ε = NBAω sin(ωt). Peak EMF and frequency calculation.

EMF Induced in a Rotating Coil — AC Generator

Question

A rectangular coil of 200 turns, each of area 0.05 m², is rotated at 50 rev/s in a uniform magnetic field of 0.04 T. The axis of rotation is perpendicular to the field direction.

(a) Find the peak EMF induced in the coil. (b) Write the expression for instantaneous EMF assuming the coil starts from the position of zero EMF. (c) Find the frequency and time period of the alternating EMF.

(JEE Main 2024 Shift 2)

Solution — Step by Step

List out: N = 200 turns, A = 0.05 m², B = 0.04 T, rotational frequency n = 50 rev/s.

The angular velocity ω = 2πn = 2π × 50 = 100π rad/s. This conversion is the first place marks get dropped — don’t confuse n (rev/s) with ω (rad/s).

The formula for peak EMF comes from Faraday’s law applied to a rotating coil:

\varepsilon_0 = NBA\omega

Substituting:

\varepsilon_0 = 200 \times 0.04 \times 0.05 \times 100\pi

\varepsilon_0 = 200 \times 0.04 \times 0.05 \times 100\pi = 40\pi \approx 125.7 \text{ V}

So peak EMF = 40π V ≈ 125.7 V.

The coil starts from the position where flux is maximum (coil plane perpendicular to B). At this position, rate of change of flux is zero, so EMF starts at zero and follows a sine function:

\varepsilon = \varepsilon_0 \sin(\omega t) = 40\pi \sin(100\pi t) \text{ V}

The sine form applies when the coil starts from the position of maximum flux (zero EMF). If it started from zero flux, you’d use cosine. Read the problem carefully.

The frequency of the alternating EMF equals the rotational frequency of the coil:

f = n = 50 \text{ Hz}

T = \frac{1}{f} = \frac{1}{50} = 0.02 \text{ s}

Frequency = 50 Hz, Time period = 0.02 s.

Why This Works

When the coil rotates in a magnetic field, the flux through it changes as $\Phi = NBA\cos(\omega t)$ . Faraday’s law says EMF = $-d\Phi/dt$ , so differentiating gives $\varepsilon = NBA\omega\sin(\omega t)$ .

The peak occurs when $\sin(\omega t) = 1$ , which is when the coil plane is parallel to B (not perpendicular). At this instant, the coil sides are cutting field lines at the maximum rate — that’s the physical reason EMF peaks here.

The frequency of the output AC equals the mechanical rotation frequency. This is exactly how power station generators work — 50 Hz in India means the turbines complete 50 rotations every second.

Alternative Method

For peak EMF, you can use the rate-of-flux approach directly without memorising ε₀ = NBAω.

At the instant of peak EMF, the coil is parallel to B. The flux changes from +NBA to −NBA in half a rotation. So:

\varepsilon_0 = N \cdot \frac{\Delta\Phi}{\Delta t} \approx N \cdot \frac{2BA}{T/2} = \frac{4NBA}{T} = NBA \cdot \frac{4}{T}

Since $\omega = 2\pi/T$ , this gives $NBA \cdot \frac{2\omega}{\pi} \cdot \frac{\pi}{2} = NBA\omega$ . Same result — the formula isn’t magic, it’s just calculus written compactly.

In MCQ format, you can also spot-check: if ω doubles, peak EMF doubles. If area doubles, peak EMF doubles. These proportionality checks help verify your calculation hasn’t gone off-track.

Common Mistake

Using n instead of ω in the formula.

Students write ε₀ = NBA × 50 instead of NBA × 100π, getting ε₀ ≈ 2 V instead of 125.7 V — a factor of 63 off.

The formula ε₀ = NBAω requires angular velocity in rad/s, not rotational frequency in rev/s. Always convert: ω = 2πn before plugging in. If the question gives “50 Hz” or “50 cycles per second”, the conversion is identical — ω = 2π × 50 = 100π rad/s.

A second trap: some students write $\varepsilon = \varepsilon_0 \cos(\omega t)$ by default. The sine vs cosine form depends entirely on the initial position of the coil. Zero EMF at t = 0 → sine. Maximum EMF at t = 0 → cosine. The problem tells you which one to use — read it.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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