NCERT Solutions for Class 12 Physics Chapter 6 – Electromagnetic Induction

Question 6.1:

Predict the direction of induced current in the situations described by the following Figs. 6.18(a) to (f ).

WhatsApp Group Join Now
Telegram Group Join Now
Instagram Group Join Now

(a)

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m48c43b10

(b)

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m30f8c01

(c)

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 273a83b2

(d)

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m559b7562

(e)

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 5c9f4db1

(f)

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m79c663fa

Answer:

The direction of the induced current in a closed loop is given by Lenz’s law. The given pairs of figures show the direction of the induced current when the North pole of a bar magnet is moved towards and away from a closed-loop respectively.

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m488c0226

Using Lenz’s rule, the direction of the induced current in the given situations can be predicted as follows:

(a) The direction of the induced current is along qrpq.

(b) The direction of the induced current is along prqp.

(c) The direction of the induced current is along yzxy.

(d) The direction of the induced current is along zyxz.

(e) The direction of the induced current is along xryx.

(f) No current is induced since the field lines are lying in the plane of the closed-loop.

Page No 230:

Question 6.2:

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.

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 66096be4

Answer:

According to Lenz’s law, the direction of the induced emf is such that it tends to produce a current that opposes the change in the magnetic flux that produced it.

(a) When the shape of the wire changes, the flux piercing through the unit surface area increases. As a result, the induced current produces an opposing flux. Hence, the induced current flows along adcb.

(b) When the shape of a circular loop is deformed into a narrow straight wire, the flux piercing the surface decreases. Hence, the induced current flows along NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 70491e69

Question 6.3:

A long solenoid with 15 turns per cm has a small loop of area 2.0 cmplaced 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:

Number of turns on the solenoid = 15 turns/cm = 1500 turns/m

Number of turns per unit length, = 1500 turns

The solenoid has a small loop of area, A = 2.0 cm2 = 2 × 10−4 m2

The current carried by the solenoid changes from 2 A to 4 A.

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 4dd19828Change in current in the solenoid, di = 4 − 2 = 2 A

Change in time, dt = 0.1 s

Induced emf in the solenoid is given by Faraday’s law as:

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m529a038a

Where,

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 2fac22b2 = Induced flux through the small loop

BA … (ii)

= Magnetic field

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 55b7b537

μ0 = Permeability of free space

4π×10−7 H/m

Hence, equation (i) reduces to:

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m6b6d0405

Hence, the induced voltage in the loop is NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 4eefdc3d

Question 6.4:

A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of the 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:

Length of the rectangular wire, l = 8 cm = 0.08 m

Width of the rectangular wire, b = 2 cm = 0.02 m

Hence, area of the rectangular loop,

A = lb

= 0.08 × 0.02

= 16 × 10−4 m2

Magnetic field strength, B = 0.3 T

Velocity of the loop, v = 1 cm/s = 0.01 m/s

(a) Emf developed in the loop is given as:

e = Blv

= 0.3 × 0.08 × 0.01 = 2.4 × 10−4 V

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m20310a2b

Hence, the induced voltage is 2.4 × 10−4 V which lasts for 2 s.

(b) Emf developed, e = Bbv

= 0.3 × 0.02 × 0.01 = 0.6 × 10−4 V

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 5c35c0c4

Hence, the induced voltage is 0.6 × 10−4 V which lasts for 8 s.

Question 6.5:

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 center and the ring.

Answer:

Length of the rod, l = 1 m

Angular frequency,ω = 400 rad/s

Magnetic field strength, B = 0.5 T

One end of the rod has zero linear velocity, while the other end has a linear velocity of lω.

The average linear velocity of the rod,NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 76745b08

Emf developed between the centre and the ring,

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m3ec7bf18

Hence, the emf developed between the centre and the ring is 100 V.

Question 6.6:

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 10Ω, 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:

Max induced emf = 0.603 V

Average induced emf = 0 V

Max current in the coil = 0.0603 A

Average power loss = 0.018 W

(Power comes from the external rotor)

Radius of the circular coil, r = 8 cm = 0.08 m

Area of the coil, A = πr2 = π × (0.08)m2

Number of turns on the coil, N = 20

Angular speed, ω = 50 rad/s

Magnetic field strength, B = 3 × 10−2 T

Resistance of the loop, R = 10 Ω

Maximum induced emf is given as:

e = Nω AB

= 20 × 50 × π × (0.08)2 × 3 × 10−2

= 0.603 V

The maximum emf induced in the coil is 0.603 V.

Over a full cycle, the average emf induced in the coil is zero.

The maximum current is given as:

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 6d7d81e5

Average power loss due to joule heating:

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m611ce8e6

The current induced in the coil produces a torque opposing the rotation of the coil. The rotor is an external agent. It must supply torque to counter this torque in order to keep the coil rotating uniformly. Hence, dissipated power comes from the external rotor.

