ROUTERA


Magnetic Field

Class 12th Concepts Of Physics Part 2 HC Verma Solution



Short Answer
Question 1.

Suppose a charged particle moves with a velocity v near a wire carrying an electric current. A magnetic force, therefore, acts on it. If the same particle is seen from a frame moving with velocity v in the same direction, the charge will be found at rest. Will the magnetic force become zero in this frame? Will the magnetic field become zero in this frame?


Answer:

Formula used:


The magnetic force is given by



Where F is magnetic force


q is electric charge =


v is velocity


Angle between magnetic force and velocity component


A Charge particle is moving near a conducting wire. And also a frame is moving in the same direction which the charge is moving. Seeing from the frame it is known that the charge is at rest. So velocity of the charge particles will be zero. Thus magnetic force will be zero.


When the charge is at rest the current will be zero. We don’t know either it is a positive or negative charge. Let us consider it is a negative charge. Seeing from the frame negative charge is at rest. But there is a positive charge flow. Due to this magnetic field will exists. So the magnetic field should not be zero.



Question 2.

Can a charged particle be accelerated by a magnetic field? Can its speed be increased?


Answer:

Formula used:



Where F is magnetic force


m is mass


a is acceleration


Here the force (magnetic force) which is acting on the charged particle will try to accelerate it. If the charged particle is in the direction perpendicular to the magnetic field, then the particle will move in a circular path. In that circular path the direction of the charged particle will change.


Formula used: Magnetic force is given by



Where F is magnetic force


q is electric charge =


v is velocity


Angle between magnetic force and velocity component


Force is always act perpendicular to the direction of velocity of a charged particle. So it can’t change the magnitude of the velocity of the charged particle. Velocity, magnetic field and force are perpendicular to each other. So it can change the direction of the velocity of a charged particle. So our answer is Yes, a charged particle can be accelerated by a magnetic filed but its speed can’t be increase.



Question 3.

Will a current loop placed in a magnetic field always experience a zero force?


Answer:

Effect of the magnetic force on a current loop is totally based on the uniformity of the magnetic field. If the field is non uniform, then current loop will not experience a force. So magnetic force on the loop will be zero. If the field is uniform, then the force experienced on the current loop either zero or non zero.



Question 4.

The free electrons in a conducting wire are in constant thermal motion. If such a wire, carrying no current, is placed in a magnetic field, is there a magnetic force on each free electron? Is there a magnetic force on the wire?


Answer:

Formula used: The magnetic force is given by



Where F is magnetic force


I is current


L is length of the wire


B is magnetic field


If the electron is in motion definitely there will be a magnetic force on it. Electrons are present in conducting wire. But current flowing in the wire is zero. When this wire is placed in a magnetic field the electrons will come to motion but the motion will be random. So we can neglect the effect of force on each electron. So total magnetic force acting on the wire will be zero.



Question 5.

Assume that the magnetic field is uniform in a cubical region and is zero outside. Can you project a charged particle from outside into the field so that the particle describes a complete circle in the field?


Answer:

Magnetic field is uniform(assumption). If the charged particle is perpendicular to the magnetic field the particle will exhibits a circular trajectory. Let us consider that the charged particle is coming into the field from outside along the positive y-axis. And Let us take the field is in x-axis. X-axis is perpendicular to the y-axis. So magnetic field will be perpendicular to the changed particle. Then the particle will make a circular path.



Question 6.

An electron beam projected along the positive x-axis deflects along the positive y-axis. If this deflection is caused by a magnetic field, what is the direction of the field? Can we conclude that the field is parallel to the z-axis?


Answer:

If the force is acting in the y-axis then the electron beam will be deflected from positive x-axis along y-axis. The electrons motion is in positive x-axis and then current will be along negative x-axis. And the force is acting along the y-axis. According to Fleming’s left hand rule the magnetic filed will be in z-axis. So we can conclude that the field is parallel to the z-axis.



Question 7.

Is it possible for a current loop to stay without rotating in a uniform magnetic field? If yes, what should be the orientation of the loop?


Answer:

Yes

Formula used: The net torque of a loop is given by



Where τ is torque


i is current


A is area vector


B is magnetic field


θ is the angle between the area vector and magnetic field


When θ=0,



When θ=180,




If θ is zero or 180° (integral multiple of π) then torque will be zero. So coil will stop its rotation. θ=nπ represents that the magnetic field is in parallel with area vector. So the loop can stay in a uniform magnetic field without rotation.



Question 8.

The net charge in a current-carrying wire is zero. Then, why does a magnetic field exert a force on it?


Answer:

Mostly positive charge on the wire is because of protons which are containing nucleus. When the protons are not in motion the force acting on them will be zero. So charge will be zero. But there is a negative charge too. Negative charge is due to electrons.

Formula used:


Where F is magnetic force


q is electric charge =


v is velocity


B is magnetic field


is angle between magnetic field and a charge


So when the electrons are in motion the wire will carry current. Magnetic force will act on the wire. That’s why magnetic field exerts a force on the wire.



Question 9.

The torque on a current loop is zero if the angel between the positive normal and the magnetic field is either θ = 0 or θ = 180°. In which of the two orientations, the equilibrium is stable?


Answer:

Formula used: Potential energy is given by



Where m is magnetic moment


B is magnetic field


Let is take B=0.1T, m = A m �


If θ is zero, then





If θ=180 then





If energy is minimum, then system will be more stable. If energy is increase, then the system will loose it’s stability. For θ=0 the energy is minimum and the equilibrium will stable. For θ=180 the energy is maximum and equilibrium will unstable.



Question 10.

Verify that the unit’s weber and volt second are the same.


Answer:

Formula used: Magnetic force is by



Where F is magnetic force


q is electric charge =


v is velocity


B is magnetic field



One of the unit that magnetic force measured is Weber/m �.






(We know that




Thus weber and the volt second are same.




Objective I
Question 1.

A positively charged particle projected towards east is deflected towards north by a magnetic field. The field may be
A. towards west

B. towards south

C. upward

D. downward.


Answer:

The positive charge is moving towards east. So the current is in the direction of east. It is deflected towards north. So the magnetic force is acting in the north direction. Motion is in east and force is acting in north. Magnetic field is always perpendicular to the both force and velocity of a particle. According to left hand Fleming’s rule the magnetic force will be in downward direction.


Question 2.

A charged particle is whirled in a horizontal circle on a frictionless table by attaching it to a string fixed at one point. If a magnetic field is switched on in the vertical direction, the tension in the string
A. will increase

B. will decrease

C. will remain the same

D. may increase or decrease


Answer:


Force is depending upon the direction of the whirl and also charge of a particle. Which type of charge particle it is not given. And We don’t know which direction the whirl is rotating. Based on these things we can say that force(tension) may increase or decrease. The magnetic field is in vertical direction and also we don’t know it is upward or downward. According to Flemings left hand rule the force will act either inward or outward. So option D is correct.


Question 3.

Which of the following particles will experience maximum magnetic force (magnitude) when projected with the same velocity perpendicular to a magnetic field?
A. Electron

B. Proton

C. He+

D. Li++


Answer:

Here the particle is experiencing a maximum magnetic filed and projected with a same velocity. There are no variations in the both. So both magnetic field and velocity are constant over here.

Formula used: The magnetic force is given by



Where F is magnetic force


q is electric charge =


v is velocity


is angle between magnetic field and a charge


In the above equation magnetic field and velocity are constant. So force will depend on the charge. In the option given Li++ has the highest charge. After Li++, He+ has highest charge. So Li++ will experience a maximum magnetic field.


Question 4.

Which of the following particles will describe the smallest circle when projected with the same velocity perpendicular to a magnetic field?
A. Electron

B. Proton

C. He+

D. Li+


Answer:

In the options given election has a lowest charge.

Electron < Proton < He+ <Li++


Formula used: The magnetic force is given by



Where F is magnetic force


q is electric charge =


v is velocity



m is mass of the charge particle


v is velocity


r is radius of the projectile motion






Here v and B are constants and q is same for all the charge particles. r is proportional to m. If radius is small radius will become small. Electron exhibits light weight among the all other charge particles. Because of light weight it will creates a circle with a small radius when it is projected. So option A is correct.


Question 5.

Which of the following particles will have minimum frequency of revolution when projected with the same velocity perpendicular to a magnetic field?
A. Electron

B. Proton

C. He+

D. Li+


Answer:

Formula used: Time period of revolution of a particle is given by




In the above options Electron is light weight particle and Li+ has highest weight compared to other particles in the options given. Frequency f is inversely proportional to mass m. So Li+ will have minimum frequency of revolution. So option D is correct.


Question 6.

A circular loop of area 1 cm2, carrying a current of 10 A, is placed in a magnetic field of 0.1 T perpendicular to the plane of the loop. The torque on the loop due to the magnetic field is
A. zero

B. 10–4 Nm

C. 10–2 Nm

D. 1 N m


Answer:

Given a circular loop area =1cm �


Current =10A


Magnetic field strength =0.1 T


Given loop is circular. On each and every point of this circular loop there exists two forces. It will happen in case of circular loop only. The forces are equal but their magnitudes are opposite. Because of the opposite magnitudes they cancel each other. Then net force will be equal to zero. If we place any loop in magnetic field it definitely conducts current. In case of circular loop, the torque due to magnetic field on loop will be zero.


Question 7.

A beam consisting of protons and electrons moving at the same speed goes through a thin region in which there is a magnetic field perpendicular to the beam. The protons and the electrons
A. will go undeviated

B. will be deviated by the same angle and will not separate

C. will be deviated by different angels and hence separate

D. will be deviated by the same angle but will separate.


Answer:

Formula used: Magnetic force is given by



Where F is magnetic force


q is electric charge =


v is velocity


Force on the electron is


Force on a the proton is


The magnitudes of force on proton and electron are different but magnetic force acting on the both are same. The force acting on the proton is in positive in direction and the force acting on the electron is in negative direction. Even same force is acting on them but because of different directions they got deviated by different angles and then they will separate. So option C is correct.


