ can be given by formula



Since area vector and magnetic field is parallel we get flux



After time interval Δt=2s



Flux through the wheel



Average induced emf in time interval Δt is given by


…(i)


Where


are flux across the cross section at time intervals respectively.


Therefore, average induced emf ϵ is



Therefore average induced emf in the wheel is



Question 27.

A 20 cm long conducting rod is st into pure translation with a uniform velocity of 10 cm s–1 perpendicular to its length. A uniform magnetic field of magnitude 0.10 T exists in a direction perpendicular to the plane of motion.

(a) Find the average magnetic force on the free electrons of the rod.

(b) For what electric field inside the rod, the electric force on a free electron will balance the magnetic force? How is this electric field created?

(c) Find the motional emf between the ends of the rod.


Answer:

Given:


Length of rod l=20cm=0.2m


Velocity of rod v


Magnetic field intensity


(a) we know that,


A charge q moving with velocity v inside a magnetic field B experiences a force F given by


…. (i)


Since velocity of rod is perpendicular to magnetic field the above eqn. reduces to



To find the force on a free electron


Putting the values of q, v and B we get,



Therefore average force experienced by a free electron is


(b) we know that,


force experienced by a charge particle having charge q in presence of an electric field E is given by



To balance this force with the magnetic force, equating above eqn. with eqn.(i),




Putting the values of v and B we get,



This electric field is created due to emf produced due to motion of conducting rod, as a result of which the free electrons in the rod experiences a force


Therefore, electric field needed to balance magnetic force is 0.01Vm-1


(c) We know that,


motional emf produced due to a conductor of length l moving with velocity v in a magnetic field B is given by



Since v⃗ and B⃗ are perpendicular and their cross product is parallel to l⃗, eqn.(i) reduces to



Putting the values of v, B and l we get,



Therefore motional emf between the ends of the rod is



Question 28.

A metallic meter stick moves with a velocity of 2 ms–1 in a direction perpendicular to its length and perpendicular to a uniform magnetic field of magnitude 0.2 T. Find the emf induced between the ends of the stick.


Answer:

Given:


Velocity of meter stick


Intensity of magnetic field


Length of stick =1m


We know that,


motional emf produced due to a conductor of length l moving with velocity v in a magnetic field B is given by



Since v⃗ and B⃗ are perpendicular and their cross product is parallel to l⃗, eqn.(i) reduces to



Putting the values of v, B and l we get,



Therefore, emf induced between the ends of a stick is 0.4V



Question 29.

A 10 m wide spacecraft moves through the interstellar space at a speed 3 × 107 m s–1. A magnetic field B = 3 × 10–10 T exists in the space in a direction perpendicular to the plane of motion. Treating the spacecraft as a conductor, calculate the emf induced across its width.


Answer:

Given:


Speed of spacecraft v=3× 107m/s


Magnetic field intensity B=


Width of spacecraft l=10m


We know that,


motional emf produced due to a conductor of length l moving with velocity v in a magnetic field B is given by



Since v⃗ and B⃗ are perpendicular and their cross product is parallel to l⃗, eqn.(i) reduces to



Putting the values of v, B and l we get,




Therefore, emf induced across the width of spacecraft is 0.09V



Question 30.

The two rails of a railway track, insulated from each other and from the ground, are connected to a millivoltmeter. What will be the reading of the millivoltmeter when a train travels on the track at a speed of 180 km h–1? The vertical component of earth’s magnetic field is 0.2 × 10–4 T and the rails are separated by 1 m.


Answer:

Given:


Vertical component of earth’s magnetic field B=0.2× 10-4T


Speed of train v=


Separation between rails l=1m


We know that,


motional emf produced due to a conductor of length l moving with velocity v in a magnetic field B is given by


…. (i)


Since v⃗ and B⃗ are perpendicular and their cross product is parallel to l⃗, eqn.(i) reduces to



Putting the values of v, B and l we get,



Therefore, the reading of millivoltmeter is 1mV



Question 31.

A right-angled triangle abc, made from a metallic wire, moves at a uniform speed v in its plane as shown in figure. A uniform magnetic field B exists in the perpendicular direction. Find the emf induced

(a) in the loop abc,

(b) in the segment bc,

(c) in the segment ac and

(d) in the segment ab.




Answer:

Given:


Speed of right-angled triangle =v


We know that,


motional emf produced due to a conductor of length l moving with velocity v in a magnetic field B is given by


…. (i)


(a) for loop abc since the flux of magnetic field through the triangle does not change hence the emf induced in loop abc is zero



Therefore, emf induced in loop abc is zero


(b) in the segment bc


bc is perpendicular to velocity. emf induced in the segment bc is given by



Therefore, emf induced in segment bc is vB(bc)


(c) in the segment ac


ac is parallel to velocity. Emf induced in the segment ac is given by



Therefore, emf induced in the segment ac is zero


(d) in the segment ab


the effective length of ab perpendicular to velocity is given by bc (-ĵ)


emf induced in the segment ab is given by



Therefore, emf induced in the segment ab is vB(bc)



Question 32.

A copper wire bent in the shape of a semicircle of radius r translates in its plane with a constant velocity v. A uniform magnetic field B exists in the direction perpendicular to the plane of the wire. Find the emf induced between the ends of the wire if

(a) the velocity is perpendicular to the diameter joining free ends,

(b) the velocity is parallel to this diameter.


Answer:

Given:


Radius of semicircular wire=r


Velocity =v


We know that,


motional emf produced due to a conductor of length l moving with velocity v in a magnetic field B is given by



In the case of semicircular wire, l denotes the effective length of wire perpendicular to velocity.


(a) when the velocity is perpendicular to diameter joining free ends



the effective length of wire perpendicular to velocity is given by length of diameter



Therefore, induced emf in the wire is given by



Therefore, induced emf in this case is 2BvR


(b) when the velocity is parallel to the diameter



The effective length of wire parallel to velocity is zero



Therefore, induced emf in the wire is given by



Therefore, induced emf in this case is zero



Question 33.

