NCERT Solutions for Class 12 Physics Chapter 6 Electromagnetic Induction: Download Free PDF

EXERCISES

F. Predict the direction of the induced current in the situations described in the following figures.

(A)

(B)

(C)

(D)

(e)

(F)

Sol. (A) Direction: qrpq

(B) Direction: prqp; yzxy

(C)Direction: yzxy

(D) Direction: zyxz.

(e) Direction: xryx

(F) Direction: No induced current

F. Use Lenz’s law to determine the direction of the induced current in the situations described in figure (a): A wire of irregular shape that turns into a circular shape; (b) A circular loop is deformed into a narrow straight wire.

Sol. According to Lenz’s law (by German physicist Heinrich Friedrich Lenz), the polarity of the induced emf is such that it tends to produce a current that counteracts the change in magnetic flux that produced it. Lenz’s law contradicts the cause that causes it.

(A) As the flowing flux increases, the direction of the induced current is, according to Lenz’s law: adcb

(B) As the flux passing through decreases, the direction of the induced current is as follows Lenz’s law is: adcba.

F. A long magnet with 15 turns per cm has a small loop with an area of ​​2.0 cm² located inside the magnet perpendicular to its axis. If the current flowing from the solenoid valve changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop as the current changes?

Sol. Given,

n = 15 turns/cm

a = 2 cm²

dt = 0.1 s

Ii = 2 A

If = 4 A

θ=0

Φ=BA cosθ…………………………(1)

Initial magnetic field, Bi =μ0nIi

Substitute,

Bi = (4π x 10-7).(1500).(2)

Or Bi = 3.77×10-³ T

Final magnetic field, Bf = (4π x 10-7).(1500).(4)

= 7.54×10-³T

Initial value of flow = Bi.A.cos0

= (3.77×10-³).(2×10-⁴)

= 7.54×10-⁷Wb

Final value of the flow = Bf.A.cos0

= (7.54×10-³). (2×10-⁴)

= 1.51×10-⁶ Wb

Hence induced emf

E = flow difference / corresponding time interval

Or E = [(1.51×10-⁶ Wb)–(7.54×10-⁷ Wb)] / 0.1

Or E = 7.56×10-⁶ V

F. A rectangular loop of wire with a side of 8 cm and 2 cm and a small cut moves from a region of a uniform magnetic field of strength 0.3 T directed perpendicular 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 and (b) shorter side of the loop? How long does the induced voltage last?

Sol. Given,

Length = 8cm

Width = 2cm

B = 0.3 T

Speed ​​= 1 cm/s

Area of ​​the loop

A = 0.0016 m²

The loop moves at the specified speed. As long as the loop is completely in the magnetic field, no emf is induced because the flux does not change. But the moment it starts sliding into space, the flux flowing through the loop decreases and an emf is induced. Since the circuit is not complete, no current flows.

Initial flow value = BAcos0

= 0.3 x 0.0016

= 0.00048 wb

Value of final flow = 0,

(A) In this case the elapsed time is

t = distance / speed

= 2 / 1

= 2 s

So induced emf

E = dФ/ dt

E = (0.00048–0) / 2

Or E = 2.4×10-⁴V

The tension lasts 2 seconds.

(B) The elapsed time is displayed here

t = 8 / 1

= 8 s

This induced emf

E = dФ/ dt

E = 0.00048 / 8

Or E = 6×10-⁵V

The tension lasts 8 seconds.

F. A 1.0 m long metal rod is rotated at an angular frequency of 400 rad s-¹ about an axis perpendicular to the rod and passing through one end of it. The other end of the rod is in contact with a circular metal ring. There is a constant and uniform magnetic field of 0.5 T parallel to the axis everywhere. Calculate the emf developed between the center and the ring.

Sol. Given,

r = 1 m

w = 400 rad/s

B = 0.5 T

Ф =BA cos 0

= 0.5 x 3.14 x 1 x 1

= 1.571 wb

Ф =BAcos 180

= -1.571 wb

w = 400

Or f = 400 / 2 x 3.14 = 63.66 / s

therefore

E = [ 1.571–(- 1.571)] / 0.0314

Or E = 100V

F. A horizontal straight wire 10 m long extending from east to west falls at a speed of 5.0 m s-¹ at right angles to the horizontal component of the Earth’s magnetic field, 0.30 x 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 has the higher electrical potential?

Sol. Given,

l = 10 m

v = 5 m/s

H = 0.3×10-⁴ Wb m²

(a) Einst = Blv

= (0.00003).(10).(5)

= 1.5×10-³ V

(b) West to East

(c)east end.

Q. The current in a circuit drops 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.

Sol. Given,

dI = 5–0 = 5 A

dt = 0.1 s

Eavg = 200V

E = L.dI/dt

Substitution exists

L = 4H

Q. A pair of adjacent coils have a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, how does the flux coupling with the other coil change?

Sol. Given,

M = 1.5H

dI = 20 A

dt = 0.5 s

d(NФ) / dt = d(MI) / dt

replace values,

d(NФ) = 30 Weber

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