Question 8.

A charged particle moves in a uniform magnetic field. The velocity of the particle at some instant makes an acute angel with the magnetic field. The path of the particle will be
A. a straight line

B. a circle

C. a helix with uniform pitch

D. a helix with non-uniform pitch


Answer:

Given that velocity of a particle is making an acute angle with the magnetic field. Velocity of a particle is resolves into two components. Magnetic field always change the direction of the velocity. One component is perpendicular to the magnetic field that is Vsinθ and another component is parallel to the magnetic field that is Vcosθ. Vertical component of a velocity is perpendicular to the magnetic field. Then force will act on the charged particle. Because of the vertical force(Vsinθ), the particle will make a circular path. Because of the two components of the velocity the particle will move in a helical path. This path will maintain the uniformity. So option C is correct.



Question 9.

A particle moves in a region having a uniform magnetic field and a parallel, uniform electric field. At some instant, the velocity of the particle is perpendicular to the field direction. The path of the particle will be
A. a straight line

B. a circle

C. a helix with uniform pitch

D. a helix with non-uniform pitch


Answer:

The region has a uniform magnetic field and an electric field.

Formula used: Lorentz’s force is given by



Where F is Lorentz’s force


q is electric charge


E is electric field strength


B is magnetic field strength


V is velocity


The electric field will accelerate the particle to move in the direction of the magnetic field. Magnetic field will always change the direction of velocity of a particle. Because of these the particle will move in a helix path with non-uniformity. So option D is correct.


Question 10.

An electric current i enters and leaves a uniform circular wire of radius a through diametrically opposite points. A charge particle q moving along the axis of the circular wire passes tough its center at speed v. The magnetic force acting on the particle when it passes through the center has a magnitude.
A.

B.

C.

D. zero


Answer:

Current is entering into the circular wire. According to right hand thumb rule the magnetic field will be in axis of the circular wire. It is given that the charge particle is moving along the axis of the wire.

Formula used: Magnetic force is given by the formula



Where F is magnetic force


q is electric charge =


v is velocity


B is magnetic filed


is angle between magnetic field and a charge


θ will be zero because the angle between the magnetic field and charge particle is 0° or 180°.




When the particle is moving along the axis of a wire the magnetic force acting on it will be zero. So option D is correct.



Objective Ii
Question 1.

If a charged particle at rest experiences no electromagnetic force,
A. the electric field must be zero.

B. the magnetic field must be zero

C. the electric field may or may not be zero

D. the magnetic field may or may not be zero


Answer:

Formula used: The magnetic force is given by



Where F is magnetic force


q is electric charge =


v is velocity


B is magnetic filed


is angle between magnetic field and charge


Here the particle is rest so velocity is zero. Then magnetic force is zero. But we don’t know whether the magnetic field is zero or not as the force exerted on a particle under the influence of magnetic field depends on the orientation of the particle with respect to the magnetic field. And the electromagnetic force on particle is also not present. Magnetic field may not be zero. So electric field must be zero. So option A and D are correct.


Question 2.

If a charged particle kept at rest experiences an electromagnetic force,
A. the electric field must not be zero

B. the magnetic field must not be zero

C. the electric field may or may not be zero

D. The magnetic field may or may not be zero.


Answer:

The electromagnetic force is acting on the charge particle.

Formula used: 1. Electric force is given by



Where F is electric force


q is electric charge


E is electric field


From the above formula we can say that electric field should not be zero. Because electric force is acting on the particle. And the particle is rest, so velocity of the charged particle is zero.


2. The magnetic force is given by



Where F is magnetic force


q is electric charge =


v is velocity


B is magnetic filed


is angle between magnetic field and charge


From the above equation we can say that magnetic fore will become zero. But we don’t know whether the magnetic filed is zero or not. We cannot make the statement about magnetic field. So magnetic field may or may not be zero. Option A and D are correct.


Question 3.

If a charged particle projected in a gravity-free room deflects,
A. there must be an electric field

B. there must be a magnetic filed

C. both fields cannot be zero.

D. both fields can be non zero.


Answer:

The particle is projected in gravity-free room. If no force is acting on the charge particle means magnetic and electric force is not acting on the particle, then the particle will not deflect. If any one the force is acting on the charged particle, then it gets deflects. And also if both the force is acting on it then also the particle will deflect. So both electric and magnetic field cannot be zero. And also they can be non zero. So options C an D are correct.


Question 4.

A charged particle moves in a gravity-free space without change in velocity. Which of the following is/are possible?
A. E = 0, B = 0

B. E = 0, B ≠ 0

C. E ≠ 0, B = 0

D. E ≠ 0, B ≠ 0


Answer:

The particle is in a gravity free space. If no force is acting on the particle, the velocity will be constant. If both the force electric and magnetic forces acting on the particle is same then both forces get canceled and force acting on the particle will be zero. And if electric force is zero and magnetic force is non zero, in this case if the magnetic field is in the direction of velocity of a particle then the magnetic force acting on it will become zero. Option A, B and D are correct.


Question 5.

A charged particle moves along a circle under the action of possible constant electric and magnetic fields. Which of the following are possible?
A. E = 0, B = 0

B. E = 0, B ≠ 0

C. E ≠ 0, B = 0

D. E ≠ 0, B ≠ 0


Answer:

charged particle is moving in a circle. If Electric force is acting on the particle, the speed of the particle will increase. Because of this the particle cannot be move in a circle. In case of magnetic field, it will only change the direction of the velocity of a particle. It can’t change speed of the charged particle. So magnetic field cannot be zero. So option b is correct.


Question 6.

A charged particle goes undeflected in a region containing electric and magnetic field. It is possible that
A.

B. is not parallel to

C. is not parallel to

D. but is not parallel to .


Answer:

Let us consider the electric field is parallel to the magnetic field. In this case the charges particle will accelerate and also it will move in same direction. Then it cannot deflect. If electric field is not parallel to magnetic field means that may be in perpendicular to each other. In this case magnetic field, electric field and velocity of a particle will be in perpendicular to each other. If both forces are equal, then force acting on the particle will be zero. Then particle will not deflect. So option A and B are correct. Option C and D are not valid conditions.


Question 7.

If a charged particle goes unaccelerated in a region containing electric and magnetic fields.
A. must be perpendicular to

B. must be perpendicular to

C. must be perpendicular to

D. E must be equal to vB.


Answer:

The charged particle must not accelerate in the region. If both the field are counter balance, then the particle won’t accelerate. If Electric force is acting along the direction of magnetic force and both are same then the particle will not effect by these fields. So particle won’t accelerate. And if the electric field is parallel to the velocity of a particle then it will definitely accelerate. So electric field must be perpendicular to the direction of the velocity. So options A and B are correct.


Question 8.

Two ions have equal masses but one is singly ionized and the other is doubly ionized. They are projected from the same place in a uniform magnetic field with the same velocity perpendicular to the field.
A. Both ions will go along circles of equal radii.

B. The circle described by the singly ionized charge will have a radius double that of the other circle.

C. The two circles do not touch each other.

D. The two circles touch each other.


Answer:

Formula used: Radius of the orbit of a particle is given by



Where r is radius of the orbit


m is mass


v is velocity


q is electric charge


B is electric field


Radius r is inversely proportional to the charge. So singly ionized charge will have a double radius than doubly ionized charge. And both the charges are projected from the same place. So they will touch each other at the beginning. So options B and D are correct.


Question 9.

An electron is moving along the positive x-axis. You want to apply a magnetic field for a short time so that the electron may reverse its direction and move parallel to the negative x-axis. This can be done by applying the magnetic field along.
A. y-axis

B. z-axis

C. y-axis only

D. z-axis only


Answer:

Electron is moving along the positive x-axis now. If the particle is moving in the direction of magnetic field, then the field does not change the direction of velocity of the electron. If we apply the magnetic field in the y-axis it can reverse the direction of velocity of the electron. And also field in z-axis also can reverse the direction of the electron. So option A and B are correct.


Question 10.

Let and denote electric and magnetic fields in a frame S. and in another frame S’ moving with respect to S at a velocity Two of the following equations are wrong. Identify them.
A.

B.

C.

D.


Answer:

Formula used:


Electric force is given by


Where F is electric force


q is electric charge =


E is electric field


Magnetic force is given by


Where F is magnetic force


q is electric charge =


v is velocity


B is magnetic force


From the above formulas and


From the dimensional analysis we can write equations as


i.


ii.


So only options A and D are correct.



Exercises
Question 1.

An alpha particle is projected vertically upward with a speed of 3.0 × 104 kg s–1 in a region where a magnetic field of magnitude 1.0 T exists in the direction south to north. Find the magnetic force that acts on the α-particle.


Answer:

Given -


Speed of the alpha particle in upward direction,


v = 3 × 104 km/s


since 1Km = 1000m


⇒ v = 3 × 107 m/s



Magnetic field, B = 1.0 T



Charge on the alpha particle is given as ,




where e is the charge of an electron.



q = 2 × 1.6 × 10−19 C,


We need to calculate the magnetic force that acts on the α-particl



The direction of magnetic force can be found using Fleming’s left-hand rule.


The direction of the magnetic field is from south to north.



We know, Lorentz force F is given by -



where,


q = charge


v = velocity of the charge


B=magnetic field


and


θ= angle between V and B


Now, magnetic force acting on the α -particle,



Substituting the values



= 2 × 1.6 × 10−19 × 3 × 107 × 1



= 9.6 × 10−12 N



Question 2.

An electron is projected horizontally with a kinetic energy of 10 keV. A magnetic filed of strength 1.0 × 10–7 T exists in the vertically upward direction.

(a) Will the electron deflect toward right or towards left of its motion?

(b) Calculate the sideways deflection of the electron in travelling through 1m. Make appropriate approximations.


Answer:

Given-
The kinetic energy of the electron when projected towards horizontal direction,


K.E = 10 keV = 1.6 × 10−15 J



Magnetic field, B = 1 × 10−7 T


Charge on an electron =1.60× 10-19


mass of an electron = 9.1× 10-31 kilograms





(a) The direction electron deflection can be found by the right-hand screw rule.