A wire of length 10 cm translates in a direction making an angle of 60° with its length. The plane of motion is perpendicular to a uniform magnetic field of 1.0 T that exists in the space. Find the emf induced between the ends of the rod if the speed of translation is 20 cm s–1.


Answer:

Given:


Length of the wire=10cm


Angle of length of wire with velocity=60°


Magnetic field intensity=1.0T


Speed of wire v=20cm/s =0.2m/s


(a) we know that,


motional emf produced due to a conductor of length l moving with velocity v in a magnetic field B is given by



It is given that plane of motion is perpendicular to electric field i.e. angle between v⃗ and B⃗ =90°


And the angle between velocity and length of wire=60°




We take only that component of length vector which is perpendicular to velocity vector


Putting the values of B, v, l we get



Therefore the emf induced in the rod is



Question 34.

A circular copper-ring of radius r translates in its plane with a constant velocity v. A uniform magnetic field B exists in the space in a direction perpendicular to the plane of the ring. Consider different pairs of diametrically opposite points on the ring.

(a) Between which pair of points is the emf maximum?

(b) Between which pair of points is the emf minimum?

What is the value of this minimum emf?


Answer:

Given:


Radius of ring =R


Velocity of ring=v



(b) we know that motional emf produced due to a conductor of length l moving with velocity v in a magnetic field B is given by


…. (i)


This value is maximum when length between the points is perpendicular to the velocity of the rod


Thus the emf is the highest between the end points of diameter perpendicular to the velocity and the value of this emf is given by



Therefore, maximum value of emf is 2BvR


(c) the value of eqn.(i) is minimum when length between the points is parallel to the velocity of the rod


Thus the emf is lowest between the end points of diameter parallel to the velocity and this value of emf is given by



Therefore, minimum value of emf is 0



Question 35.

Figure shows a wire sliding on two parallel, conducting rails placed at a separation ℓ. A magnetic field B exists in a direction perpendicular to the plane of the rails. What force is necessary to keep the wire moving at a constant velocity v?




Answer:

Given:


Velocity of wire=v


Separation between rail=l



We know that,


Force experienced by a wire of length l carrying current I in a magnetic field B and placed perpendicular to magnetic field is given by



Now since here there is no formation of closed circuit the circuit is open and the current flowing in the wire =0


Therefore, force experienced by wire due to magnetic field =0


Hence net force on wire becomes zero and it moves with constant velocity v


Therefore, no external force is needed to move the wire with velocity v



Question 36.

Figure shows a long U-shaped wire of width ℓ placed in a perpendicular magnetic field B. A wire of length ℓ is slid on the U-shaped wire with a constant velocity v towards right. The resistance of all the wires is r per unit length. At t = 0, the sliding wire is close to the left edge of the U-shaped wire. Draw an equivalent circuit diagram, showing the induced emf as a battery. Calculate the current in the circuit.




Answer:

Current will flow from the left edge to the right, that is, in the clockwise direction. Therefore, the induced emf will also flow along the clockwise direction.


Diagram showing induced emf as a battery:



Given:


Resistance per unit length = r


Length of wire = l


Velocity with which the wire moves = v


Formula used:


We know, that … (i), where E = emf, B = magnetic field, v = velocity with which the wire moves, l = length of wire.


Now, total resistance … (ii), where r = resistance per unit length, l’ = total length of loop


Horizontal length of loop = vt, where v = velocity, t = time.


Hence, total length of loop


Therefore, R(total resistance) = … (iii), where r = resistance per unit length, l = length of wire


By Ohm’s law, we know that where E = emf, I = current, R = total resistance. Hence, from (iii), we get, ……….(iv)


Equation (i) and (iv), we get:



=>


Current in the circuit (Answer)



Question 37.

Consider the situation of the previous problem.

(a) Calculate the force needed to keep the sliding wire moving with a constant velocity v.

(b) If the force needed just after t = 0 is F0, find the time at which the force needed will be F0/2.


Answer:

Formula used:


(a) Magnetic force on a current carrying wire where I = current, l = length of wire, B = magnetic field.


Since l and B are perpendicular to each other, magnetic force F becomes … (i)


Now, from the previous problem, … (ii)


Now, the force needed to keep the sliding wire from moving will be equal to the magnetic force, but in the opposite direction.


Let this force be F’.


Hence, , where F = magnetic force, I = current, L = length of wire, B = magnetic field.


Substituting the value of I from (ii):


=


Hence, force required to keep the wire from sliding = (Ans)


(b) Now, just after time t = 0, the force required to stop the wire from sliding will be … (i) (substituting t = 0), from the previous part of this question.


Now, let the time taken for the required force to be be t = T.


Hence, from the previous question, substituting t = T,


… (ii)


Substituting the value of F0 from (i), we get





Time taken for the force to reduce to (Ans)



Question 38.

Consider the situation shown in figure. The wire PQ has mass m, resistance r and can slide on the smooth, horizontal parallel rails separated by a distance ℓ. The resistance of the rails is negligible. A uniform magnetic field B exists in the rectangular region and a resistance R connects the rails outside the field region. At t = 0, the wire PQ is pushed towards right with a speed v0. Find

(a) the current in the loop at an instant when the speed of the wire PQ is v,

(b) the acceleration of the wire at this instant,

(c) the velocity v as a function of x and

(d) the maximum distance the wire will move.




Answer:

Given:


Mass of PQ = m


Resistance of PQ = r


Length of PQ between the two rails = l


Magnetic field = B


Resistance connected to the rails = R


Velocity with which PQ is pushed towards right at t=0 = v0


Formula used:


(a) By Ohm’s law, E = IR’, where E = emf, I = current, R’ = total resistance.


Hence, current … (i)


Now, emf induced due to the moving road in the magnetic field … (ii), where B = magnetic field, l = length of rod, v= velocity of rod


Also, total resistance … (iii), where r = resistance of PQ, R = resistance attached to the rails.


