Electromagnetic Induction
Class 12th Physics Part I CBSE Solution
Exercises
Question 1.Predict the direction of induced current in the situations described by the following Figs. 6.18(a) to (f).
Answer:
The direction of induced current will be given by Lenz’s law.
According to the Lenz’s law: The direction of induced current by change in magnetic field (Faraday’s induction) will be such that it opposes the change that caused it.
As seen from the given figures, the direction of the induced current is shown when the north pole of the magnet is moved towards the closed loop or away from the closed loop.
Question 2.
Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.19:
(a) A wire of irregular shape turning into a circular shape;
(b) A circular loop being deformed into a narrow straight wire.
Answer:
According to the Lenz’s law: The direction of induced current by change in magnetic field (Faraday’s induction) will be such that it opposes the change that caused it.
(a) In the figure given below, the direction of magnetic field is perpendicular into the paper.
Now, when we change the shape of irregular loop into a circular loop, then the area of the loop will increase. Because, area of circle is considered to be greatest.
Since, area of loop increases, ∴ the magnetic flux will also increase.
Now, the current will be induced in the loop such that the magnetic flux decreases. And, magnetic field will decrease when the area will decrease i.e. the wires should be pulled in the inwards directions, to decrease the net magnetic field.
∴ the current will be along “adcba”.
(b) In the figure given below, the direction of magnetic field is perpendicular coming out of the paper.
Now, when circular loop is transformed into a narrow straight wire, then the area of the loop will decrease. Since, area of loop decreases, ∴ the magnetic flux will also decrease.
Now, the current will be induced in the loop such that the net magnetic flux increases in the wire. ∴ the current flow will be along “adcba” which produces the induced magnetic field outwards the paper.
Question 3.
A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?
Answer:
Given:
No. of turns in the solenoid coil, N = 15 turns/cm = 15000 turns/m
(∵ 1m = 100 cm)
∴ no. of turns in the solenoid coil per unit length, n = 15000 turns
Area of the solenoid coil = 2.0 cm2
In m2, Area will be 2 × 10 - 4 m2
Since current carried by the solenoid coil changes from 2.0 to 4.0 A in 0.1 seconds
Therefore, change in the current in the coil of solenoid = final current – initial current
di = 4.0 – 2.0 = 2.0 A
The change in time “dt” is given as ,dt = 0.1 seconds
Applying Faraday’s law, induced e.m.f can be calculated as follows:
…(1)
Where Ф is flux induced in the loop
And, flux is given as, Ф = BA
Where “B” is the magnetic field and “A” is the area of the loop.
For a solenoid, B= μ0nI
Therefore, the equation (1) becomes:
Or 
Or E = Aμ0 n (dI/dt)…(2)
∵ A, μo and n are constants
Substituting the values in equation (2), we get,
⇒ E = 7.54 × 10 - 6 V
Or E = 7.54 μV
(∵ 10-6 = μ)
Hence, the induced voltage in the loop is 7.54 μV
Question 4.
A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s–1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?
Answer:
Given:
Length of rectangular wire = 8 cm
Breadth of rectangular wire = 2 cm
The area of the rectangular wire can be calculated as follows:
A = l × b
A = 8 cm × 2 cm = 16 cm2
In m2, area will be A = 16 × 10 - 4 m2
Magnitude of magnetic field B = 0.3T
Velocity of the loop, v =1 cms - 1 = 0.01ms - 1
(a) The emf developed across the cut can be calculated as follows:
e = Blv …(1)
where B is the magnetic field
l is the length of the loop and,
v is the velocity of the rectangular loop
substituting the values in equation (1), we get,
e = (0.3 T × 0.08m × 0.01ms - 1)
e = 2.4 × 10 - 4 V
Time taken for travelling along the width of the loop can be calculated as follows:
T =distance travelled/velocity with which distance is travelled
T = b(width of the loop)/v (velocity)
T = 0.02 m/0.01 ms - 1
T= 2 seconds
This means that the induced voltage will last for very short direction of 2 seconds.