Given the direction of magnetic field is vertically upward.


So, the electron will be deflected towards left.



(b) Kinetic energy-




Where


m is the mass of the electron


v is the velocity of the electron


Solving for v, we get



We know, Lorentz force F is given by -



where,


q = charge


v = velocity of the charge


B=magnetic field


and


θ= angle between V and B


And Newton's second law of motion



where m = mass


a= acceleration


we know






From (1) and (2) -



Applying 2nd equation of motion




where,


a= acceleration


u = initial velocity


t= Time taken to cross the magnetic field


Since, the initial velocity is zero,


from (1),(2) and (3)




substituting the values -





From (1)



substituting the values-





Question 3.

A magnetic field of (4.0 x 10-3 vector k)T exerts a force of (4.0 vector i + 3.0 vector j) x 10-10N on a particle having a charge of 1.0 x 10-9C and going in the X-Y plane. Find the velocity of the particle.


Answer:

Given -




Magnetic field, B = 4.0 × 10–3 T



Electric charge on the particle, q = 1 × 10−9 C


Also given that the charge is going in the X-Y plane,



Therefore, the x-component of force, Fx = 4 × 10−10 N



and the y-component of force, Fy = 3 × 10−10 N



We know, Lorentz force F is given by -



where,


q = charge


v = velocity of the charge


B=magnetic field


and


θ= angle between V and B




On putting the respective values, we get


Let’s take motion along x-axis-


From (1)





Similarly, motion along y-axis



from (1)




Hence. velocity of the particle, = m/s



Question 4.

An experimenter’s diary reads as follow: “a charged particle is projected in a magnetic field of The acceleration of the particle is found to be The number to the left of in in the last expression was not readable. What can this number be?


Answer:

Given-


Magnetic field,


Acceleration of the particle,


Let denoted the unidentified number as


Since, magnetic force always acts perpendicular to the motion of


the particle, so, B and a are perpendicular to each other.


So, the dot product of the two quantities should be zero.



That is,








Hence acceleration of the particle is m/s2.



Question 5.

A 10 g bullet having a charge of 4.00 μC is fired at a speed of 270 ms–1 in a horizontal direction. A vertical magnetic field of 500 μT exists in the space. Find the deflection of the bullet due to the magnetic field as it travels through 100 m. Make appropriate approximations.


Answer:

Given-



Mass of the bullet, m = 10g = 10-3 Kg



Charge of the bullet, q = 4.00 μC =10-6 C



Speed of the bullet in horizontal direction, v = 270 m/s



Vertical magnetic field, B = 500 μT = 500× 10-6 T



Distance travelled by the bullet, d = 100 m



Magnetic force,


We know, Lorentz force F is given by -



where,


q = charge


v = velocity of the charge


B=magnetic field


and


θ= angle between V and B


Also,



And Newton's second law of motion



where m = mass


a= acceleration


Using equation (1) –





we know





Substituting the values, time taken by the bullet to travel 100 m horizontally,



Applying 2nd equation of motion




where,


a= acceleration


u = initial velocity


t= Time taken to cross the magnetic field


Since, the initial velocity is zero,


from (1),(2) and (3)



Now, the deflection caused by the magnetic field in this time


interval,






Question 6.

When a proton is released from rest in a room, it starts with an initial acceleration a0 towards west. When it is projected towards north with a sped v0, it moves with an initial acceleration 3a0 towards west. Find the electric field and the minimum possible magnetic field in the room.


Answer:

Given-


Proton is released from rest in a room, it starts with an initial acceleration a0 towards west


Now, we know coulomb’s force F given by –




where


q = charge


E= electric field


Also,



And Newton's second law of motion



Where


m = mass


a= acceleration






From (1) and (2)



Electric field,



which acts towards west.


We know, Lorentz force F is given by -



where,


q = charge


v = velocity of the charge


B=magnetic field


and


and θ = the angle between B and l


Now, given that when the proton is projected towards north with a speed v0, it moves with an initial acceleration 3a0 towards west.





An electric force will act on the proton in the west direction, which produces an acceleration a0 on the proton.


Initially the proton was moving with velocity v, so a magnetic force was also acting on the proton.


So, magnetic force acting will be the only cause behind the change in acceleration of the proton



Change in acceleration towards west due to the magnetic force acting on it is -



∆a = 3a0a0 = 2a0



So from (2), the force will become -




Hence, required magnetic field




Question 7.

Consider a 10-cm long portion of a straight wire carrying a current of 10 A placed in a magnetic field of 0.1 T making an angle of 53° with the wire. What magnetic force does the wire experience?


Answer:

Given-
Length of wire, l = 10 cm



Electric current passing through the wire, I = 10 A



Magnetic field, B = 0.1 T



Angle between the wire and magnetic field, θ = 53°


Magnetic Force on a Current carrying wire is given by



where,


B= magnetic field


I = current


L = length of the wire


and θ = the angle between B and l



hence, magnetic force,









The direction of force can be found using Fleming’s left-hand rule –


which states that –



whenever a current carrying conductor is placed inside a magnetic field, a force acts on the conductor, in a direction perpendicular to both the directions of the current and the magnetic field



Therefore, the direction of magnetic force is perpendicular to the wire as well as the magnetic field.



Question 8.

A current of 2A enters at the corner d of a square frame abcd of side 20 cm and leaves at the opposite corner b. A magnetic field B = 0.1 T exists in the space in a direction perpendicular to the plane of the frame as shown in figure. Find the magnitude and direction of the magnetic force on the four sides of the frame.




Answer:

Given-


square of side, l = 20 cm


Electric current passing through the wire, I = 2 A



Magnetic field, B = 0.1 T



The direction of magnetic field is perpendicular to the plane of the frame, coming out of the plane.



Given in the question, that current enters at the corner d of the square frame and leaves at the opposite corner b.



Hence, angle between the frame and magnetic field, θ = 90°


Now, we know-


Magnetic Force on a Current carrying wire is given by



where,


B= magnetic field


I = current


L = length of the wire


and θ = the angle between B and l


Hence, magnetic force,




For wire along the sides da and cb,







The direction of force can be found using Fleming’s left-hand


rule.


which states that –



whenever a current carrying conductor is placed inside a magnetic field, a force acts on the conductor, in a direction perpendicular to both the directions of the current and the magnetic field



Thus, the direction of magnetic force will be towards the left.



now, for wires along sides, dc and ab,










Again, here the direction of force can be found using Fleming’s left-hand rule.



Thus, the direction of magnetic force will be downwards.



Question 9.

A magnetic field of strength 1.0 T is produced by a strong electromagnet in a cylindrical region of radius 4.0 cm as shown in figure. A wire, carrying a current of 2.0 A, is placed perpendicular to and intersecting the axis of the cylindrical region. Find the magnitude of the force acting on the wire.




Answer:

Given-


Magnetic field, (B) = 1 T


Radius of the cylindrical region, r = 4.0 cm


Electric current through the wire, I = 2 A


The wire is placed perpendicular the direction of magnetic field


So, angle between wire and magnetic field, θ = 90


Magnetic Force on a Current carrying wire is given by



where,


B= magnetic field


I = current


l = length of the wire


and θ = the angle between B and l


Magnetic force,




Here, length l which is under the base of the cylinder is of radius 2r










Question 10.

A wire of length ℓ carries a current i along the x-axis. A magnetic field exits which is given as Find the magnitude of the magnetic force acting on the wire.


Answer:

Given-



length of wire l cm



Electric current through the wire = A



Magnetic field in vector form is given as –




Given in the question, that the current is passing along the X-axis.


Magnetic Force on a Current carrying wire is given by



where,


B= magnetic field


I = current


l = length of the wire


and θ = the angle between B and l


Magnetic force-




Substituting the values –




since current I is along x- axis, from vector laws, cross product of same vectors is zero –







The magnitude of the magnetic force,






Question 11.

A current of 5.0 A exists in the circuit shown in figure. The wire PQ has a length of 50 cm and the magnetic field in which it is immersed has a magnitude of 0.20 T. Find the magnetic force acting on the wire PQ.




Answer:

Given-
Length of the wire PQ inside the magnetic field, l = 50 cm
Electric current , I = 5 A
Magnetic field, B = 0.2 T



From fig since magnetic field is displayed as “cross”, we can say that the direction of magnetic field is perpendicular to the plane and it is going into the plane.




Angle between the plane of the wire and the magnetic field, θ = 900




where,


B= magnetic field


I = current


L = length of the wire


and θ = the angle between B and L








The direction of force can be found using Fleming’s left-hand rule.


whenever a current carrying conductor is placed inside a magnetic field, a force acts on the conductor, in a direction perpendicular to both the directions of the current and the magnetic field


Thus, the direction of magnetic force is upwards in the plane of the paper.



Question 12.

A circular loop of radius a, carrying a current i, is placed in a two-dimensional magnetic field. The centre of the loop coincides with the centre of the field figure. The strength of the magnetic field at the periphery of the loop is B. Find the magnetic force on the wire.




Answer:

Given –


circular loop of radius = a


So, the length of the loop is its circumference ,




Electric current through the loop = i



Given that the loop is placed in a two-dimensional magnetic field.


Also the centre of the loop coincides with the centre of the field.


The strength of the magnetic field at the periphery of the loop is B


Hence, the direction of the magnetic is radially outwards.



Here, the angle between the length of the loop and the magnetic field, θ = 900




where,


B= magnetic field


I = current


l = length of the wire


and θ = the angle between B and l





The direction of force can be found using Fleming’s left-hand rule -



whenever a current carrying conductor is placed inside a magnetic field, a force acts on the conductor, in a direction perpendicular to both the directions of the current and the magnetic field


Thus, the direction of magnetic force lies perpendicular to the plane pointing inwards.



Question 13.

A hypothetical magnetic field existing in a region is given by , where denotes the unit vector along the radial direction. A circular loop of radius a, carrying a current i, is placed with its plane parallel to the x-y plane and the centre at (0, 0, d). Find the magnitude of the magnetic force acting on the loop.