Hence, substituting the values of E and R’ from (ii) and (iii) in (i), we get



Therefore, current in the loop when the speed of the wire PQ is . (Ans)


(b) Now, magnetic force on a current carrying wire … (i), where I = current, l = length of wire, B = magnetic field.


From the previous part, the value of current at an instant when velocity = v is … (ii)


Therefore, from (i) and (ii), magnetic force … (iii)


According to Newton’s second law of motion, … (iv), where F = force, m = mass, a = acceleration.


Hence, equating (iii) and (iv):



Therefore, acceleration of the wire at this instant = (Ans)


(c) Velocity v’ can be expressed as … (i), where v0 = initial velocity, a = acceleration, t = time. We put a negative sign before at since the force is opposite to velocity, and it


Now, from the previous part, we can write acceleration … (ii)


Hence, from (i) and (ii), we can write


But, distance travelled x = vt, where v = velocity, t = time.


Therefore, velocity v as a function of x is (Ans)


(d) We know that,, where a = acceleration,


v = velocity, x = distance, t = time.


Now, from the part (b), acceleration as a function of time




Now, the wire can travel maximum distance when its velocity is v0.


Hence, integrating on both sides, we get



Therefore, maximum distance travelled by the wire


(Ans)



Question 39.

A rectangular frame of wire abcd has dimensions 30 cm × 80 cm and a total resistance of 2.0 Ω. It is pulled out of a magnetic field B = 0.020 T by applying a force of 3.2 × 10–6 N (figure). It is found that the frame moves with constant speed. Find

(a) this constant speed,

(b) the emf induced in the loop,

(c) the potential difference between the points a and b and (d) the potential difference between the points c and d.




Answer:

Given:


Length ab = cd = 30 cm = 0.3 m


Length bc = ad = 80 cm = 0.8 m


Total resistance R = 2 Ω


Magnetic field B = 0.02 T


Force F = 3.2 × 10–6 N


Formula used:


(a) Magnetic force on a current carrying wire … (i),where I = current, l = length of wire, B = magnetic field.


Hence, current … (ii)


Now, emf … (iii), where B = magnetic field, l = length of wire, v = constant velocity with which it is moving


Also, by Ohm’s law, … (iv), where I = current,


R = resistance.


Hence, equating (iii) and (iv) and substituting I from (ii), we get


, where B = magnetic field, v = velocity, F = force, R = resistance, l = length of wire.


Here, since the force is applied on the side cd, we consider


l =30 cm = 0.3 m (the shorter length).


Hence,


Substituting the given values, we get


ms-1 = 0.18 ms-1


Constant speed with which the frame moves = 0.18 ms-1(Ans)


(b) Emf induced in the loop , where B = magnetic field, l = length of the wire which is moving, v = velocity


Hence, = 0.001 V


Emf induced in loop = 0.001 V (Ans)



Question 40.

Figure shows a metallic wire of resistance 0.20 Ω sliding on a horizontal, U-shaped metallic rail. The separation between the parallel arms is 20 cm. An electric current of 2.0 μA passes through the wire when it is slid at a rate of 20 cm s–1. If the horizontal component of the earth’s magnetic field is 3.0 × 10–5 T, calculate the dip at the place.




Answer:

Given:


Resistance of wire(R) = 0.2 Ω


Length of wire(l) = 20 cm = 0.2 m


Current(I) = 2 μA = 2 x 10-6 A


Velocity with which the wire moves(v) = 20 cms-1 = 0.2 ms-1


Horizontal component of earth’s magnetic field(BH) = 3.0 × 10–5 T


Formula used:


Angle of dip ) … (i), where BV = vertical component of earth’s magnetic field, BH = horizontal component of earth’s magnetic field.


Now, emf induced in the wire … (ii), where BV = vertical component of earth’s magnetic field, l = length of wire, v = velocity with which the wire moves


Also, by Ohm’s law … (iii), where E = emf, I = current, v = velocity.


Equating (ii) and (iii) we get


= IR ⇒


Substituting the given values, we get


= 10-5 T


Hence, angle of dip ⇒ (Ans)



Question 41.

A wire ab of length ℓ, mass m and resistance R slides on a smooth, thick pair of metallic rails joined at the bottom as shown in figure. The plane of the rails makes an angle θ with the horizontal. A vertical magnetic field B exists in the region. If the wire slides on the rails at a constant speed v, show that






Answer:

Given:


Length of ab = l


Mass = m


Resistance of ab = R


Angle between the plane of the rails and the horizontal = θ


Magnetic field = B


Velocity with which the wire slides along the rails = v


Diagram:



Formula used:


Emf induced in the wire ab … (i), where


E = emf, B = magnetic field, l’ = component of l perpendicular to the magnetic field = l cosθ, v = velocity


Magnetic force on ab F = I (l X B), where I = current, l = length of ab, B = magnetic field


Now, the angle between l and B is 900 - θ.


Therefore, F = IlBsin (900 - 𝛉) = IlBcosθ … (ii)


This force will be equal to the sinθ component of the weight of the wire.


Hence, … (iii)


Now, = … (iv), where I = current, E = emf, R = resistance, B = magnetic field, v = velocity, l = length of wire, = angle between plane of rod and horizontal


Therefore, substituting this in (ii), we get


F = … (v)


Hence, equating (iii) and (v), we get



⇒ Magnetic field (proved)



Question 42.

Consider the situation shown in figure. The wires P1Q1 and P2Q2 are made to slide on the rails with the same speed 5 cm s–1. Find the electric current in the 19 Ω resistor if

(a) both the wires move towards right and

(b) If P1Q2 moves towards left but P2Q2 moves towards right.




Answer:

Given:


Velocity with which the wires move(v) = 5 cm s-1 = 0.05 m s-1


Resistance(R) = 19 Ω


Resistance of each of the wires(r) = 2 Ω


Length of wire(l) = 4 cm = 0.04 m


Magnetic field(B) = 1 T


Formula used:


Emf induced… (i) where B = magnetic field, l = length of wire, v = velocity


Also, by Ohm’s law, emf … (ii), where I = current, R = total resistance.