(b) The emf developed across the cut if the velocity of the loop is 1 cms–1 in a direction normal to the shortest side of the loop can be calculated as follows:
e = Bbv…(2)
where B is the magnetic field
b is the width of the loop and,
v is the velocity of the rectangular loop
substituting the values in equation (2), we get,
e = (0.3 T × 0.02 m × 0.01 ms - 1)
e = 0.6 × 10 - 4 V
Time taken for travelling along the width of the loop can be calculated as follows:
T =distance travelled/velocity with which distance is travelled
T = l(length of the loop)/v
T = 0.08 m/0.01ms - 1
T = 8 seconds
This means that the induced voltage will last for 8 seconds.
Question 5.
A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s–1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.
Answer:
Given:
Length of the metallic rod = 1.0 m
Angular frequency = 400 rads - 1
Strength of magnetic field B= 0.5 T
Since it is clear that one end of the metallic rod has zero linear velocity whereas the other end of the rod has linear velocity of lω
∴ The average linear velocity of the metallic rod can be calculated as follows:
v = (lω + 0)/2
⇒ v = lω /2
The e.m.f. developed between the centre and the ring can be calculated as follows:
e = Blv ...(1)
where B is the magnetic field
l is the length of the loop and,
v is the velocity of the rectangular loop
⇒ Substituting the value of v in above equation, we get
e = B × l × (lω/2)
e = (Bl2ω)/2
e = (0.5 T × 1m2 × 400 rads - 1)/2
On calculating, we get
e = 100 V
Hence, the emf developed across the ring is 100 V.
Question 6.
A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s–1 in a uniform horizontal magnetic field of magnitude 3.0 × 10–2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 W, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating.
Where does this power come from?
Answer:
Given:
Radius of the circular coil r = 8.0 cm = 0.08 m
No. of turns in the coil, N =20
Angular frequency ω = 50 rad s - 1
Magnitude of magnetic field B =3.0 × 10–2 T
Resistance of the closed loop = 10 Ω
The area of the coil can be calculated using the formula
A =πr2
A = 3.14 × (0.08 m)2
The maximum induced e.m.f is calculated as follows:
e = N ω A B
e = 20 × 50 rad s - 1 × 3.14 × (0.08 m)2 × 3.0 × 10–2 T
On calculating we get
e = 0.603 V
Since for a full cycle, the average induced emf will be zero.
The maximum current for the circular coil can be calculated as follows:
I = e/R
Substituting the values
I = 0.603 V/10 Ω
I = 0.0603A
The average power loss due to Joule’s heating effect in the circular coil will be given by:
P = (eI)/2
P = (0.603 V × 0.0603 A)/2
On calculating we get
P = 0.018 W
A torque was produced due to current induced in the coil. This torque opposes the rotation of the circular coil. Since a rotor is the external agent in the coil and it should apply the torque so that the coil keeps rotating in the uniform manner. Therefore, the dissipated power comes from the external rotor.
Question 7.
A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s–1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 × 10–4 Wb m–2.
(a) What is the instantaneous value of the emf induced in the wire?
(b) What is the direction of the emf?
(c) Which end of the wire is at the higher electrical potential?
Answer:
Given:
Length of the wire, l = 10 m
Speed of wire, v = 5.0 ms - 1
Magnetic field B = 0.30 × 10–4 Wb m–2
The diagram is:
(a) The e.m.f induced in the wire can be calculated as follows:
e = Blv ...(1)
where B is the magnetic field
l is the length of the loop and,
v is the velocity of the rectangular loop
substituting the values in above equation, we get,
e = 0.30 × 10–4 Wb-m - 2 × 10 m × 5ms-1
e = 1.5 × 10–3 V
(b) The direction of the induced emf will be from west to east in accordance with the Fleming’s right - hand rule.
According to Fleming’s right - hand rule: When the right hand is held with the thumb, first finger and second finger mutually perpendicular to each other, as shown below.
Then, the thumb points in the direction in which the conductor moves; the first finger gives the direction of the magnetic field (N ⇒ S); and the second finger gives the direction of the induced current.
(c) The east end of the straight wire will be having high electrical potential.
Question 8.
Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self - inductance of the circuit.
Answer:
Given: Initial current I1 = 5.0 A
Final current I2 = 0.0 A
Time = 0.1 seconds
Average e.m.f = 200 V
Change in the current can be calculated as:
dI = I1 - I2 = 5.0A – 0.0A
dI = 5 A
The self-conductance of the circuit is calculated using the formula:
or 
substituting the values, we get
∴ L = 4 Henry
Question 9.
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?
Answer:
Given: Mutual conductance, μ = 1.5 H
Initial current I1= 0
Final current I2 = 20 A
Time taken t = 0.5 seconds
Change in the current can be calculated as:
dI = I1 - I2 = 20 – 0.0
dI = 20 A
The mutual conductance of the circuit is calculated using the formula:
or 
As we know that 
Substituting it in equation, we get
Question 10.
A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 × 10–4 T and the dip angle is 30°.
Answer:
Given: Speed of the jet plane towards west = 1800 km/h
In m/s, speed of the jet plane towards west = 500 m/s
Length of the span of the jet plane = 25 m
Magnetic field = 5 × 10–4 T
Dip angle =30°
The vertical component of Earth’s magnetic field can be calculated as follows:
Bv= B sinδ
On substituting the values, we get,
Bv= 5 × 10–4 T × sin30°
⇒ Bv= 5 × 10–4 T × 1/2
∴ Bv = 2.5 × 10–4 Tesla
The voltage difference developed between the ends of the jet can be calculated using the formula:
e = Bv × l × v
On substituting the values, we get
e =2.5×10–4 T× 25m × 500m/sec
on calculating, we get
e = 3.125 V
∴ the voltage difference developed between the ends of the wings is approximately 3V.
Additional Exercises
Question 1.Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T s–1. If the cut is joined and the loop has a resistance of 1.6Ω , how much power is dissipated by the loop as heat? What is the source of this power?
Answer:
Given: Length of rectangular wire = 8 cm
Breadth of rectangular wire = 2 cm
The area of the rectangular wire can be calculated as follows:
A = l × b
A = 8cm × 2cm = 16 cm2= 16 × 10-4 m2
Initial Magnitude of magnetic field B = 0.3T
Velocity of the loop, v =1 cms-1
The rate of decrease of the magnetic field i.e. 
The e.m.f developed in the rectangular loop is given as:
dФ is the change in the flux across the loop = AB
⇒ 
Or 
Therefore, substituting the values in above equation, we get:
e = 16×10-4 m2× 0.02 Ts-1
⇒ e = 0.32 × 10-4V
The current induced in the loop can be calculated as follows:
i = e/r
i = (0.32 × 10 - 4)/1.6
⇒ i = 2 × 10 - 5A
The Power loss or dissipation can be calculated as follows:
P = i2R
Where, R is resistance = 1.6Ω
⇒ P = (2 × 10-5A)2 × 1.6Ω
⇒ P = 6.4×10-10 W
The source of the power is an external agent. This is due to change in the magnetic field with time.
Question 2.
A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s–1 in the positive x - direction in an environment containing a magnetic field in the positive z - direction. The field is neither uniform in space nor constant in time. It has a gradient of 10–3 T cm–1 along the negative x - direction (that is it increases by 10–3 T cm–1 as one moves in the negative x - direction), and it is decreasing in time at the rate of 10–3 T s–1. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mW.
Answer:
Given: side of square loop = 12 cm
In metre, the side of square loop = 0.12m
Area of the square loop = 0.12 × 0.12 = 144 × 10 - 4m2
Velocity of the loop = 8 cms-1
In metre/s, the velocity of the loop, v= 0.08m/s
Since gradient of electric field is along negative x - direction
In Tm - 1, 
∴ The rate of decrease of magnetic field in time is given as:
Resistance of the square loop = 4.50 mΩ =4.5 × 10 - 3Ω
Due to change in the motion of the square loop in presence of non - uniform magnetic field, the decrease in the magnetic flux is given as:
Due to explicit time variation in magnetic field, the rate of change of flux is given by:
Substituting values in above, we get
The total emf induced in the square loop is :
e =1.44 × 10–5 V+ 11.52× 10-5 V
e = 12.96 × 10-5 V
The induced current can be written as:
i =e/R
⇒ 
⇒ i = 2.88 × 10 - 2A
The current direction will be in the way that there will be increase in the flux along positive z direction.