Answer:

Given-
hypothetical magnetic field exists in a region,



Where


, is a unit vector along the radial direction a.


It is a circular loop of radius a


So, the length of the loop,


Electric current through loop = i



Also, the loop is placed with its plane parallel to the XY plane and its centre lies at (0,0, d).




The angle between the length of the loop l and the magnetic field B is θ


Magnetic force is given by



where,


B= magnetic field


I = current


l = length of the wire


and θ = the angle between B and l



Substituting the values -



Looking into the fig,





Question 14.

A rectangular wire-loop of width a is suspended from the insulated pan of a spring balance as shown in figure. A current i exists in the anticlockwise direction in the loop. A magnetic field B exists in the lower region. Find the change in the tension of the spring if the current in the loop is reversed.




Answer:


Given-


width of wire loop = a


Electric current through the loop = i


Direction of the current is anti-clockwise.


Strength of the magnetic field in the lower region = B
From fig, we can say that direction of the magnetic field is into the plane of the loop.


Here, angle between the length of the loop and magnetic field, θ = 900


Magnetic force is given by



where,


B= magnetic field


I = current


l = length of the wire


and θ = the angle between B and l


The magnetic force will act only on side AD and BC.



As side AD is outside the magnetic field, so F = 0
Magnetic force on side BC is





Direction of force can be found using Fleming’s left-hand rule.



whenever a current carrying conductor is placed inside a magnetic field, a force acts on the conductor, in a direction perpendicular to both the directions of the current and the magnetic field


Thus, the direction of the magnetic force will be upward.


Similarly altering the direction of current to clockwise, the force


along BC-




Thus, the change in force is equal to the change in tension






Question 15.

A current loop of arbitrary shape lies in a uniform magnetic field B. Show that the net magnetic force acting on the loop is zero.


Answer:

Let’s


assume a square magnetic loop


Let,
Uniform magnetic field existing in the region of the arbitrary loop = B
current flowing through the loop be i.



Length of each side of the loop be l.



Assuming the direction of the current clockwise.



Direction of the magnetic field is going towards the plane of the loop.


Magnetic force is given by



where,


B= magnetic field


I = current


l = length of the wire


and θ = the angle between B and l





Here, θ = 900


Direction of force can be found using Fleming’s left-hand rule.



Force F1 acting on side AB is


= directed upwards



Force F2 acting on side DC =


directed downwards



Since, F1 and F2 are equal and in opposite direction, they will cancel each other.



Similarly,


Force F3 acting on AD =


directed outwards pointing


And



Force F4 acting on BC = directed outwards



Since, F3 and F4 are equal and in opposite direction, they will cancel each other



Hence, the net force acting on the arbitrary loop is 0.



Question 16.

Prove that the force acting on a current carrying wire, joining two fixed points a and b in a uniform magnetic field, is independent of the shape of the wire.


Answer:


Magnetic force acting on a current carrying wire in an uniform magnetic field is given by



where,


B= magnetic field


I = current


l = length of the wire


and θ = the angle between B and l


Since


is vector


Also, the length of the wire is fixed from A to B, so force is independent of the shape of the wire.



Question 17.

A semicircular wire of radius 5.0 cm carries a current of 5.0 A. A magnetic field B of magnitude 0.50 T exists along the perpendicular to the plane of the wire. Find the magnitude of the magnetic force acting on the wire.


Answer:


Given-



Radius of semicircular wire, r = 5.0 cm



Thus, the length of the wire = 2r= 10 cm



Electric current flowing through wire = 5.0 A



Magnetic field, B = 0.50 T



Direction of magnetic field is perpendicular to the plane of the.


Hence angle between length of the wire and magnetic field, θ = 900


As we know the magnetic force is given by








Question 18.

A wire, carrying a current i, is kept in the x-y plane along the curve A magnetic field B exists in the z-direction. Find the magnitude of the magnetic force on the portion of the wire between x = 0 and x = λ.


Answer:

Given-



Current passing through the wire = I A



The wire is kept in the xy plane along the curve,



Also given that the magnetic field exists in the z direction.


To find the magnetic force on the portion of the wire between x = 0 and x = λ.


We know ,magnetic force acting on a current carrying wire in an uniform magnetic field is given by



where,


B= magnetic field


I = current


l = length of the wire


and θ = the angle between B and l


For a differential length dl,



The effective force on the whole wire due to force acting on the wire of length λ placed along the x axis.



So,





Question 19.

A rigid wire consists of a semicircular portion of radius R and two straight sections figure. The wire is partially immersed in a perpendicular magnetic field B as shown in the figure. Find the magnetic force on the wire if it carries a current i.




Answer:

Given-



Radius of the semi-circular portion = R



Perpendicular Magnetic field = B



Electric current flowing through the wire = I A



Given in the question, the wire is partially immersed in a perpendicular magnetic field.




As AB and CD are straight wires of length l each and strength of the magnetic field is also same on both the wires, the force acting on these wires will be equal in magnitude


Direction of force can be found out using Fleming’s left hand rule.


So, their directions will be opposite to each other.



So, the magnetic force on the wire AB and the force on the wire CD are equal and opposite to each other. Both the forces cancel out each other.



Therefore, only the semicircular loop RC will experience a net magnetic force.



Here, angle between the length of the wire and magnetic field, θ = 900



We know ,magnetic force acting on a current carrying wire in an uniform magnetic field is given by



where,


B= magnetic field


I = current


l = length of the wire


and θ = the angle between B and l


Here, length l = 2R






Question 20.

A straight horizontal wire of mass 10 mg and length 1.0 m carries a current of 2.0 A. What minimum magnetic field B should be applied in the region so that the magnetic force on the wire may balance its weight?


Answer:

Given-



Mass of the wire, M = 10 mg = 10−5 Kg



Length of the wire, l = 1.0 m



Electric current flowing through wire, I = 2.0 A



It is said in the question, the weight of the wire should be balanced by the magnetic force acting on the wire.


Also angle between the length of the wire and magnetic field is


90°.


We k now weight of an object is given by



where,


m = mass


g = acceleration due to gravity =9.8 m/s2


We know,magnetic force acting on a current carrying wire in an uniform magnetic field is given by



where,


B= magnetic field


I = current


l = length of the wire


and θ = the angle between B and l


Thus,







Question 21.

Figure shows a rod PQ of length 20.0 cm and mass 200 g suspended through a fixed point O by two threads of lengths 20.0 cm each. A magnetic field of strength 0.500 T exists in the vicinity of the wire PQ as shown in the figure. The wires connecting PQ with the battery are loose and exert no force on PQ.

(a) Find the tension in the threads when the switch S is open.

(b) A current of 2.0 A is established when the switch S is closed. Find the tension in the threads now.




Answer:


Given-



Length of the rod PQ = 20.0 cm



Mass of the rod, M = 200 g



Length of the two threads, l = 20.0 cm



Magnetic field applied, B = 0.500 T


(a) When the switch is open-



The weight of the rod is balanced by the tension in the rod.


So,








(b)When switch is closed and current flowing through the circuit = 2 A



Then, there exists a magnetic force due to presence of current given by –




where,


B= magnetic field


I = current


l = length of the wire


and θ = the angle between B and l


Hence,


substituting the values












Question 22.

Two metal strips, each of length ℓ, are clamped parallel to each other on a horizontal floor with a separation b between them, A wire of mass m lies on them perpendicularly as shown in figure. A vertically upward magnetic field of strength B exists in the space. The metal strips are smooth but the coefficient of friction between the wire and the floor is μ. A current i is established when the switch S is closed at the instant t = 0. Discuss the motion of the wire after the switch is closed. How far away from the strips will the wire reach?




Answer:


Given-



Length of the two metallic strips = l



Distance between the strips = b



Mass of the wire = m



Strength magnetic field = B



Coefficient of friction between the wire and the floor = μ



Let the wire moved by a distance x.



The magnetic field present,will act on the wire towards the right.



As coefficient of friction is zero as the space between the wire and strip is smooth .


Due to the influence of magnetic force, the wire firstly will travel a distance equal to the length of the strips.


After this, it travels a distance x and then ,a frictional force will act opposite to its direction of motion on the wire.



So work done by the magnetic force and the frictional


force will be equal.



where


μ is the coefficient of friction for the two surfaces


W is the weight of the object


= mass × acceleration due to gravity


= mg


Magnetic force due to presence of current given by –




where,


B= magnetic field


I = current


l = length of the wire


and θ = the angle between B and l



Thus,







Question 23.

A metal wire PQ of mass 10 g lies at rest on two horizontal metal rails separated by 4.90 cm figure. A vertically downward magnetic field of magnitude 0.800 T exists in the space. The resistance of the circuit is slowly decreased and it is found that when the resistance goes below 20.0 Ω, the wire PQ starts sliding on the rails. Find the coefficient of friction.




Answer:


Given-



Mass of the metal wire, M = 10 g



Distance between the two metallic strips, l = 4.90 cm



Magnetic field acting vertically-downward, B = 0.800 T



Given in the question that when the resistance of the circuit is slowly decreased below 20.0Ω the wire end PQ starts sliding on the metallic rails.


At that time, current I from ohm’s law becomes




where,


v= applied voltage and


R = resistance of the circuit




From Fleming’s left-hand rule, we can confer that the magnetic force will act towards the right.


This magnetic force will make the wire glide on the rails.


The frictional force present at the surface of the metallic rails will try to oppose this motion of the wire.


When the wire starts sliding on the rails, the frictional force acting on the wire present between the wire and the metallic rail will just balance the magnetic force acting on the wire due to current flowing through it.


Thus,



,


where



μ is the coefficient of friction



R is the normal reaction force and



F is the magnetic force


frictional force will be equal.



where


μ is the coefficient of friction for the two surfaces


W is the weight of the object


= mass × acceleration due to gravity


= mg


Also,


Magnetic force due to presence of current given by –




where,


B= magnetic field


I = current


l = length of the wire


and θ = the angle between B and l


Hence,




substituting values







Question 24.