(a) When the wires slide in the same direction, we have two parallel sources of emf with current flowing in the same direction.


Hence, from (i),net emf


Net parallel resistance of the 2 Ω wires = Ω = 1 Ω


Hence, total resistance(R) = (1 + 19) Ω = 20 Ω


Therefore, substituting these values in (ii), we get


Current A = 10-4 A (Ans)


(b) When the wires slide in opposite directions, the two parallel sources of emf have opposing directions. Hence, the net emf is 0.


Therefore, the net current is also 0. (Ans)



Question 43.

Suppose the 19Ω resistor of the previous problem is disconnected. Find the current through P2Q2 in the two situations.

(a) and (b) of that problem.


Answer:

(a) When the 19Ω resistor is removed and the wires in move in the same direction, their polarity remains the same. Hence, the circuit remains incomplete and the current through P2Q2 is 0. (Ans)


(b) When the wires slide in opposite directions, the polarity of one of the wires reverses and current flows.


In this case, emf , where B = magnetic field, l = length, v = velocity.


Hence, V =


But here, resistance R = 2 Ω only


Therefore, Current flowing through the wire = 10-3 A (Ans)



Question 44.

Consider the situation shown in figure. The wire PQ has a negligible resistance and is made to slide on the three rails with a constant speed of 5 cm s–1. Find the current in the 10Ω resistor when the switch S is thrown to (a) the middle rail (b) the bottom rail.




Answer:

Given:


Speed(v) = 5 cms-1 = 0.05 ms-1


External resistance(R) = 10Ω


Magnetic field(B) = 1 T


Formula used:


Induced emf … (i), where B = magnetic field, l = length of sliding wire, v = velocity


(a) When the switch S is thrown to the middle rail, length of sliding wire l = 2 cm = 0.02 m


Hence, induced emf in this case from (i) is


E = (1 x 0.02 x 0.05) V = 10-3 V


Given resistance R = 10Ω


Therefore, current flowing through the resistor


where E = emf, R = resistance.


= 10-4 A = 0.1 mA (Ans)


(b) When the switch S is thrown to the bottom rail, length of sliding wire(l’) = 4 cm = 0.04 m


Hence, induced emf = (1 x 0.04 x 0.05) V = 2 x 10-3 V, where B = magnetic field, l’ = length of sliding wire, v = velocity


Resistance R = 10Ω


Therefore, current flowing through the resistor I’ = E’/R, where E’ = emf, R = resistance


A = 2 x 10-4 A= 0.2 mA (Ans)



Question 45.

The current generation Ig, shown in figure, sends a constant current i through the circuit. The wire cd is fixed and ab is made to slide on the smooth, thick rails with a constant velocity v towards right. Each of these wires has resistance r. Find the current through the wire cd.




Answer:

Given:


Initial current passing through the circuit = i


Velocity with which the wire ab moves = v


Resistance of each wire = r


Formula used:


Induced emf due to moving of wire ab … (i), where B = magnetic field, l = length of sliding wire, v = velocity


Initial emf in the wire E0 = ir, where i = current, r = resistance of wire ab.


Hence, net emf … (ii)


Now, net resistance = 2r


Hence, current passing through the wire cd = (Ans)



Question 46.

The current generator Ig, shown in figure, sends a constant current i through the circuit. The wire ab has a length ℓ and mass m and can slide on the smooth, horizontal rails connected to I4. The entire system lies in a vertical magnetic field B. Find the velocity of the wire as a function of time.




Answer:

Given:


Initial current = i


Length of sliding wire ab = l


Mass = m


Magnetic field = B


Formula used:


Magnetic force on the wire ab … (i), where i = current, l = length of sliding wire, B = magnetic field


Now, velocity v can be written as … (ii), where u = initial velocity = 0(in this case), a = acceleration, t = time.


Hence, acceleration a = v/t … (iii)


Now, according to Newton’s 2nd law of motion, … (iv), where F = force, m = mass, a = acceleration.


Substituting (iii) in (iv) and equating (i) and (iv), we get



Hence, velocity of the wire as a function of time is (Ans)



Question 47.

The system containing the rails and the wire of the previous problem is kept vertically in a uniform horizontal magnetic field B that is perpendicular to the plane of the rails figure. It is found that the wire stays in equilibrium. If the wire ab is replaced by



another wire of double its mass, how long will it take in falling through a distance equal to its length?


Answer:

Given:


Initial mass = m


Magnetic field = B


Length of sliding wire = l


Formula used:


Magnetic force … (i), where i = current, l = length of sliding wire, B = magnetic field.


At equilibrium, this magnetic force balances the weight of the wire mg acting downward, where m = mass, g = acceleration due to gravity.


Therefore, … (ii)


Now, when the wire ab is replaced by another wire of mass 2m, the weight acting downward will be 2mg, where g = acceleration due to gravity.


Hence, net force = 2mg - ilB … (iii) where 2m = mass, g = acceleration due to gravity, i = current, l = length of sliding wire, B = magnetic field.


According to Newton’s law of motion, … (iv), where F = net force, m’ = mass, acceleration


In this case, m’ = 2m.


Therefore, equating (iii) and (iv), we get


2mg - ilB = 2ma


… (v)


Now, distance travelled can be expressed as … (vi), where s = distance travelled, u = initial velocity = 0(in this case), t = time, a = acceleration.


Now, distance travelled = l (given)


Therefore, from (v), (vi) becomes:




But, from (i),


Therefore,


Required time taken = (Ans)



Question 48.

The rectangular wire-frame, shown in figure has a width d, mass m, resistance R and a large length. A uniform magnetic field B exists to the left of the frame. A constant force F starts pushing the frame into the magnetic field at t =0.

(a) Find the acceleration of the frame when its speed has increased to v.

(b) Show that after some time the frame will move with a constant velocity till the whole frame enters into the magnetic field. Find this velocity v0.

