HC Verma Chapter 38 Problem 44

Problem Statement

A circular coil of $N = 60$ turns and radius $R = 25\,\text{cm}$ is in a field that changes by $\Delta B = 0.5\,\text{T}$ in $\Delta t = 0.5\,\text{s}$. The plane is perpendicular to the field. Find the induced emf.

Given Information

$N = 60$ turns
$R = 0.25\,\text{m}$, area $A = \pi R^2 = 0.19635\,\text{m}^2$
$\Delta B = 0.5\,\text{T}$
$\Delta t = 0.5\,\text{s}$

Physical Concepts & Formulas

Faraday’s law: $\mathcal{E} = -N\,d\Phi/dt = -NA\,dB/dt$. The magnitude of induced emf equals $N$ times the rate of flux change through one turn. Lenz’s law gives the direction opposing the change.

$|\mathcal{E}| = NA\,|dB/dt|$ — Faraday’s law for N-turn coil

Step-by-Step Solution

Step 1 — Area of coil: $$A = \pi R^2 = \pi\times(0.25)^2 = 0.19635\,\text{m}^2$$

Step 2 — Rate of field change: $$dB/dt = 0.5/0.5 = 1.00\,\text{T s}^{-1}$$

Step 3 — Induced emf: $$|\mathcal{E}| = NA\cdot|dB/dt| = 60\times0.19635\times1.00 = 11.7810\,\text{V}$$

Worked Calculation

$$|\mathcal{E}| = 60\times0.19635\times1.00 = 11.7810\,\text{V}$$

Answer

$$\boxed{\mathcal{E} = 11.7810\,\text{V} \approx 11780.97\,\text{mV}}$$

The 11780.97 mV emf is induced by the changing flux. More turns, larger area, or faster field change all increase the emf — the design parameters of transformers and generators.

Physical Interpretation