A straight wire of length ℓ can slide on two parallel plastic rails kept in a horizontal plane with a separation d. The coefficient of friction between the wire and the rails is μ. If the wire carries a current i, what minimum magnetic field should exist in the space in order to slide the wire on the rails.


Answer:

Given-
Length of the wire = l



Distance between the plastic rails = d



The coefficient of friction between the wire and the rails = μ



Electric current flowing through the wire = i


The magnetic force will make the wire glide on the rails.


The frictional force present at the surface of the metallic rails will try to oppose the motion of the wire.


The minimum magnetic field required in the space, in order to slide the wire on the rails, will be such that this magnetic force acting on the wire should be able to balance the frictional force on the wire.



Thus,



,


where



μ is the coefficient of friction



R is the normal reaction force and



F is the magnetic force


Frictional force will be equal.



where


μ is the coefficient of friction for the two surfaces


W is the weight of the object


= mass × acceleration due to gravity


= mg


Also,


Magnetic force due to presence of current given by –




where,


B= magnetic field


I = current


l = length of the wire


and θ = the angle between B and l


Hence,






Question 25.

Figure shows a circular wire-loop of radius a, carrying a current i, placed in a perpendicular magnetic field B.

(a) Consider a small part dl of the wire. Find the force on this part of the wire exerted by the magnetic field.

(b) Find the force of compression in the wire.




Answer:

Given-



Radius of the circular wire = a



Electric current passing through the loop = i



Magnetic field Perpendicular to the plane = B



(a) Magnetic force due to presence of current on a small differential length dl given by –




where,


B= magnetic field


I = current


dl =differential length of the wire


and θ = the angle between B and dl



The direction of magnetic force,using Fleming’s left-hand rule is towards the centre for any differential length dl of the wire.


Also, dl and B are perpendicular to each other


(b) Suppose a part of loop subtends a small angle 2θ at the centre of a circular loop as shown in fig.



Then, looking into the fig. we can say



We know length l of an arc –



where ,


r is radius of the circle and the angle subtended by the arc at the center


here, the arc is subtending an angle 2θ





Since is small, sinθ will become negligible








Question 26.

Suppose that the radius of cross section of the wire used in the previous problem is r. Find the increase in the radius of the loop if the magnetic field is switched off. The Young modulus of the material of the wire is Y.


Answer:

Given-


Radius of cross section of the wire used in the previous


problem = r



Young’s modulus of the metallic wire = Y.



It said in the question that, when the applied magnetic field is switched off, the tension in the wire increased and so iyd length is increased.



We know Young’s modulus,



we know, stress is, s is –



and strain



Young’s modulus,




Area of circle



Let, a be the radius of loop



from (1)



Now the tensile force will be the magnetic force acting on a conductor of length l given by –



where,


B= magnetic field


I = current


a =differential length of the wire


and θ = the angle between B and a






Question 27.

The magnetic field existing in a region is given by



A square loop edge ℓ and carrying a current i, is placed with its edges parallel to the x-y axes. Find the magnitude of the net magnetic force experienced by the loop.


Answer:

Given-


A square loop with-


Magnetic field,


Length of the edge of a square loop = l



Electric current flowing through it = i



Given in the question that the loop is lies with its edges parallel to the XY axes.



In the fig, arrow shows the direction of force on different sides of the square.



Magnetic force acting due to presence of current on a small length l given by –




where,


= magnetic field


I = current


= length of the wire


Now,


Force on side AB -



Consider a differential element of length dx at a distance x from the origin along line AB



Force on this small element,



Force on the full length of AB,





Force on AB will be acting downwards.



Similarly, force on CD,



The net force acting vertically will be -





Force on AD,




Force on BC



Then, the net horizontal force





Question 28.

A conducting wire of length ℓ, lying normal to a magnetic field B, moves with velocity v as shown in figure.

(a) Find the average magnetic force on a free electron of the wire.

(b) Due to this magnetic force, electrons concentrate at one end resulting in the electric field inside the wire. The redistribution stops when the electric force on the free electrons balances the magnetic force. Find the electric field developed inside the wire when the redistribution stops.

(c) What potential difference is developed between the ends of the wire?




Answer:

Given-



Length of the conducting wire = l



Inward magnetic field = B



Velocity of the conducting wire = v



As the wire is moving with velocity v, we can take this as the motion of free electrons present inside the wire with velocity v.



(a) The average magnetic force on a free electron of the wire



We know, Lorentz force F is given by -



where,


e = charge on an electron


v = velocity of the electron


B=magnetic field



(b)The redistribution of electrons stops when the electric force is just balanced by the magnetic force.



Electric force coulomb’s law,



where


e = charge


E=electric field of the charge


and also magnetic force, we know, Lorentz force F is given by -



where,


e = charge on an electron


v = velocity of the electron


B=magnetic field



On equating these two forces, we get-






(c) The potential difference developed between the ends of the wire , V is -




where l is length of wire and E is applied electric field


From (1)




Question 29.

A current i is passed through a silver strip of width d and area of cross section A. The number of free electrons per unit volume is π.

(a) Find the drift velocity v of the electrons.

(b) If a magnetic field B exists in the region as shown in figure, what is the average magnetic force on the free electrons?

(c) Due to the magnetic force, the free electrons get accumulated on one side of the conductor along its length. This produces a transverse electric field in the conductor which opposes the magnetic force on the electrons. Find the magnitude of the electric field which will stop further accumulation of electrons.

(d) What will be the potential difference developed across the width of the conductor due to the electron-accumulation? The appearance of a transvers emf, when a current carrying wire is placed in a magnetic field, is called Hall effect.




Answer:

Given-
Width of the silver strip = d


Area of cross-section = A


Electric current flowing through the strip = i


The number of free electrons per unit volume = n



(a) We know the relation between the drift velocity of electrons and current through any wire,




where


e is the charge of an electron


vd is the drift velocity.


A is the area of the conductor


On solving for the drift velocity, we get



(b)The magnetic field existing in the region is B


.
Magnetic force acting due to presence of current on a small length l given by –




where,


= magnetic field


I = current


= length of the wire


So, the force on a free electron





which acts towards upward direction


(c) Let us consider, electric field as E.


Now, the accumulation of electrons will stop when magnetic force just balances the electric force.


Electric force, coulomb’s law,



where


e = charge


E=electric field of the charge


From (1) and (2)




(d)The potential difference , V developed across the width d of


the conductor due to the electron-accumulation is given by-



where


d is length of wire


and E is applied electric field




Question 30.

A particle having a charge of 2.0 × 10–8 C and a mass of 2.0 ×10–10 g is projected with a speed of 2.0 × 103 ms–1 in a region having a uniform magnetic field of 0.10 T. The velocity is perpendicular to the field. Find the radius of the circle formed by the particle and also the time period.


Answer:

Given-



Charge on the particle, q = 2.0 × 10−8 C



Mass of the particle, m = 2.0 × 10−10 g



velocity of the particle when projected, v = 2.0 × 103 m s−1



Magnetic field, B = 0.10 T.



Given in the question that, the velocity is perpendicular to the field.



So, for the particle to move in a circle, the centrifugal force comes into acts which is provided by the magnetic force acting on it.


Also magnetic force, we know, Lorentz force F is given by -



where,


e = charge on an electron


v = velocity of the electron


B=magnetic field



Using the formula for centrifugal force



where,


v= velocity of the particle


r= radius of circle form


m =mass of the electron


Equating the two forces, we will get-






Now,


Time period,






Question 31.

A proton describes a circle of radius 1 cm is a magnetic field of strength 0.10 T. What would be the radius of the circle described by an α-particle moving with the same speed in the same magnetic field?


Answer:

Given-
Radius of the circle, r = 1 cm



Magnetic field , B= 0.10 T



We know that the charge on a proton is e and that of an alpha


particle is 2e.


Also, the mass of a proton is m


Mass of an alpha particle is 4m.


let assume that both the particles are moving with speed v.


So, for the particle to move in a circle, the centrifugal force comes into acts which is provided by the magnetic force acting on it.


Also magnetic force, we know, Lorentz force F is given by -



where,


e = charge on an electron


v = velocity of the electron


B=magnetic field



Using the formula for centrifugal force



where,


v= velocity of the particle


r= radius of circle form


Equating the two forces, we will get-





Then, we can confer from question that-



Where


rp is the radius of the circle described by the proton which is 0.01



For alpha particle, radius is given by –




On dividing equation (1) by (2), we get:






Question 32.

An electron having a kinetic energy of 100 eV circulates in a path of radius 10 cm in a magnetic field. Find the magnetic field and the number of revolutions per second made by the electron.


Answer:

Given-
Kinetic energy of an electron = 100 eV



Radius of the circle = 10 cm


We now kinetic energy is given by



Where


From question , we can confer that




Here,
m is the mass of an electron


v is the velocity of an electron.


We know that 1 eV = 1.6 × 10−19 J


Thus,






Now, radius r -








Therefore, magnetic field applied is 3.4 × 10−4 T



Number of revolutions per second of the electron is nothing but the frequency given by ,




From (1)












Question 33.

Protons having kinetic energy K emerge from an accelerator as a narrow beam. The beam is bent by a perpendicular magnetic field so that it just misses a plane target kept at a distance ℓ in front of the accelerator. Find the magnetic field.


Answer:

Given-
Kinetic energy of proton = K



Distance between the target and the accelerator = l



Therefore, radius of the circular orbit ≤ l



From question it is given that, the beam is bent by a


perpendicular magnetic field, let it be B.



We know-


Now, radius r -




where,


m is the mass of a proton


v= velocity of the particle


B = magnetic force


e= charge on the particle


For a proton, the above equation can be written as-



where r = l


Kinetic energy, K



Substituting the values of v in the equation (1)






Question 34.

A charged particle is accelerated through a potential difference of 12 kV and acquires a speed of 1.0 × 108ms–1. It is then injected perpendicularly into a magnetic filed of strength 0.2 T. Find the radius of the circle described by it.