(c) Show that the velocity at time t is given by




Answer:

Given:


Length of sliding wire = width of frame = d


Mass = m


Resistance = R


Magnetic field = B


Initial force = F


Formula used:


(a) Induced emf(when it attains a speed v) … (i), where B = magnetic field, d = width of frame, v = velocity


Therefore, induced current , where E = induced emf, R = resistance ⇒ … (ii)


Now, magnetic force acting on the wire … (iii), where I = current, d = length of sliding wire = width of frame, B = magnetic field


Substituting (ii) in (iii), … (iv)


Now, as the magnetic force is in opposite direction to applied force, net force = … (v)


But, from Newton’s 2nd law of motion, net force = ma … (vi), where m = mass, a = acceleration


Equating (v) and (vi):



Acceleration of the frame at speed (Ans)


(b) For the velocity to be constant, acceleration needs to be 0.


Hence, from previous part,


where F = external force, m = mass, B = magnetic field, d = width of frame, v0 = constant velocity, R = resistance


=>


Constant velocity (Ans)


(c) From part (a), acceleration


Now, acceleration a = dv/dt, where v= velocity, t = time


Hence,


=>


Integrating with proper limits, we get



=>>


=>


=>


=>


But, from previous part (b), we found out that



Hence,


=> (proved)



Question 49.

Figure shows a smooth pair of thick metallic rails connected across a battery of emf ϵ having a negligible internal resistance. A wire ab of length ℓ and resistance r can slide smoothly on the rails. The entire system lies in a horizontal plane and is immersed in a uniform vertical magnetic field B. At an instant t, the wire is given a small velocity v towards right.

(a) Find the current in it at this instant. What is the direction of the current?

(b) What is the force acting on the wire at this instant?

(c) Show that after some time the wire ab will slide with a constant velocity. Find this velocity.




Answer:

Given:


Emf of the battery = ϵ


Length of sliding wire = l


Resistance = r


Magnetic field = B


Velocity = v


Formula used:


(a) Induced emf … (i), where B = magnetic field, l = length of sliding wire, v = velocity


Therefore, net emf =


Hence, current in the wire(Ans)


(b) Magnetic force acting on the wire … (ii), where i = current, l = length of sliding wire, B = magnetic field.


Hence, from (a),


Therefore, force (Ans)


(c) At constant velocity, net force will be 0.


Hence, = 0



Hence, value of velocity = (Ans)



Question 50.

A conducting wire ab of length ℓ, resistance r and mass m starts sliding at t = 0 down a smooth, vertical, thick pair of connected rails as shown in figure. A uniform magnetic field B exists in the space in a direction perpendicular to the plane of the rails.

(a) Write the induced emf in the loop at an instant t when the speed of the wire is v.

(b) What would be the magnitude and direction of the induced current in the wire?

(c) Find the downward acceleration of the wire at this instant.

(d) After sufficient time, the wire starts moving with a constant velocity. Find this velocity vm.

(e) Find the velocity of the wire as a function of time.

(f) Find the displacement of the wire as a function of time.

(g) Show that the rate of heat developed in the wire is equal to the rate at which the gravitational potential energy is decreased after steady state is reached.




Answer:

Given:


Length of sliding wire ab = l


Resistance = r


Mass = m


Magnetic field = B


Speed = v


Formula used:


(a) Induced emf in the loop (Ans), where B = magnetic field, l = length, v = velocity


(b) Induced current in the wire = (Ans) , where E = emf, r = resistance


As the wire is moving, the magnetic flux is increasing. Hence, the direction of the current will be such as to oppose the increase in flux. Hence, the current will move from b to a.


(c) Magnetic force on the wire(upward) … (i), where I = current, l = length of sliding wire, B = magnetic field.


Weight acting downward = mg


Hence, net downward force = … (ii)


According to Newton’s 2nd law of motion, net force = ma … (iii), where m = mass, a = acceleration.


Hence, equating (ii) and (iii),


But, I = E/r, where I = current, E = emf, r = resistance => I = Blv/r, where B = magnetic field, l = length of sliding wire, v = velocity


=> acceleration (Ans)


(d) When the wire will move with constant velocity (let it be v0), acceleration will be 0.


Hence, from part (d) of this question,


= 0


=> constant velocity (Ans)


(e) From part (c),



But, a = dv/dt, where v = velocity, t = time


Therefore,



Integrating with proper limits, we get



>



But, from the previous part,


Therefore, velocity as a function of time : (Ans)


(f) Now, v can be written as dx/dt, where x = position, t = time


Therefore, from previous part,


=>


Integrating with suitable limits, we get:



=>


=> Displacement as a function of time (Ans)


(g) Then,




After steady state,





Hence after steady state,



So, the rate of heat developed in the wire is equal to the rate at which the gravitational potential energy is decreased after steady state is reached.



Question 51.

A bicycle is resting on its stand in the east-west direction and the rear wheel is rotated at an angular speed of 100 revolutions per minute. If the length of each spoke is 30.0 cm and the horizontal component of the earth’s magnetic field is 2.0 × 10–5 T, find the emf induced between the axis and the outer end of a spoke. Neglect centripetal force acting on the free electrons of the spoke.


Answer:

Given:


Angular speed(w) = 100 revolutions/minute x 2π = 100 revolutions/60 sec x 2π= 10π/3 revolutions/sec


Length of each spoke(l) = 30 cm = 0.3 m


Magnetic field(B) = 2.0 × 10–5 T


Formula used:


Induced emf … (i), where B = magnetic field, l = length of spoke, v = velocity


Now, linear speed of the spoke v = ωr, where ω = angular speed, r = distance from the axis to the outer end.


Here, , where l = length of spoke


Hence,


Therefore, emf induced (from (i))


V = 9.42 x 10-6 V (Ans)



Question 52.

A conducting disc of radius r rotates with a small but constant angular velocity ω about its axis. A uniform magnetic field B exists parallel to the axis of rotation. Find the motional emf between the center and the periphery of the disc.


