Electromagnetic Induction: Numerical Problems Set (9)

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Question

A rectangular coil of 200200 turns and area 0.05 m20.05\text{ m}^2 rotates at 50 rev/s50\text{ rev/s} in a uniform magnetic field of 0.4 T0.4\text{ T}. The coil’s axis of rotation is perpendicular to the field. Find (a) the peak EMF, (b) the RMS EMF, and (c) the EMF at the instant the plane of the coil is parallel to the field.

Solution — Step by Step

ω=2πf=2π×50=100π rad/s\omega = 2\pi f = 2\pi \times 50 = 100\pi\text{ rad/s}.

For a rotating coil, ε(t)=NBAωsin(ωt)\varepsilon(t) = NBA\omega \sin(\omega t). Peak value:

ε0=NBAω=200×0.4×0.05×100π\varepsilon_0 = NBA\omega = 200 \times 0.4 \times 0.05 \times 100\pi ε0=400π1256.6 V\varepsilon_0 = 400\pi \approx 1256.6\text{ V}
εrms=ε02=400π2=200π2888.6 V\varepsilon_{\text{rms}} = \frac{\varepsilon_0}{\sqrt{2}} = \frac{400\pi}{\sqrt{2}} = 200\pi\sqrt{2} \approx 888.6\text{ V}

When the plane of the coil is parallel to the field, the area vector (normal to the plane) is perpendicular to the field. So flux is zero — but rate of change of flux is maximum here. So the EMF is at its peak value, ε01256.6 V\varepsilon_0 \approx 1256.6\text{ V}.

ε01256.6 V\varepsilon_0 \approx 1256.6\text{ V}, εrms888.6 V\varepsilon_{\text{rms}} \approx 888.6\text{ V}, EMF at parallel-to-field 1256.6 V\approx 1256.6\text{ V}.

Why This Works

Faraday’s law says induced EMF is dΦ/dt-d\Phi/dt. With Φ=NBAcos(ωt)\Phi = NBA\cos(\omega t), differentiation gives ε=NBAωsin(ωt)\varepsilon = NBA\omega\sin(\omega t). Peak occurs when sin=1\sin = 1, which is exactly when cos=0\cos = 0 — when flux is zero, its rate of change is largest.

This is the geometric reason students sometimes get backwards: max flux \ne max EMF.

Alternative Method

Energy argument: the kinetic energy of rotation is converted to electrical energy at the rate at which the coil “cuts” field lines. The cutting rate is proportional to the velocity of the wire perpendicular to the field, which peaks when the coil’s plane lies along the field direction.

Common Mistake

Forgetting to multiply by NN (number of turns). ε0=BAω\varepsilon_0 = BA\omega is the EMF for a single loop. With 200 turns in series, the total EMF multiplies by 200. NEET 2024 had exactly this trap — three options used the single-turn value.

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