Answer:

Given-


Applied potential difference to the charged particle , V = 12 kV = 12 × 103 V


Speed acquire by the charged particle, v =1.0 × 106 m s−1


Perpendicular Magnetic field B = 0.2 T


We know


The kinetic energy acquired by the particle is produced due to


applied potential difference, hence






and



where,


m is the mass of a proton


v= velocity of the particle


B = magnetic force


q= charge on the particle






Question 35.

Doubly ionized helium ions are projected with a speed of 10 km s–1 in a direction perpendicular to a uniform magnetic field of magnitude 1.0 T. Find

(a) the force acting on an ion,

(b) the radius of the circle in which it circulates and

(c) the time taken by an ion to complete the circle.


Answer:

Given-


Speed of the helium ions, v = 10 km s−1 = 104 m/s


Uniform magnetic field, B = 1.0 T


Charge on the helium ions = 2e


Mass of a helium ion, m = 4×1.6×10-27 Kg


(a) The force acting on an ion –


Magnetic force, we know, Lorentz force F is given by -



where,


e = charge on an electron


v = velocity of the electron


B=magnetic field


θ= angle between B and v








(b) The radius of the circle is given by,



where,


m is the mass of a proton


v= velocity of the particle


B = magnetic force


q= charge on the particle





(c) The time taken by an ion to complete the circle,


we know





Since the total time taken for a complete cycle will be its circumference and the velocity is V -






Question 36.

A proton is projected with a velocity of 3×106 m–1perpendicular to a uniform magnetic field of 0.6 T. Find the acceleration of the proton.


Answer:

Given-



Velocity of the proton, v = 3 × 106 m s−1



Uniform magnetic field, B = 0.6 T



As per the question, the proton is projected perpendicular to a


uniform magnetic field.



We know, Newton’s second law, Force F is given by –



where


m is mass of the object


a=acceleration


For proton




Magnetic force, we know, Lorentz force F is given by -



where,


e = charge on an electron


v = velocity of the electron


B=magnetic field


θ= angle between B and v


Equating (1) and (2),




As θ = 900






Question 37.

(a) An electron moves along a circle of radius 1 m in a perpendicular magnetic field of strength 0.50 T. What would be its sped? Is it reasonable?

(b) If a proton moves along a circle of the same radius in the same magnetic field, what would be its speed?


Answer:

Given-



(a) For electron


Radius of the circle = 1 m



Magnetic field strength = 0.50 T


Now,



The radius of the circle is given by,



where,


m is the mass of a electron


v= velocity of the particle


B = magnetic force


q= charge on the particle = C






Speed of light is 3x108 m/s2


Here, the speed of the electron moving along the circle is greater than the speed of light, so it is not reasonable.


(b) For a proton,


The radius of the circle is given by,



where,


m is the mass of a proton


v= velocity of the particle


B = magnetic force


q= charge on the particle = C







Question 38.

A particle of mass m and positive charge q, moving with a uniform velocity v, enters a magnetic field B as shown in figure.

(a) Find the radius of the circular arc it describes in the magnetic field.

(b) Find the angle subtended by the arc at the centre.

(c) How long does the particle stay inside the magnetic field?

(d) Solve the three parts on the above problem if the charge q on the particle is negative.




Answer:

Given-
Mass of the particle = m
Positive charge on the particle = q
velocity of the particle = v
Magnetic field = B


(a)The radius of the circular arc described by the particle in the magnetic field-


We know,


The radius of the circle is given by,



where,


m is the mass of a proton


v= velocity of the particle


B = magnetic force


q= charge on the particle = C


(b) The angle subtended by the arc at the centre



Line MAB is the tangent to arc ABC,


When the particle enters into the magnetic field, it follows a path in the form of arc as shown in fig.


Now, the angle described by the charged particle is nothing but the angle ∠ MAO is ,



∠MAO = 90°



Also from fig , ∠NAC = 90°



∠OAC = ∠ OCA = θ
Then, by angle-sum property of a triangle, sum of all angles of a triangle is 1800


∠AOC =180°− (θ + θ)


=


(c)The time for which the particle stay inside the magnetic field is given by-


Distance covered by the particle inside the magnetic field,is the length of arc subtended by angle θ and the radius



l = rθ


from (1)



l= r


time taken for a complete cycle will be its circumference and


the velocity is V -



Also The radius of the circle is given by,



where,


m is the mass of a proton


v= velocity of the particle


B = magnetic force


q= charge on the particle = C



(d) If the charge q on the particle is negative, then



(i) Radius of circular arc is given by



(ii) The centre of the arc lies within the magnetic field


Therefore, the angle subtended by the arc =



(iii) The time taken by the particle to cover the path inside the magnetic field




Question 39.

A particle of mass m and charge q is projected into a region having a perpendicular magnetic field B. Find the angle of deviation figure of the particle as it comes out of the magnetic field if the width d of the region is very slightly smaller than

(a) (b).

(c)




Answer:

Given-



Mass of the particle = m



Charge of the particle = q



Perpendicular magnetic field = B


To find the angle of deviation figure of the particle as it comes out of the magnetic field when -



(a) When the width,




d is equal to the radius


θ is the angle between the radius and tangent drawn to the circle , which is equal to.



(b) When the width,



Now, in this case, the width of the region in where magnetic field is applied is half of the radius of the circular path.



As the magnetic force is acting only along the y –axis,the velocity of the particle will remain constant along the x-axis.


So, if d distance is travelled along the x axis,


then,






where,


v=velocity


t=time



for constant velocity the acceleration along the x direction is zero.


Hence the force will act only along the y direction.



Using the 3rd equation of motion along the y axis-



where


u = initial velocity


a = acceleration


t= time taken


since, initial velocity is 0


Also, as θ = 900





From (1)



We know



From above fig.








(c) When the width, d =




From above fig., it can be concluded that the angle between the initial direction and final direction of velocity is .




Question 40.

A narrow beam of singly charged carbon ions, moving at a constant velocity of 6.0 × 104ms–1, is sent perpendicularly in a rectangular region having uniform magnetic field B = 0.5 T figure. It is found that two beams emerge from the field in the backward direction, the separations from the incident beam being 3.0 cm and 3.5 cm. Identify the isotopes present in the ion beam. Take the mass of an ion = A(1.6 × 10–27) kg, where A is the mass number.




Answer:

Given-



Velocity of a narrow beam of singly-charged carbon ions,


v = 6.0 × 104 m s−1



Strength of magnetic field B = 0.5 T



Separation between the two beams from the incident beam are 3.0 cm and 3.5 cm.



Mass of an ion = A(1.6 × 10−27) kg



The radius of the curved path taken by the beam 1 -




The radius of the curved path taken by the beam 2-



The radius of the circle is given by,



where,


m is the mass of a proton


v= velocity of the particle


B = magnetic force


q= charge on the particle = C


For the first beam




where


m1 = mass of the first isotope


q = s the charge.


For the second beam




where


m2 = mass of the second isotope


q = the charge.



from (1) and (2),




Now, we know,








Also, from (3)





Looking at the mass of the material obtained, we can infer that these are the two isotopes of carbon used are 12C6 and 14C6.



Question 41.

Fe+ions are acceleration through a potential difference of 500 V and are injected normally into a homogeneous magnetic field B of strength 20.0 mT. Find the radius of the circular paths followed by the isotopes with mass numbers 57 and 58. Take the mass of an ion = A(1.6 × 10–27) kg where A is the mass number.


Answer:

Given-
Potential difference through which the Fe+, V = 500 V



Strength of the homogeneous magnetic field


B = 20.0 mT= 20 × 10−3 T



Mass numbers of the two isotopes ,


m1 =57 and m2 =58.



Mass of an ion = A (1.6 × 10−27) kg



The radius of the circular path described by a particle in a magnetic field,



where,


m is the mass of a proton


v= velocity of the particle


B = magnetic force


q= charge on the particle = C


For calculating the radius of isotope 1-



For isotope 2,




Since isotopes are accelerated from the same potential V, the Kinetic energy gained by the two particles will be same for both the particles.


We, know the force developed by potential difference –



where


v = applied potential


q=charge





Also,we have



From (1)







For calculating the radius of 2 isotope-


From (2)






Question 42.

A narrow beam of singly charged potassium ions of kinetic energy 32 keV is injected into a region of width 1.00 cm having a magnetic field of strength 0.500 T as shown in figure. The ions are collected at a screen 95.5 cm away form the field region. If the beam contains isotopes of atomic weights 39 and 41, find the separation between the points where these isotopes strike the screen. Take the mass of a potassium ion = A(1.6 × 10–27) kg where A is the mass number.




Answer:

Given -


Kinetic energy of singly-charged potassium ions = 32 keV



Width of the magnetic region = 1.00 cm



Magnetic field’s strength, B = 0.500 T



Distance between the screen and the region = 95.5 cm



Atomic weights of the two isotopes are m1 =39 and m2 =41


.Mass of a potassium ion = A (1.6 × 10−27) kg


For a singly-charged potassium ion K-39, separation between the points can be calculated as follows –


Mass of K-39 = 39 × 1.6 × 10−27 kg


Charge, q = 1.6 × 10−19 C


Given in the question that the narrow beam of singly-charged potassium ions is injected into a region of magnetic field.


As, given that kinetic energy K.E is



K.E = 32 keV






We know that throughout the motion, the horizontal velocity remains constant.


So, the time taken to cross the magnetic field,





Now, the acceleration in the magnetic field region-








Now, velocity in the vertical direction can be found from Newton’s 2nd law as



Since, initial velocity is 0,



Substituting the values-






Time taken by the ion to reach the screen-





Now, distance moved by the ion vertically in this given time -









Vertical distance travelled by the particle inside magnetic field


can be calculated by using 3rd equations of motion –



where,


v= final velocity


u=initial velocity


a= acceleration acting on the ion


S=distance travelled


Since initial velocity is zero,







Now, time taken t,






Now,for the potassium ion K-41






And, acceleration a is given by-





t = time taken by the ion to exit the magnetic field is given by -




From newton’s law velocity in vertical direction-








Time to reach the screen, t -



.