Answer:

Given:


Radius = r


Angular velocity = w


Magnetic field = B


Diagram:



Formula used:


In this case, the velocity will increase radially.


Let us consider a strip of width dx at a distance x from the centre.


Hence, induced emf of this portion will be , where B = magnetic field, dx = width of the element, x = distance of the element from the centre, w = angular velocity


Hence, integrating on both sides using proper limits, we get



=> Total motional emf (Ans)



Question 53.

Figure shows a conducting disc rotating about its resistance R is connected between the centre and the rim. Calculate the current in the resistor. Does it enter the disc or leave it at the centre? The radius of the disc is 5.0 cm, angular speed ω = 10 rad/s, B = 0.40 T and R = 10 Ω.




Answer:

Given:


Resistance(R) = 10Ω


Radius(r) = 5 cm = 0.05 m


Angular speed(w) = 10 rad/s


Magnetic field(B) = 0.4 T


Formula used:


We consider a rod of length 5 cm from the centre and rotating with same w.


Hence, length of sliding rod(l) = 5 cm = 0.05 m.


Now, induced emf … (i), where B = magnetic field, l = length of sliding rod, v = velocity


Now, velocity … (ii), where l = length or rod, w = angular velocity


From Ohm’s law, current through resistor R(I) = E/R = Bl2w/2R (from (i) and (ii)), where E = emf, R = resistance


Hence, A = 0.5 mA (Ans)


Since the disc is rotating anticlockwise, the emf induced is such that the centre is at a higher potential than the periphery. Hence, the current leaves from the centre.



Question 54.

The magnetic field in a region is given by where L is a fixed length. A conducting rod of length L lies along the Y-axis between the origin and the point (0, L, 0). If the rod moves with a velocity v = v0, find the emf induced between the ends of the rod.


Answer:

Given:



Where B = magnetic field, y = distance from origin on y axis, L = fixed length.


Velocity of rod = v0 i


Diagram:



Formula used:


Now, we consider a small element dy at a distance y from the origin.


Emf induced in the element dE = Bvdy, where B = magnetic field, v = velocity, dy = length of element



Integrating on both sides with proper limits, we get



=> Total emf


(Ans)



Question 55.

Figure shows a straight, long wire carrying a current i and a rod of length ℓ coplanar with the wire and perpendicular to it. The rod moves with a constant velocity v in a direction parallel to the wire. The distance of the wire from the centre of the rod is x. Find the motional emf induced in the rod.




Answer:

Given:


Current = i


Length of rod = l


Velocity = v


Distance of centre of rod from wire = x


Hence, the two ends of the rod are at distances and from the wire


Diagram:



Formula used:


We consider an element of length da at a distance of ‘a’ from the wire.


Now, magnetic field due to an infinite current carrying wire at a distance a … (i), where μ0 = magnetic permeability of vacuum, i = current, a = distance from wire.


Therefore, emf induced in the element … (ii), where B = magnetic field, da = element, v = velocity


Hence, putting (i) in (ii), we get



Integrating on both sides and putting suitable limits, we get



(ans)



Question 56.

Consider a situation similar to that of the previous problem except that the ends of the rod slide on a pair of thick metallic rails laid parallel to the wire. At one end the rails are connected by resistor of resistance R.

(a) What force is needed to keep the rod sliding at a constant speed v?

(b) In this situation what is the current in the resistance R?

(c) Find the rate of heat developed in the resistor.

(d) Find the power delivered by the external agent exerting the force on the rod.


Answer:

Given:


Resistance = R


Constant velocity = v


Formula used:


From previous question, induced emf


w, induced current = … (i), where E = emf, R = resistance, μ0 = magnetic permeability of vacuum, i = current in the wire, v = velocity of sliding rod, x = distance of centre of rod from wire, l = length of rod.


Now, magnetic force on element … (ii), where I = induced current, da = element, B = magnetic field due to infinitely straight wire


… (iii), where μ0 = magnetic permeability of vacuum, i = current in wire, a = distance from wire.


Hence, (ii) becomes


dF = integrating with suitable limits, we get



=>Force needed to keep the wire sliding at constant velocity v ns)


(b) Current I = E/R = here E = emf, R = resistance, μ0 = magnetic permeability of vacuum, i = current in the wire, v = velocity of sliding rod, x = distance of centre of rod from wire, l = length of rod. (Ans)


(c) Rate of heat developed in the resistor = Power(P) = I2R, where I = current, R = resistance


From previous part, I = therefore, rate of heat developed = (ans)


(d) Power delivered by external agent = rate of heat developed in resistor = (ans)



Question 57.

Figure shows a square frame of wire having a total resistance r placed co-planarly with long, straight wire. The wire carries a current I given by i = i0 sin ωt. Find

(a) the flux of the magnetic field through the square frame,

(b) the emf induced in the frame and

(c) the heat developed in the frame in the time interval 0 to .




Answer:

Given:


Current in wire i = i0 sin ωt


Length of each side of square loop = a


Distance of one edge from wire = b


Diagram:



Formula used:


Magnetic flux … (i), where = magnetic flux, B = magnetic field, da = area element


Magnetic field due to a long current carrying wire at distance x … (ii), where μ0 = magnetic permeability of vacuum, i = current, x = distance from wire


We consider a strip of width dx at a distance x from the wire.


Now, area element da = a dx, where a = length of loop, dx = width element


Hence, from (i) and (ii),


Flux (ans)


(b) Emf induced in frame where = flux, t = time


From previous part, re,


= (Ans)


(c)Heat developed in wire, where i = current through frame, r = resistance, t = time


From previous i = E/r =


where E = emf, r = resistance


Hence H =


Now, Given:


Hence, H =



Question 58.

A rectangular metallic loop of length ℓ and width b is placed

coplanarly with a long wire carrying a current i figure. The loop is moved perpendicular to the wire with a speed v in the plane containing the wire and the loop. Calculate the emf induced in the loop when the rear end of the loop is at a distance a from the wire. Solve by using Faraday’s law for the flux through the loop and also by replacing different segments with equivalent batteries.