Question 3.
It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area 2 cm2 with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region.
Equivalently, one can give it a quick 90° turn to bring its plane parallel to the field direction). The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50
. Estimate the field strength of magnet.
Answer:
Given: Area of small flat search coil = 2 cm2
Total number of turns in the coil n = 25
Total charge measured by the ballistic galvanometer = 7.5 mC=7.5 ×10-3C
combined resistance of the coil and the galvanometer R= 0.50
The current induced in the coil can be calculated as follows:
i = e/R
also,
substituting in the equation we get
⇒ 
…………..(i)
Initial flux through the coil, Ф1 = AB
Final flux through the coil, Ф2= 0
Integrating equation (i) on both sides, we get
Total charge 
Therefore, 
⇒ Q = (NBA)/R
Therefore, B = (QR)/NA
So, substituting all the values in above equation, We get
On solving, we get
B = 0.75 T
Question 4.
Figure 6.20 shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 9.0 mW. Assume the field to be uniform.
(a) Suppose K is open and the rod is moved with a speed of 12 cm s–1 in the direction shown. Give the polarity and magnitude of the induced emf.
(b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed?
(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain.
(d) What is the retarding force on the rod when K is closed?
(e) How much power is required (by an external agent) to keep the rod moving at the same speed (=12cm s–1) when K is closed? How much power is required when K is open?
(f) How much power is dissipated as heat in the closed circuit? What is the source of this power?
(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?
Answer:
Given: Length of the rod = 15 cm
Length of rod in metres l = 0.15 m
Magnetic field strength B = 0.50 T
Resistance of the loop R = 9 mΩ = 9 × 10 - 3 Ω
Emf induced in the loop = 9 mV
Speed of the rod, v= 12 cm/s
Speed of the rod in m/s, v= 0.12m/s
It is shown as below:
(a) Induced emf can be calculated as follows:
e = Bvl
e = 0.50T× 0.12ms-1× 0.15m
e = 9× 10-3V = 9mV
The polarity of the induced emf is such that the end P is positive and end Q is negative.
(b) Yes, there is an excess charge built up at the ends of the rods when K is open. When K is closed, the excess charge present will be maintained by the flow of current continuously.
(c) The net force on the rod PQ is zero still they experience magnetic force due to motion of the rod magnetic force is cancelled by the electric force set up. This happens because of the excess charge on both the opposite ends of the rod.
(d) The retarding force acting on the rod when K is closed can be calculated as follows:
F = iBl……………(i)
Where i= induced emf(e)/Resistance (RΩ)
I = 9mV/9×10-3Ω
I= 1A
Substituting the values in equation (i), we get
F = 1A× 0.50T× 0.15m
F = 75×10-3 N
(e) Power required when K is closed:
Power = Fv
P = 75×10-3 N× 0.12ms-1
P = 9×10 - 3 W
Power = 9 mW
When key K is open, there is no power required.
(f) Power dissipated H = I2R
H = 1A 2× 9×10-3Ω = 9 mW
9mW power is dissipated as heat in the closed circuit and the source of this power is external agent.
(g) The induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular is zero because the motion of the rod doesn’t cross across the lines and therefore induced emf will be zero.
Question 5.
An air - cored solenoid with length 30 cm, area of cross - section 25 cm2 and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10–3 s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.
Answer:
Given: Length of the solenoid = 30 cm=0.30 m
Area of cross - section = 25 cm2 = 25m× 10-4m-2
No. of turns in the solenoid = 500
Current, i = 2.5A
Time for which the current flows t = 10–3 s
The average back emf induced across the ends of the open switch in the circuit is as follows:
………………(i)
where dФ is the change in the flux and is equal to NAB
B is the magnetic field = (μ0Ni)/l
Substituting in equation (i), we get
Substituting the values in above equation , we get