Revision notes, practice numericals, a question bank, assertion and reason questions and answers to the NCERT questions. Tap a group to open it; opening one closes the others.
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A loop of 10 cm × 20 cm is in a field of 0.5 T with its normal at 60° to the field. Find the flux.
Given: A = 0.02 m²; B = 0.5 T; θ = 60°
Use: Φ = B A cos θ
Φ = 0.5 × 0.02 × 0.5 = 5 × 10⁻³ Wb
Answer: 5 × 10⁻³ Wb
The flux through a coil of 100 turns falls from 0.02 Wb to 0.005 Wb in 0.1 s. Find the induced emf.
Given: N = 100; ΔΦ = 0.015 Wb; Δt = 0.1 s
Use: e = N ΔΦ / Δt
e = 100 × 0.015 / 0.1 = 15 V
Answer: 15 V
A coil of 50 turns and area 0.02 m² has its plane perpendicular to a field that falls from 0.8 T to 0.2 T in 0.3 s. Find the emf.
Given: N = 50; A = 0.02 m²; ΔB = 0.6 T; Δt = 0.3 s
Use: e = N A ΔB / Δt
e = 50 × 0.02 × 0.6 / 0.3 = 2 V
Answer: 2 V
The flux through a coil of resistance 10 Ω changes by 0.05 Wb. Find the charge that flows through the coil.
Given: R = 10 Ω; ΔΦ = 0.05 Wb
Use: q = ΔΦ / R
q = 0.05 / 10 = 5 × 10⁻³ C
Answer: 5 × 10⁻³ C, whatever the time taken
A field B = 2t² (tesla) acts normally on a loop of area 0.01 m². Find the emf at t = 2 s.
Given: A = 0.01 m²; B = 2t²
Use: e = A dB/dt
dB/dt = 4t = 8 T/s at t = 2 s
e = 0.01 × 8 = 0.08 V
Answer: 0.08 V
A small loop of area 5 cm² lies inside a long solenoid of 1000 turns per metre, normal to its axis. The current rises from 0 to 5 A in 0.5 s. Find the emf in the loop.
Given: n = 1000 m⁻¹; A = 5 × 10⁻⁴ m²; dI/dt = 10 A/s
Use: e = A μ₀ n dI/dt
dB/dt = 4π × 10⁻⁷ × 1000 × 10 = 1.26 × 10⁻² T/s
e = 5 × 10⁻⁴ × 1.26 × 10⁻² = 6.3 × 10⁻⁶ V
Answer: 6.3 × 10⁻⁶ V
A rod 0.5 m long moves at 4 m/s perpendicular to a field of 0.2 T. Find the emf across its ends.
Given: l = 0.5 m; v = 4 m/s; B = 0.2 T
Use: e = B l v
e = 0.2 × 0.5 × 4 = 0.4 V
Answer: 0.4 V
The rod of the previous problem slides on rails joined by a 2 Ω resistor. Find the current, the force needed to move it uniformly and the power.
Given: e = 0.4 V; R = 2 Ω
Use: I = e/R; F = BIl; P = F v
I = 0.4 / 2 = 0.2 A
F = 0.2 × 0.2 × 0.5 = 0.02 N
P = F v = 0.02 × 4 = 0.08 W (= I²R = 0.04 × 2 ✓)
Answer: 0.2 A; 0.02 N; 0.08 W
A rod 0.5 m long rotates about one end at 50 rad/s in a field of 0.4 T normal to the plane of rotation. Find the emf between its ends.
Given: l = 0.5 m; ω = 50 rad/s; B = 0.4 T
Use: e = ½ B ω l²
e = 0.5 × 0.4 × 50 × 0.25 = 2.5 V
Answer: 2.5 V
An aeroplane with a wingspan of 40 m flies at 250 m/s where the vertical component of the earth's field is 5 × 10⁻⁵ T. Find the emf across its wing tips.
Given: l = 40 m; v = 250 m/s; B_v = 5 × 10⁻⁵ T
Use: e = B l v
e = 5 × 10⁻⁵ × 40 × 250 = 0.5 V
Answer: 0.5 V
The rails of a track are 1.5 m apart. A train moves at 36 km/h where the vertical field is 4 × 10⁻⁵ T. Find the emf across the axle.