Now distance moved in vertical direction is-






Now,
Vertical distance travelled by the particle inside magnetic field can be found out by using 3rd equation of motion given by



where,


v= final velocity


u=initial velocity


a= acceleration acting on the ion


S=distance travelled


Since initial velocity is zero,







Net distance travelled by K-41 potassium ion






Net gap between K-39 and K-41 potassium ions is given by-



From (1) and (2)






Question 43.

Figure shows a convex lens of focal length 12 cm lying in a uniform magnetic field B of magnitude 1.2 T parallel to tis principal axis. A particle having a charge 2.0 × 10–3 C and mass 2.0 × 10–6 kg is projected perpendicular to the plane of the diagram with a speed of 4.8 ms–1. The particle moves along a circle with its centre on the principal axis at a distance of 18 cm from the lens. Show that the image of the particle goes along a circle and find the radius of that circle.




Answer:

Given-
Focal length of the convex lens = 12 cm



Uniform magnetic field, B = 1.2 T



Charge of the particle, q = 2.0 × 10−3 C



an mass, m = 2.0 × 10−5 kg



Speed of the particle, v = 4.8 m s−1



Distance between the particle and the lens = 18 cm


Given in the question that the object is projected perpendicularly on the plane of the paper.


The radius of the circular path described by a particle in a magnetic field r,



where,


m is the mass of a proton


v= velocity of the particle


B = magnetic force


q= charge on the particle = C





Given that, the object distance, u = -18 cm



Using the lens formula –



where,


v=distance of image formed from lens


u=distance of the object from lens


f =focal length of the lens


substituting the values-




Let the radius of the circular path of image be r’.



Hence magnification -





Therefore, the radius of the circular path in which the image of the object formed from the lens moves is 8 cm.



Question 44.

Electrons emitted with negligible speed from an electron gun are acceleration through a potential difference V



Along the x-axis. These electrons emerge form a narrow hole into a uniform magnetic field B directed along this axis. However, some of the electrons emerging from the hole make slightly divergent angles as shown in figure. Show that these paraxial electrons are refocused on the x-axis at a distance.




Answer:

Given-
Electrons are accelerated by applying a potential difference = V



Let the mass of an electron = m


charge of an electron= e



Electric field,



Force experienced by the electron by coulomb’s law is given by,




Acceleration a, of the electron is given by,



where,


e= electronic charge


V= applied potential difference


r= radius of the curve


m= mass of the object


Using the 3rd equation of motion




where,


v= final velocity


u=initial velocity


a= acceleration acting on the ion


S=distance travelled


Since initial velocity is zero,





Here, s = r which is the radius of the curve


From (1)




We know that time taken by electron to cover the curved path is given as,



As the acceleration of the electron is along the y axis only, it travels along the xaxis with uniform velocity.


Velocity of the electron moving along the field remains v.



Therefore, the distance at which the beam is


d = velocity × Time





Question 45.

Two particles, each having a mass m are placed at a separation d in a uniform magnetic filed B as shown in figure. They have opposite charges of equal magnitude q. At time t = 0, the particles are projected towards each other, each with a speed v. Suppose the Coulomb force between the charges is switched off.

(a) Find the maximum value vm of the projection speed so that the two particles do not collide.

(b) What would be the minimum and maximum separation between the particles if v = vm/2 ?

(c) At what instant will a collision occur between the particles if v = 2vm?

(d) Suppose v = 2vm and the collision between the particles is completely inelastic. Describe the motion after the collision.




Answer:

Given-
Mass of two particles = m
Distance between the particles = d
Also magnitude of charges of both the particles are equal but opposite polarity equal = q.
From the question we can confer that, both the particles are projected towards each other with equal speed v.



Assuming that Coulomb force between the charges is switched off.


(a) The maximum value vm of the projection speed so that the two particles do not collide-


The particles will not collide with each other if



d = r1 + r2


where, r1 = r2 = radius of circular orbit followed by the charged particles


The radius of the circular path described by a particle in a magnetic field r,



where,


m is the mass of a proton


v= velocity of the particle


B = magnetic force


q= charge on the particle = C




So,



(b) The minimum and maximum separation between the particles if v = vm/2-


Let the radius of the curved path taken by the particles in the condition when vm/2, be r



So, minimum separation between the particles will be






Now, the maximum distance of separation between the particles will be = (d + 2r)




(c) The instant at which the collision occurs between the particles if v=2vm



The particles along the horizontal direction will collide at a distance , d/2


Let they collide after time t.


Velocity of the particles before and after collision along the horizontal direction will remain the same.



Therefore,



From (1)




(d) The motion of the two particles after collision when the collision is completely inelastic and v = 2vm


Let the P be t he point of particles collision.


And at point P, both the particles will have motion in upwards


As the collision is inelastic these particles will stick together.



Distance between centers = d


Velocity of the particles before and after collision along the horizontal direction will remain the same.



At point P, velocities along the horizontal direction are equal in magnitude and opposite in direction.


So, they will cancel out each other.


So, the velocity along the vertical upward direction will add up.


Magnetic force acting along the vertical direction,


Magnetic force, we know, Lorentz force F is given by -



where,


q = charge on an electron


v = velocity of the electron


B=magnetic field


θ= angle between B and v



Now, from Newton’s second law, the acceleration along the vertical direction,




So, from 1st equation of motions-



where


v= final velocity


u=initial velocity


a= acceleration due to gravity


t=time taken


since initital velocity u =0,


Velocity of the combined mass at point P is along the vertical


direction v’.





Hence, the particles will behave as a combined mass and move with same velocity vm.



Question 46.

A uniform magnetic field of magnitude 0.20 T exists in space from east to west. With what speed should a particle of mass 0.010 g and having a charge 1.0 × 10–5 C projected from south to north so that it moves with a uniform velocity?


Answer:

Given-


Magnetic field, B = 0.20 T


Mass of the particle, m = 0.010 g = 1 × 10−5 kg


Charge of the particle, q = 1.0 × 10−5 C


Given in the question that, if the particle has to move with


uniform velocity in the region of the applied field,


the gravitational force experienced by the particle must be


equal to the magnetic force experienced by the particle.


Gravitational force,



where


m is the massof theobject


g=acceleration due to gravity


And


Magnetic force, we know, Lorentz force F is given by -



where,


q = charge on an electron


v = velocity of the electron


B=magnetic field


θ= angle between B and v


So,







Question 47.

A particle moves in a circle of diameter 1.0 cm under the action of a magnetic field of 0.40 T. An electric field of 200 V m–1 makes the path straight. Find the charge/mass ratio of the particle.


Answer:

Given-
Diameter of the circle = 1.0 cm



Thus, radius of circle, r = = 0.5 × 10−2 m



Magnetic field, B = 0.40 T



Electric field, E = 200 V m−1


From the question we can confer that, the particle is moving in a circle under the action of a magnetic field.


But when an electric field is applied on the particle, it moves in a straight line.



So, we can say that the electric field is balanced by the magnetic field ie,



Magnetic force, we know, Lorentz force F is given by -



where,


q = charge on an electron


v = velocity of the electron


B=magnetic field


θ= angle between B and v






The radius of the circular path described by a particle in a magnetic field r,



where,


m is the mass of a proton


v= velocity of the particle


B = magnetic force


q= charge on the particle = C






Question 48.

A proton goes undeflected in a crossed electric and magnetic field (the fields are perpendicular to each other) at a speed of 2.0 × 106 ms–1. The velocity is perpendicular to both the fields. When the electric field is switched off, the proton moves along a circle of radius 4.0 cm. Find the magnitudes of the electric and the magnetic fields. Take the mass of the proton = 1.6 × 10–27 kg.


Answer:

Given-
Mass of the proton, m = 1.6 × 10−27 kg



Speed of the proton inside the crossed electric and magnetic field, v = 2.0 × 105 ms−1



Given from question we can confer that , the proton is not deflected under the combined action of the electric and magnetic fields.


Thus, the forces applied by both the fields are equal and opposite to each other


Magnetic force, we know, Lorentz force F is given by -



where,


q = charge on an electron


v = velocity of the electron


B=magnetic field


θ= angle between B and v


Also,


We, know the Columb’s force–



where


E = applied electric field


e=charge



Now,






But when the electric field is switched off, the proton moves follows circular path due to the presence of force of the magnetic field.



The radius of the circular path described by a particle in a magnetic field r,



where,


m is the mass of a proton


v= velocity of the particle


B = magnetic force


q= charge on the particle = C





Substituting the value of B in equation (1), we get





Hence magnitudes of the electric and the magnetic fields are



respectively



Question 49.

A particle having a charge of 5.0 μC and a mass of 5.0 × 10–12 kg is projected with a speed of 1.0 km s–1 in a magnetic field of magnitude 5.0 mT. The angle between the magnetic field and the velocity is sin–1(0.90). Show that the path of the particle will be a helix. Find the diameter of the helix and its pitch.


Answer:

Given-
Charge on the particle, q = 5 μC = 5 × 10−6 C



Magnetic field, B = 5 × 10−3 T



Mass of the particle, m = 5 × 10−12 kg



Velocity of projection, v = 1 Km/s = 103 m/s



Angle between the magnetic field and velocity,


θ= sin−1(0.9)


Since there are no forces in the horizontal direction ie there is no force in the direction of magnetic field, so, the particle moves with uniform velocity in horizontal direction.


Thus, it moves in a helical form.In helical motion we have two components of velocity.



Component of velocity which is perpendicular to the magnetic field is given by -



Similarly component of velocity in the direction of magnetic field will be the parallel component given by -



The velocity has a vertical component along which it accelerates with an acceleration a and moves in a circular cross-section.


We know motion in helical direction which is centripetal force and is given by –



Now, this force is balanced by Lorentz force acting due to presence of magnetic field ,








Hence, diameter of the helix can be calculated as ,




And the Pitch of the helix is ,







Question 50.