Question 6.7:

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:

Length of the wire, l = 10 m

Falling speed of the wire, v = 5.0 m/s

Magnetic field strength, B = 0.3 × 10−4 Wb m−2

(a) Emf induced in the wire,

e = Blv

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m604dd834

(b) Using Fleming’s right-hand rule, it can be inferred that the direction of the induced emf is from West to East.

(c) The eastern end of the wire is at a higher potential.

Question 6.8:

Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V is induced, give an estimate of the self-inductance of the circuit.

Answer:

Initial current, I1 = 5.0 A

Final current, I2 = 0.0 A

Change in current,NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m2ef75ded

Time taken for the change, t = 0.1 s

Average emf, e = 200 V

For self-inductance (L) of the coil, we have the relation for average emf as:

e = L NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 5637e516

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m626a791d

Hence, the self-induction of the coil is 4 H.

Question 6.9:

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:

Mutual inductance of a pair of coils, µ = 1.5 H

Initial current, I1 = 0 A

Final current I2 = 20 A

Change in current, NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m48fa83bb

Time is taken for the change, t = 0.5 s

Induced emf, NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 6f066621

Where NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 4b1b6ff1 is the change in the flux linkage with the coil?

Emf is related with mutual inductance as:

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m4aa3af85

Equating equations (1) and (2), we get

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m5a49709a

Hence, the change in the flux linkage is 30 Wb.

Question 6.10:

A jet plane is traveling towards the west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing 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:

Speed of the jet plane, v = 1800 km/h = 500 m/s

Wing spanof jet plane, l = 25 m

Earth’s magnetic field strength, B = 5.0 × 10−4 T

Angle of dip, NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m153b1308

Vertical component of Earth’s magnetic field,

BV = B sinNS 18 11 08 Sravana 12 Physics 6 17 NRJ html m14aad32c

= 5 × 10−4 sin 30°

= 2.5 × 10−4 T

The voltage difference between the ends of the wing can be calculated as:

e = (BV) × l × v

= 2.5 × 10−4 × 25 × 500

= 3.125 V

Hence, the voltage difference developed between the ends of the wings is

3.125 V.

Page No 231:

Question 6.11:

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 1.6 Ω how much power is dissipated by the loop as heat? What is the source of this power?

Answer:

Sides of the rectangular loop are 8 cm and 2 cm.

Hence, area of the rectangular wire loop,

A = length × width

= 8 × 2 = 16 cm2

= 16 × 10−4 m2

Initial value of the magnetic field, NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m6b268676

Rate of decrease of the magnetic field, NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 5dce4393

Emf developed in the loop is given as:

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 1b4eeaed

Where,

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 4b1b6ff1 = Change in flux through the loop area

AB

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m6e73503d

Resistance of the loop, R = 1.6 Ω

The current induced in the loop is given as:

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 3595d1b0

Power dissipated in the loop in the form of heat is given as:

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 54933e20

The source of this heat loss is an external agent, which is responsible for changing the magnetic field with time.

Question 6.12:

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 mΩ.

Answer:

Side of the square loop, s = 12 cm = 0.12 m

Area of the square loop, A = 0.12 × 0.12 = 0.0144 m2

Velocity of the loop, v = 8 cm/s = 0.08 m/s

Gradient of the magnetic field along negative x-direction,

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m2d2d01a6

And, rate of decrease of the magnetic field,

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m5a73dc17

Resistance of the loop, NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 1ba8f9d8

Rate of change of the magnetic flux due to the motion of the loop in a non-uniform magnetic field is given as:

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m125770aa

Rate of change of the flux due to explicit time variation in field is given as:

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 6f184a2e

Since the rate of change of the flux is the induced emf, the total induced emf in the loop can be calculated as:

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 6f914e12

∴Induced current, NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 4e54e1c9

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m996e69d

Hence, the direction of the induced current is such that there is an increase in the flux through the loop along positive z-direction.

Question 6.13:

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 Ω. Estimate the field strength of magnet.

Answer:

Area of the small flat search coil, A = 2 cm= 2 × 10−4 m2

Number of turns on the coil, N = 25

Total charge flowing in the coil, Q = 7.5 mC = 7.5 × 10−3 C

Total resistance of the coil and galvanometer, R = 0.50 Ω

Induced current in the coil,

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m2bd4e13c

Induced emf is given as:

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m7955ad8f

Where,

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 4b1b6ff1 = Charge in flux

Combining equations (1) and (2), we get

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 71bd420e

Initial flux through the coil, NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 2286445a = BA

Where,

B = Magnetic field strength

Final flux through the coil, NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m28b5c945

Integrating equation (3) on both sides, we have

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 57f82f

But total charge, NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m6a3cffc7

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m3f560629

Hence, the field strength of the magnet is 0.75 T.