Given: l = 1.5 m; v = 10 m/s; B = 4 × 10⁻⁵ T
Use: e = B l v
36 km/h = 10 m/s
e = 4 × 10⁻⁵ × 1.5 × 10 = 6 × 10⁻⁴ V
Answer: 0.6 mV
A rod of mass 0.05 kg and length 0.4 m slides down smooth vertical rails joined by a 1 Ω resistor in a horizontal field of 0.5 T. Find its terminal speed. (g = 9.8 m/s²)
Given: m = 0.05 kg; l = 0.4 m; R = 1 Ω; B = 0.5 T
Use: At terminal speed m g = B²l²v / R
v = m g R / (B² l²) = 0.05 × 9.8 × 1 / (0.25 × 0.16) = 0.49 / 0.04 = 12.25 m/s
Answer: 12.25 m/s
An emf of 4 V is induced in a coil when the current changes from 0 to 2 A in 0.1 s. Find its self-inductance.
Given: e = 4 V; ΔI = 2 A; Δt = 0.1 s
Use: L = e Δt / ΔI
L = 4 × 0.1 / 2 = 0.2 H
Answer: 0.2 H
A solenoid has 500 turns, length 0.5 m and area 4 × 10⁻⁴ m². Find its self-inductance.
Given: N = 500; l = 0.5 m; A = 4 × 10⁻⁴ m²
Use: L = μ₀ N² A / l
L = 4π × 10⁻⁷ × 250000 × 4 × 10⁻⁴ / 0.5 = 2.5 × 10⁻⁴ H
Answer: 0.25 mH
An inductor of 0.5 H carries a current of 4 A. Find the energy stored.
Given: L = 0.5 H; I = 4 A
Use: U = ½ L I²
U = 0.5 × 0.5 × 16 = 4 J
Answer: 4 J
An emf of 3 V is induced in a coil when the current in a neighbouring coil changes at 6 A/s. Find the mutual inductance.
Given: e = 3 V; dI/dt = 6 A/s
Use: M = e / (dI/dt)
M = 3 / 6 = 0.5 H
Answer: 0.5 H
The mutual inductance of two coils is 0.8 H. The current in one changes from 2 A to 7 A in 0.25 s. Find the emf induced in the other.
Given: M = 0.8 H; ΔI = 5 A; Δt = 0.25 s
Use: e = M ΔI / Δt
e = 0.8 × 5 / 0.25 = 16 V
Answer: 16 V
Two coaxial solenoids each have 1000 turns per metre. The inner one has radius 2 cm and length 0.5 m. Find their mutual inductance.
Given: n₁ = n₂ = 1000 m⁻¹; r₁ = 0.02 m; l = 0.5 m
Use: M = μ₀ n₁ n₂ π r₁² l
M = 4π × 10⁻⁷ × 10⁶ × π × (0.02)² × 0.5 = 7.9 × 10⁻⁴ H
Answer: 0.79 mH
The current in a 2 H inductor rises at 5 A/s. Find the emf across it.
Given: L = 2 H; dI/dt = 5 A/s
Use: e = L dI/dt
e = 2 × 5 = 10 V
Answer: 10 V (opposing the rise)
A coil of 200 turns and area 0.05 m² rotates at 50 Hz in a field of 0.2 T. Find the peak emf and the rms value.
Given: N = 200; A = 0.05 m²; B = 0.2 T; f = 50 Hz
Use: e₀ = N B A ω; e_rms = e₀/√2
ω = 2π × 50 = 314 rad/s
e₀ = 200 × 0.2 × 0.05 × 314 = 628 V
e_rms = 628 / 1.414 = 444 V
Answer: 628 V; 444 V
A generator coil rotates at 3000 revolutions per minute. Find the frequency of the emf.
Given: 3000 rev/min
Use: Frequency = revolutions per second
f = 3000 / 60 = 50 Hz
Answer: 50 Hz
A generator has e₀ = 628 V and f = 50 Hz. Find the emf at t = 1/300 s.