A proton projected in a magnetic field of 0.020 T travels along a helical path of radius 5.0 cm and pitch 20 cm. Find the components of the velocity of the proton along and perpendicular to the magnetic field. Take the mass of the proton = 1.6 × 10–27 kg.


Answer:

Given


Mass of the proton, mp = 1.6 × 10−27 kg



Magnetic field intensity, B = 0.02 T



Radius of the helical path, r = 5 cm = 5 × 10−2 m



Pitch of the helical path, p = 20cm = 2 × 10−1 m


In helical motion we have two components of velocity.



Component of velocity which is perpendicular to the magnetic field is given by -



Similarly component of velocity in the direction of magnetic field will be the parallel component given by -




Now,


Now, this force is balanced by Lorentz force acting due to presence of magnetic field ,








Now, pitch of the helix is calculated as –








Question 51.

A particle having mass m and charge q is release from the origin in a region in which electric field and magnetic field are given by



Find the speed of the particle as a function of its z-coordinate.


Answer:

Given-



Mass of the particle = m



Charge of the particle = q



Electric field and magnetic field are given by




Velocity,



Magnetic force, we know, Lorentz force F is given by -



where,


q = charge on an electron


v = velocity of the electron


B=magnetic field


θ= angle between B and v


Also, coulomb’s force experienced by the electron is given by,



where e= charge on the electron and


E= electric field applied


So, total force on the particle,




=q


= q



Now, since


,



So, acceleration is given by



From 3rd equation for motion



where


u = initial velocity


v= final velocity


s=distance travelled


and a = acceleration of the particle



So,



Here, z is the distance along the z-direction.



Question 52.

An electron is emitted with negligible speed from the negative plate of a parallel plate capacitor charged to a potential difference V. The separation between the plates is d and a magnetic field B exists in the space as shown in figure. Show that the electron will fail to strike the upper plate if.




Answer:

Given-
Potential difference applied across the plates of the capacitor = V
Separation between the plates = d
Magnetic field intensity = B


The electric field applied across the plates of a capacitor


,



Also, coulomb’s force experienced by the electron is given by,



where e= charge on the electron and


E= electric field applied


Hence, the force experienced by the electron due to this electric field,



Now, acceleration a is given by-




where


e = charge of the electron


me = mass of the electron



From 3rd equation for motion



where


u = initial velocity


v= final velocity


s=distance travelled


and a = acceleration of the particle


substituting the value of a-




The electron will move in a circular path due to the presence of the magnetic field.


The radius of the circular path described by a particle in a magnetic field r,



where,


m is the mass of a proton


v= velocity of the particle


B = magnetic force


q= charge on the particle = C


Radius of the circular path followed by the electron is ,



And the electron will fail to strike the upper plate of the


capacitor if and only if the radius of the circular path will be less


than d,


i.e. d>r




Thus, the electron will fail to strike the upper plate if




Question 53.

A rectangular coil of 100 turns has length 5 cm and width 4 cm. It is placed with its plane parallel to a uniform magnetic field and a current of 2A is sent through the coil. Find the magnitude of the magnetic field B, if the torque acting on the coil is 0.2 N m–1.


Answer:

Given-
No. of turns in the coil, n = 100



Area of the coil, A = 5 × 4 cm2 = 20 × 10−4 m2



Magnitude of current = 2 A



Torque acting on the coil, τ = 0.2 N m−1


We know that torque acting on a rectangular coil having n turns is given by



Where


B= applied magnetic field


A= area of rectangular loop


I = current flowing through coil


n = number of turns











Question 54.

A 50 turn circular coil of radius 2.0 cm carrying a current of 5.0 A is rotated in a magnetic field of strength 0.20 T.

(a) What is the maximum torque that acts on the coil?

(b) In a particular position of the coil, the torque acting on it is half of this maximum. What is the angle between the magnetic field and the plane of the coil?


Answer:

Given-
No. of turns of the coil, n = 50



Magnetic field intensity, B = 0.20 T = 2 × 10−1 T



Radius of the coil, r = 0.02 m = 2 × 10−2 m



Magnitude of current =5 A



We know that torque acting on a rectangular coil having n turns is given by



Where


B= applied magnetic field


A= area of rectangular loop


I = current flowing through coil


= angle between the area vector and magnetic field


Torque is maximum if


θ = 90°.





Given that ,





So, the angle between the magnetic field and the plane of coil is given by -




Question 55.

A rectangular loop of sides 20 cm and 10 cm carries a current of 5.0 A. A uniform magnetic field of magnitude 0.20 T exists parallel to the longer side of the loop.

(a) What is the force acting on the loop?

(b) What is the torque acting on the loop?


Answer:

Given-
No. of turns of the coil, n = 50



Magnetic field, B = 0.20 T = 2 × 10−1 T



Magnitude of current, I = 5 A



Length of the loop, l = 20 cm = 20 × 10−2 m,



Breadth of the loop, w = 10 cm = 10 × 10−2 m,



So, area of the loop, A = length ×width = 0.02 m2,


Let ABCD be the rectangular loop.



(a)There is no force acts along the the sides ab and cd, as they are parallel to the magnetic field.




But the force acting on the sides ad and bc are equal in magnitude but opposite


So, they cancel out each other


.
Hence, the net force on the loop is zero.


(b) Torque acting on the coil,




Where


B= applied magnetic field


A= area of rectangular loop


I = current flowing through coil


= angle between the area vector and magnetic field







So, the torque acting on the loop is 0.02 Nm and is parallel to the shorter side



Question 56.

A circular coil of radius 2.0 cm ahs 500 turns in it and carries a current of 1.0 A. Its axis makes an angle of 30° with the uniform magnetic field of magnitude 0.40 T that exist in the space. Find the torque acting on the coil.


Answer:

Given-


No. of turns of the coil, n = 500


Magnetic field intensity, B = 0.40 T = 4 × 10−1 T


Radius of the coil, r = 2 cm = 2 × 10−2 m


Magnitude of current, i = 1 A


Angle between the area vector and magnetic field, θ = 30°


Torque acting on the coil,




Where


B= applied magnetic field


A= area of rectangular loop


I = current flowing through coil


= angle between the area vector and magnetic field








Question 57.

A circular loop carrying a current i has wire of total length L. A uniform magnetic field B exists parallel to the plane of the loop.

(a) Find the torque on the loop.

(b) If the same length of the wire is used to form a square loop, what would be the torque? Which is larger?


Answer:

Given-
Magnetic field intensity = B


Magnitude of current = i


Circumference, ,


where


r is the radius of the coil



So, area of the coil,



Torque acting on the coil,




Where


B= applied magnetic field


A= area of rectangular loop


I = current flowing through coil


= angle between the area vector and magnetic field




(b) Let s be the length of the square loop.


Then given in the question that







So, the torque in case of the circular loop is larger.



Question 58.

A square coil of edge ℓ having n turns carries a current i. It is kept on a smooth horizontal plate. A uniform magnetic field B exists in a direction parallel to an edge. The total mass of the coil is M. What should be the minimum value of B for which the coil will start tipping over?


Answer:

Given-
Number of turns in the coil = n



Edge of the square loop = l



Magnetic field intensity = B



Magnitude of current = i



Angle between area vector and magnetic field, θ = 90°



Torque acting on the coil,




Where


B= applied magnetic field


A= area of rectangular loop


I = current flowing through coil


= angle between the area vector and magnetic field





Torque produced due to weight is


given by,



where m is the mass


g is the acceleration due to gravity


For the coil to start tipping towards downwards,





For minimum value of B,




From (1) and(2)





Question 59.

Consider a non-conducting ring of radius r and mass m which has a total charge q distributed uniformly on it. The ring is rotated about its axis with an angular speed ω.

(a) Find the equivalent electric current in the ring.

(b) Find the magnetic moment μ of the ring.

(c) Show that ℓ where ℓ is the angular momentum of the ring about its axis of rotation.


Answer:

Given-
Radius of the ring = r



Mass of the ring = m



Total charge enclosed on the ring = q



(a) Angular speed-


We know that angular speed is given by-



now frequency f





Current in the ring,





(b)For a ring of area A with current i flowing through it, magnetic moment,




where,


A= area of cross section


i=current flowing through it


n = number of turns


for number of turns n = 1



From (1)



× =



(d) Angular momentum l,



where


I is moment of inertia of the ring about its axis of rotation.


is angular velocity



Where


m is the mass


r is the radius of gyration.



So,



Putting this value in equation (2), we get-




Question 60.

Consider a non-conducting plate of radius r and mass m which has a charge q distributed uniformly over it. The plate is rotated about its axis with an angular speed ω. Show that the magnetic moment μ and the angular momentum ℓ of the plate are related as


Answer:

Given-
Radius of the ring = r



Mass of the ring = m



Total charge of the ring = q


Angular speed,


Where


F is the frequency


Now frequency f



Where,


T is the time period




Magnetic moment,


where,


A= area of cross section


i=current flowing through it


n = number of turns


for number of turns n = 1



Current in the ring,





For the ring of area A with current i, magnetic moment,m is given by



Where A is the area of the loop





Angular momentum,



where


I is the moment of inertia of the ring about its axis of rotation.


is the angular moment


Moment of inertia, I is given by-



where,


m is the mass of the object


r is the radius of the circular object


So, substituting in (2)




Putting this value in equation (1),




Question 61.

Consider a solid sphere of radius r and mass m which has a charge q distributed uniformly over its volume. The sphere is rotated about a diameter with an angular speed ω. Show that the magnetic moment μ and the angular momentum ℓ of the sphere are related as


Answer:

Given


radius of solid sphere = r


mass of sphere = m


charge on the sphere =q


angular speed of the sphere = ω


magnetic moment and the angular momentum ℓ of the sphere are related –



Lets consider a differential strip of width dx at a distance x from the centre of the sphere.



Area of the strip is given as,




now, current i is given by-



Angular speed,




Now frequency f





current i becomes



Magnetic moment is given by-


where,


A= area of cross section


i=current flowing through it


n = number of turns


for number of turns n = 1



Integrating over the sphere-