Answer:

Given: Length of the rectangular loop=l


Breadth of the rectangular loop=b


Current in the wire=i


Speed of loop=v


We have to find the emf induced in the loop by Faradays’s Law.


Faraday’s law states that whenever the magnetic flux through


a closed surface changes, there will be an induced emf prodced in


the loop that encloses the surface. The mathematical relation is



where is the magnitude of the induced emf, ϕB is the the magnetic


flux through the surface. The negative sign arises because the


induced emf will be produced such that it will oppose the change of


magnetic flux. The magnetic flux is given by



where B is the magnetic field and dS is a small area element on the


surface.


The wire which is near the loop is responsible for the magnetic field .


The wire is carrying current in the upward direction so by Fleming’s


right hand thumb rule, the magnetic field will be perpendicular to the


plane of the paper in the inward direction.



We have to now calculate the magnetic field acting across the loop.


Let us consider a small rectangular element of length b and width dx


at a distance of x from wire. The magnteic field on this element due


to the current carrying wire is given by Ampere circuital law.




(cos θ=1 because the vector and vector are both acting in ward in the plane of paper so θ=0°, here dl is a small current carrying element of the circular amperian loop)





The magnetic flux through this element will be





(cosθ=1 because the magnetic field across the area and the normal


vector of this area element both point in the same direction so θ=0°


so cosθ=1)


The total magnetic flux through the loop will be the flux through


Infinite such elementss from x=a to x=a+l





(taking constants out of the integral)





The flux through the loop is


The emf induced will be



(taking constants out of the differential)




(because as the rate of change of Distance from wire is the loop speed)




The magnitude of the induced emf is .



Question 59.

Figure shows a conducting circular loop of radius a placed in a uniform, perpendicular magnetic field B. A thick metal rod OA is pivoted at the centre O. The other end of the rod touches the loop at A. Tec entre O and a fixed point C on the loop are connected by a wire OC of resistance R. A force is applied at the middle point of the rod OA perpendicularly, so that the rod rotates clockwise at a uniform angular velocity ω. Find the force.




Answer:

Given:


Radius = a


Magnetic field = B


Resistance = R


Angular velocity = w


Formula used:


Let us consider an element of length dr at a distance r from the centre.


Hence, induced emf on this portion … (i), where B = magnetic field, dr = length of element, w = angular velocity, r = distance from centre (since v = ωr)


Hence, integrating on both sides with suitable limits, we get



Now, current , where E = emf, resistance = R


=>


Hence, force on the rod (where I = current, a = length of rod, B = magnetic field) ⇒ (Ans)



Question 60.

Consider the situation shown in the figure of the previous problem. Suppose the wire connecting O and C has zero resistance but the circular loop has a resistance R uniformly distributed along its length. The rod OA is made of rotate with a uniform angular speed ω as shown in the figure. Find the current in the rod when ∠AOC = 90°.


Answer:

Given:


Resistance of circular loop = R


∠AOC = 90°


Angular velocity = w


Formula used:


From the previous problem, emf … (i), where B = magnetic field, w = angular velocity, a = radius


Now, since ∠AOC = 90°, the major and minor segments of the arc AC consist of parallel combination of resistances of R/4 and 3R/4 respectively (since the resistance is divided in the ratio of the angle at the centre).


Hence, equivalent resistance =


Therefore, current through the rod , where E = emf, R’ = equivalent resistance


= (Ans)



Question 61.

Consider a variation of the previous problem figure. Suppose the circular loop lies in a vertical plane. The rod has a mass m. The rod and the loop have negligible resistances but the wire connecting O and C has a resistance R. The rod is made to rotate with a uniform angular velocity ω in the clockwise direction by applying a force at the midpoint of OA in a direction perpendicular to it. Find the magnitude of this force when the rod makes an angle θ with the vertical.


Answer:

When the circular loop is in the vertical plane, it tends to rotate in the clockwise direction because of its weight.

Let the force applied be F and its direction be perpendicular to the rod.
The component of mg along F is mg sin θ.
The magnetic force is in perpendicular and opposite direction to mg sin θ.



Now, Current in the rod will be



The force on the rod will be



So, the net force will be



The net force passes through the centre of mass of the rod.
Net torque on the rod about the centre O will be



Because the rod rotates with a constant angular velocity, the net torque on it is zero.


Thus,





Question 62.

Figure shows a situation similar to the previous problem. All parameters are the same except that a battery of emf ϵ and a variable resistance R are connected between O and C. Neglect the resistance of the connecting wires. Let θ be the angle made by the rod from the horizontal position (shown in the figure), measured in the clockwise direction. During the part of the motion 0 < θ < π/4 the only forces acting on the rod are gravity and the forces exerted by the magnetic field and the pivot. However, during the part of the motion, the resistance R is varied in such a way that the rod continues to rotate with a constant angular velocity ω. Find the value of R in terms of the given quantities.




Answer:

Given:


Emf = ϵ


Resistance = r


Angle made by rod = θ


Angular velocity = w


Formula used:


From the previous questions, induced emf … (i), where B = magnetic field, w = angular velocity, a = radius


Hence, Total emf =


Total current i = total emf/resistance = … (ii), where R = resistance


‘Now, net force on the rod … (iii),


where mgcosθ = component of weight along rod, where m = mass, g = acceleration due to gravity, θ = angle made by rod with horizontal, and ilB = Magnetic force, where i = current, l = length of rod, B= magnetic field.


Since the rod rotates with uniform angular velocity, net torque about O = 0.


Hence, torque = net force x distance from line of action = , where a = radius of rod


Therefore, … (iii)


Hence, (Ans)



Question 63.

A wire of mass m and length ℓ can slide freely on a pair of smooth, vertical rails figure. A magnetic field B exists in the region in the direction perpendicular to the plane of the rails. The rails are connected at the top end by a capacitor of capacitance C. Find the acceleration of the wire neglecting any electric resistance.