Given: e₀ = 628 V; f = 50 Hz
Use: e = e₀ sin ωt
ωt = 2π × 50 / 300 = π/3 = 60°
e = 628 × sin 60° = 544 V
Answer: 544 V
A coil of 50 turns and area 0.1 m² rotates at 20 revolutions per second in a field of 0.5 T. Find the peak emf.
Given: N = 50; A = 0.1 m²; B = 0.5 T; f = 20 Hz
Use: e₀ = N B A 2π f
e₀ = 50 × 0.5 × 0.1 × 2π × 20 = 314 V
Answer: 314 V
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In every case use Lenz's law: the induced current opposes the change of flux. Directions below are the sense of the induced current as seen along the common axis; from them you can mark the direction on the labelled points of the figure.
(a) The N pole approaches the coil, so the flux through it increases. The coil's nearer end must become a north pole (to repel the magnet): current anticlockwise as seen from the magnet's side.
(b) The magnet (S at the left end) moves left. For the left coil the S pole approaches, so its right end becomes S: current clockwise seen from the right. For the right coil the magnet is moving away; the flux (directed to the right) decreases, so the coil tries to keep it: current anticlockwise as seen from the right.
(c) When the key is just closed, the current in the first coil rises from zero, so the flux it sends through the second coil increases. The induced current in the second coil is in the opposite sense to the current in the first coil.
(d) The induced current in the second coil depends on how the rheostat is moved. If the current in the first coil is increasing, the induced current is in the opposite sense; if the current is decreasing, the induced current is in the same sense as the current in the first coil.
(e) When the key is released, the current in the first solenoid falls to zero, so the flux through the second decreases. The induced current in the second is in the same sense as the original current in the first (it tries to maintain the flux).
(f) The straight wire passes through the centre of the loop and perpendicular to its plane, so the magnetic field lines of the wire (circles about it) lie in the plane of the loop. The flux through the loop is zero all the time, so no current is induced.
(a) A wire of irregular shape turning into a circle encloses a larger area for the same perimeter, so the flux (out of the page) increases. The induced current produces a field into the page: the current is clockwise.
(b) A circle being deformed into a narrow straight wire encloses a smaller area, so the flux out of the page decreases. The induced current produces a field out of the page: the current is anticlockwise.
n = 15 per cm = 1500 per m. The rate of change of current is 2.0 / 0.1 = 20 A/s.
B = μ₀ n I, so dB/dt = μ₀ n dI/dt = 4π × 10⁻⁷ × 1500 × 20 = 3.77 × 10⁻² T/s.
e = A dB/dt = 2.0 × 10⁻⁴ × 3.77 × 10⁻² = 7.5 × 10⁻⁶ V.
v = 1 cm/s = 0.01 m/s. The emf comes from the side that moves out of the field, e = B l v.
(a) Velocity normal to the longer side: the longer side (8 cm) cuts the field. e = 0.3 × 0.08 × 0.01 = 2.4 × 10⁻⁴ V. The loop moves out through its width of 2 cm, which takes t = 0.02 / 0.01 = 2 s.
(b) Velocity normal to the shorter side: the 2 cm side cuts the field. e = 0.3 × 0.02 × 0.01 = 6 × 10⁻⁵ V. The loop moves out through 8 cm, so t = 0.08 / 0.01 = 8 s.
e = ½ B ω l² = 0.5 × 0.5 × 400 × (1.0)² = 100 V.
(a) e = B l v = 0.30 × 10⁻⁴ × 10 × 5.0 = 1.5 × 10⁻³ V.
(b) The horizontal component of the earth's field points north, the velocity is downward. v × B is towards the east, so the emf is directed from west to east.
(c) Positive charges are pushed towards the east end, so the eastern end is at the higher potential.
e = L |ΔI/Δt|, so L = e Δt / ΔI = 200 × 0.1 / 5.0 = 4 H.
Flux linkage in the second coil is Φ = M I. The change is ΔΦ = M ΔI = 1.5 × 20 = 30 Wb.
(The average induced emf is ΔΦ/Δt = 30 / 0.5 = 60 V.)