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Physics XII: Electromagnetic Induction
Solutions to Numerical Problems (Unit 14)
Problem 3
Given:
Resistance of coil (R coil
Resistance of galvanometer (R g
Number of turns (N) = 100
Initial magnetic flux (Φ 1
) = 1 mWb = 1 × 10⁻³ Wb
Final magnetic flux (Φ 2
) = 0.2 mWb = 0.2 × 10⁻³ Wb
Time (Δt) = 0.1 s
1. Average e.m.f. (ε):
ε = N * |ΔΦ| / Δt = 100 * |1x10⁻³ - 0.2x10⁻³| / 0.
Result: ε = 0.8 V
2. Induced Current (I):
I = ε / (R coil
+ R
g
Result: I = 1.6 mA
Problem 4
Given:
Length (l) = 10 m, Speed (v) = 5.0 m/s
Magnetic field (B) = 0.30 × 10⁻⁴ T
(a) Instantaneous e.m.f.:
ε = Blv = (0.30 × 10⁻⁴) * 10 * 5.
Result: 1.5 × 10⁻³ V
(b) Direction: By Fleming's Right-Hand Rule, current flows East to West.
(c) Higher Potential: The East end is at a higher electrical potential.
Problem 5
Given:
ΔI = 5.0 A, Δt = 0.1 s, ε = 200 V
Self-Inductance (L):
L = (ε * Δt) / ΔI = (200 * 0.1) / 5.
Result: L = 4 H
Problem 6
Given:
n = 15 turns/cm = 1500 turns/m
Loop Area (A) = 2.0 cm² = 2.0 × 10⁻⁴ m²
ΔI = 2.0 A, Δt = 0.1 s
Induced e.m.f. (ε):
ε = (μ₀ * n * ΔI * A) / Δt
Result: ε ≈ 7.54 × 10⁻⁶ V