Answer:

Given:


Mass = m


Length = l


Magnetic field = B


Capacitance = C


Formula used:


Induced emf … (i), where B = magnetic field, l = length, v = velocity


Also, we know that … (ii), where q = charge, C = capacitance.


Hence, .. (ii)


Therefore, current , where q = charge, t = time


=> … (iii), where a = acceleration


Therefore, net force on rod = weight - magnetic force = mg - ilB .. (iv), where m = mass, g = acceleration due to gravity, i = current, l = length, B = magnetic force


From newton’s second law of motion, F = ma … (v), where f= force, m = mass, a = acceleration


Therefore, (from (iii))



=>acceleration (Ans)



Question 64.

A uniform magnetic field B exists in a cylindrical region, shown dotted in figure. The magnetic field increases at a constant rate dB/dt. Consider a circle of radius r coaxial with the cylindrical region.

(a) Find the magnitude of the electric field E at a point on the circumference of the circle.

(b) Consider a point P on the side of the square circumscribing the circle. Show that the component of the induced electric field at P along ba is the same as the magnitude forum in part (a).




Answer:

Given:


Magnetic field = B


Rate of increase of magnetic field = dB/dt


Radius = r


Formula used:


(a) Induced emf … (i), where = magnetic flux, t = time


Now, = B.A where B = magnetic field, A = area


Hence, … (ii)


For the circular loop, … (iii), where A = area, r = radius


Let the electric field be E


Hence, … (iv) ,where dr = element of length, E’ = emf


Hence, for this loop, , where r = radius



(Ans)


(b) When the square is considered, A = (2r)2 = 4r2, where A = area, r = radius


In this case, (perimeter of square)


Hence, from , where E = electric field, dr = length element, E’ = emf, we get



=> electric field (Ans)



Question 65.

The current in an ideal, long solenoid is varied at a uniform rate of 0.01 A s–1. The solenoid has 2000 turns/m and its radius is 6.0 cm.

(a) Consider a circle of radius 1.0 cm inside the solenoid with its axis coinciding with the axis of the solenoid. Write the change in the magnetic flux through this circle in 2.0 seconds.

(b) Find the electric field induced at a point on the circumference of the circle.

(c) Find the electric field induced at a point outside the solenoid at a distance 8.0 cm from its axis.


Answer:

Given:


Rate of variation of current = 0.01 A s–1.


No of turns/m (n) = 2000


Radius(r) = 6 cm = 0.06 m


Formula used:


(a) Radius of circle(r’) = 1 cm = 0.01 m


Time(t) = 2 s


For two seconds, change of current = (2 x 0.01 A.) = 0.02 A


Magnetic flux , where B = magnetic field, A = area


Area of circle


Magnetic field of a solenoid , where μ0 = magnetic permeability of vacuum, n = number of turns per unit length, Δi = change in current


Hence, flux =


Wb


Hence, in 1 second = Wb (Ans)


(b) = , where E = electric field, dr = line element, E’ = emf, = flux, t = time


Hence, in this case, this becomes


, where r = radius of circle


Vm-1(Ans)


(c) For the point located outside,


Wbs-1


= flux, t = time, μ0 = magnetic permeability of vacuum, n = number of turns per unit length, di/dt = rate of change in current


, where E = electric field, r = radius of circle(since )


Hence,


Vm-1 (Ans)



Question 66.

An average emf of 20 V is induced in an inductor when the current in it is changed from 2.5 A in one direction to the same value in the opposite direction in 0.1 s. Find the self-inductance of the inductor.


Answer:

Given:


Emf(E) = 20 V


Rate of change of current = A/s = 50 As-1


Formula used:


Emf , where L = self inductance, di/dt = rate of change of current


Substituting the values, we get



=> self inductance (Ans)



Question 67.

A magnetic flux of 8 × 10–4 weber is linked with each turn of a 200-turn coil when there is an electric current of 4A in it. Calculate the self-inductance of the coil.


Answer:

Given:


Magnetic flux(Φ) = 8 × 10–4 Wb


Number of turns(n) = 200


Current(i) = 4 A ω


Formula used:


, where L = self inductance, n = number of turns, Φ = flux, i = current


Putting the values, we get


(Ans)



Question 68.

The current in a solenoid of 240 turns, having a length of 12 cm and a radius of 2 cm, changes at a rate of 0.8 A s–1. Find the emf induced in it.


Answer:

Given:


Number of turns(n) = 240


Length of solenoid(l) = 12 cm = 0.12 m


Radius(r) = 2 cm = 0.02 m


Rate of change of current(di/dt) = 0.8 As-1


Formula used:


L = μ0n2A/l, where L = self inductance, μ0 = magnetic permeability of vacuum n = number of turns, A = area, l = length


Now, emf , where L = self inductance, = rate of change of current


Putting the values:


= V (Ans)



Question 69.

Find the value of t/τ for which the current in an LR circuit builds

up to

(a) 90%

(b) 99%

(c) 99.9%

of the steady-state value.


Answer:

For a series LR circuit, the current across the inductor varies as a


function of time. The current across the inductor at time t will


be


…(i)


where i0 is the current at time t=0(also called the steady state value), R is the resistance of the resistor and L is the inductance of the inductor.


We can define a quantity called the time constant for a series LR circuit. It is given as



So equation(i) becomes


…(ii)


We have to find the values of for three different values of current


(a)- when i is 90% of i0


90% of i0 is i0, so


Putting these values in eq(ii)






Taking natural logarithm on both sides




(as ln (1) is equal to 0)



The value of for which the current is 90% of steady state


value is 2.3.


(b)- when i is 99% of i0


99% of i0 is i0, so


Putting these values in eq(ii)






Taking natural logarithm on both sides




(as ln (1) is equal to 0)



The value of for which the current is 99% of steady state


value is 4.6.


(c)- when i is 90% of i0


99.9% of i0 is i0, so


Putting these values in eq(ii)






Taking natural logarithm on both sides




(as ln (1) is equal to 0)



The value of for which the current is 99.9% of steady state


value is 6.9.