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NCERT Class 12 Physics Chapter 6 Electromagnetic Induction
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Electromagnetic Induction
Chapter: 6
| Part – I |
EXERCISE
1. Predict the direction of induced current in the situations described by the following Figs. 6.15(a) to (f).
Ans: (a) At point Q, where the south pole develops, the induced current must be clockwise. Therefore, in the coil, the induced current flows from point P to point Q.
(b) In this scenario, coil P will develop a south pole at Q. Similarly, coil XY will develop a south pole at X. Consequently, the induced current in coil P will flow from Q to P, and the induced current in coil XY will flow from X to Y.
(c) The induced current in the right loop will circulate along the path XYZ.
(d) The induced current in the left loop will circulate along the path ZYX when viewed from the front.
(e) In the right coil, the induced current flows from X to Y.
(f) No current is induced in the coil if the magnetic lines of force lie within the plane of the loop, as there is no change in magnetic flux through the loop.
2. Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.16:
(a) A wire of irregular shape turning into a circular shape.
Ans: If a wire with an irregular shape is reshaped into a circular loop, the area of the loop increases because a circle encloses a larger area than an irregular shape. As a result, the magnetic flux through the loop also increases. To counteract this increase in flux, the induced current will flow in a direction that creates a magnetic field opposing the change. In this case, the current will flow in an anticlockwise direction around the loop. This direction is such that the wire appears to be pulling inward, reducing the area to counteract the increase in flux. Therefore, the induced current will flow clockwise around the loop, not anticlockwise. This will create a magnetic field that opposes the increasing external magnetic flux.
(b) A circular loop being deformed into a narrow straight wire.
Ans: Conversely, if the loop deforms into a narrow straight wire, its area- and hence the magnetic flux linked with it-decreases. To oppose this decrease in flux, the induced current will flow in a direction that creates a magnetic field to counteract the reduction. Therefore, the induced current will flow clockwise around the wire, not anticlockwise. This will create a magnetic field that tries to maintain the original magnetic flux through the loop.
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?
Ans: Given:
Σ = dØB / dt
= ØB = B × A
B = μ0nl
μ0 = 4𝜋 × 10-7
n = 15 turns/cm = 1500 turns/m
l = 2.0 A to 4.0 over 0.1s
Binitial = μ0nlinitial = (4𝜋 × 10-7) × 1500 turns/m × 2.0 A
= μ0nfinal = (4𝜋 × 10-7) × 1500 turns/m × 4.0 A
Magnetic Flux (ΔØB)
ΔØB = ΔB × A = (μ0nlinitial – μ0nfinal) × A
A = 2.0cm2 = 2.0 × 19-4 m2
Calculate the Induced:
Σ = − ΔØB/Δt
Binitial = (4𝜋 × 10-7) × 1500 × 2.0 = 3.77 × 10-3 T.
Bfinal = (4𝜋 × 10-7) × 1500 × 4.0 = 7.54 × 10-3 T.
= (7.54 × 10-3 T) – (3.77 × 10-3 T) = 3.77 × 10-3 T.
ΔØB = ΔB × A = (3.77 × 10-3 T) × (2.0 × 10-4 m2)
= 7.54 × 10-7 Wb.
Σ = 7.54 × 10-7 Wb / 0.1s
= – 7.54 × 10-6 V.
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?
Ans: Given:
B = 0.3 T
v = 1 cm/s = 0.01 m/s
Dimensions of the loop = 8 cm = 0.08m, 2 cm = 0.02 m
Here:
Σ = B × l × v
Σ = 0.3 T × 0.02 m × 0.01 m/s = 6 × 10-5 V = 60 μV
t = 0.08m / (0.01m/s) = 8 s.
(b) Normal to the shorter side:
l = 0.08 m
Σ = 0.3 T × 0.08 m × 0.01 m/s = 2.4 × 10-4 V = 240 μV
= 0.02 m / (0.01 m/s)
= 2s.
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.
Ans: Given:
ɭ = 1 m
ω = 400 s-1
B = 0.5 T,
e = ?
Here:
v = 0 + ɭω / 2 = lω / 2
= e = Bɭv
= Bɭ2ω / 2
= 0.5 × ɭ 2× 400 / 2
= 100 V.
6. 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?
Ans: Given:
L = 10m
v = 5.0 m/s
B = 0.30 × 10-4 Wb/m2 = 3.0 × 10-5 T.
Σ = B × L × v
= 3.0 × 10-5 T × 10 m × 5.0 m/s
= 1.5 × 10-3 V
= 1.5 m V.
(b) What is the direction of the emf?
Ans: According to Fleming’s right-hand rule, if the force on the wire is downward, the induced emf will be directed from west to east.
(c) Which end of the wire is at the higher electrical potential?
Ans: Since the induced emf is directed from west to east, the west end of the wire is at a higher potential, because electric current naturally flows from a point of higher potential to a point of lower potential.
7. 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.
Ans: Given:
ε = – L (dI/dt)
Rearranging Formula:
L = -ε / (dI/dt)
Change of current:
dI/dt = (5.0 A – 0.0 A) / 0.1 s
= 50 A/s
L = 200 V / 50 A/s
= 4.0 H.
8. 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?
Ans: Given:
M = 1.5 H
Ⅰ1 = 0
Ⅰ2 = 20 A
t = 0.5 s
change in current (20 A – 0 A = 20 A)
dØ = LdⅠ
= 1.5 H × 20 A
= 30 Wb-turns.