Question 6.14:

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, = 0.50 T, resistance of the closed loop containing the rod = 9.0 mΩ. 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.

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m32b06605

(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 (=12 cm 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:

Length of the rod, l = 15 cm = 0.15 m

Magnetic field strength, B = 0.50 T

Resistance of the closed loop, R = 9 mΩ = 9 × 10−3 Ω

(a) Induced emf = 9 mV; polarity of the induced emf is such that end P shows positive while end Q shows negative ends.

Speed of the rod, v = 12 cm/s = 0.12 m/s

Induced emf is given as:

e = Bvl

0.5 × 0.12 × 0.15

= 9 × 10−3 v

= 9 mV

The polarity of the induced emf is such that end P shows positive while end Q shows negative ends.

(b) Yes; when key K is closed, excess charge is maintained by the continuous flow of current.

When key K is open, there is excess charge built up at both ends of the rods.

When key K is closed, excess charge is maintained by the continuous flow of current.

(c) Magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rod.

There is no net force on the electrons in rod PQ when key K is open and the rod is moving uniformly. This is because magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rods.

(d) Retarding force exerted on the rod, F = IBl

Where,

I = Current flowing through the rod

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 5584cce3

(e) 9 mW; no power is expended when key K is open.

Speed of the rod, v = 12 cm/s = 0.12 m/s

Hence, power is given as:

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 43714a44

When key K is open, no power is expended.

(f) 9 mW; power is provided by an external agent.

Power dissipated as heat = I2 R

= (1)2 × 9 × 10−3

= 9 mW

The source of this power is an external agent.

(g) Zero

In this case, no emf is induced in the coil because the motion of the rod does not cut across the field lines.

Page No 232:

Question 6.15:

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:

Length of the solenoid, l = 30 cm = 0.3 m

Area of cross-section, A = 25 cm2 = 25 × 10−4 m2

Number of turns on the solenoid, N = 500

Current in the solenoid, I = 2.5 A

Current flows for time, t = 10−3 s

Average back emf, NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m8ed7b45

Where,

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 4b1b6ff1 = Change in flux

NAB … (2)

Where,

B = Magnetic field strength

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 68bac341

Where,

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m144f80ba = Permeability of free space = 4π × 10−7 T m A−1

Using equations (2) and (3) in equation (1), we get

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m22b0c4a9

Hence, the average back emf induced in the solenoid is 6.5 V.

Question 6.16:

(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side as shown in Fig. 6.21.

(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, = 10 m/s.

Calculate the induced emf in the loop at the instant when = 0.2 m.

Take = 0.1 m and assume that the loop has a large resistance.

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m1b235f67

Answer:

(a) Take a small element dy in the loop at a distance y from the long straight wire (as shown in the given figure).

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m1a91c65e

Magnetic flux associated with elementNS 18 11 08 Sravana 12 Physics 6 17 NRJ html 1d20d240

Where,

dA = Area of element dy a dy

B = Magnetic field at distance y

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m77c6303b

I = Current in the wire

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m144f80ba = Permeability of free space = 4π × 10−7 T m A−1

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 1be19844

y tends from x to NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 3ee29444.

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 2b18dba8

(b) Emf induced in the loop, e = B’avNS 18 11 08 Sravana 12 Physics 6 17 NRJ html m7e043ee4

Given,

= 50 A

x = 0.2 m

a = 0.1 m

v = 10 m/s

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m322406d1

Question 6.17:

A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Fig. 6.22). A uniform magnetic field extends over a circular region within the rim. It is given by,

= − B0 (≤ aR)

= 0 (otherwise)

What is the angular velocity of the wheel after the field is suddenly switched off?

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 258849d

Answer:

Line charge per unit length NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 37b54236

Where,

= Distance of the point within the wheel

Mass of the wheel = M

Radius of the wheel = R

Magnetic field, NS 18 11 08 Sravana 12 Physics 6 17 NRJ html 8ea7a1b

At distance r,themagnetic force is balanced by the centripetal force i.e.,

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m425cefb8

∴Angular velocity, NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m3856a32c

NS 18 11 08 Sravana 12 Physics 6 17 NRJ html m62ba1bc0


I hope you like the information and notes given by us. This will help you a lot in your upcoming exams. Here we are giving ncert solutions to all of you students, through which all of you can score very well in your exams. In this post, we are giving you ncert solutions. hope you benefit from it. You can download the pdf file of this ncert solutions.

Leave a Comment

WhatsApp Group Join Now
Telegram Group Join Now
Instagram Group Join Now
X
Vada Pav Girl Net Worth Post Office KVP Yojana में 5 लाख के मिलते है 10 लाख रूपये, जाने पैसा कितने दिनों में होगा डबल SSC GD 2024 Result, Merit List Cut-Off What is the Full Form of NASA?
Copy link
Powered by Social Snap