AC generator — derive EMF equation ε = NBA ω sin(ωt)

Question

A rectangular coil of $N$ turns, area $A$ , rotates with angular velocity $\omega$ in a uniform magnetic field $B$ . Derive the expression for the instantaneous EMF induced in the coil. What is the peak EMF?

(NCERT Class 12, Chapter 6 — this derivation is asked almost every year in boards)

Solution — Step by Step

At time $t$ , the coil makes angle $\theta = \omega t$ with the magnetic field. The magnetic flux through one turn:

\Phi = BA\cos\theta = BA\cos(\omega t)

For $N$ turns: $\Phi_{total} = NBA\cos(\omega t)$

By Faraday’s law, the induced EMF is:

\varepsilon = -\frac{d\Phi_{total}}{dt} = -NBA\frac{d}{dt}[\cos(\omega t)]

\varepsilon = NBA\omega\sin(\omega t)

The maximum value of $\sin(\omega t)$ is 1, so:

\varepsilon_0 = NBA\omega

The complete EMF equation:

\boxed{\varepsilon = \varepsilon_0 \sin(\omega t) = NBA\omega\sin(\omega t)}

This is an alternating EMF — it oscillates between $+\varepsilon_0$ and $-\varepsilon_0$ with frequency $f = \omega/(2\pi)$ .

EMF is maximum when $\sin(\omega t) = \pm 1$ , i.e., the coil plane is parallel to $\vec{B}$ (flux is zero but changing fastest).

EMF is zero when $\sin(\omega t) = 0$ , i.e., the coil plane is perpendicular to $\vec{B}$ (flux is maximum but momentarily not changing).

Why This Works

The key insight: it is the rate of change of flux that produces EMF, not the flux itself. When flux is at its peak (coil perpendicular to $B$ ), the flux is momentarily constant — EMF is zero. When flux passes through zero (coil parallel to $B$ ), it is changing most rapidly — EMF is maximum.

This is why the EMF equation has $\sin(\omega t)$ while the flux equation has $\cos(\omega t)$ . The derivative of cosine is negative sine — the 90° phase difference between flux and EMF is fundamental.

Alternative Method

Start from the motional EMF on each side of the coil. The two sides of length $l$ (perpendicular to the rotation axis) move with velocity $v = r\omega$ where $r$ is the distance from the axis. The EMF in each side is $Blv\sin(\omega t)$ . With $N$ turns and both sides contributing: $\varepsilon = 2NBlr\omega\sin(\omega t) = NBA\omega\sin(\omega t)$ (since $A = 2rl$ for the rectangular coil).

For CBSE boards, always mention Faraday’s law by name and show the differentiation step explicitly — examiners look for this. For JEE, remember: peak EMF $\varepsilon_0 = NBA\omega$ and RMS EMF $\varepsilon_{rms} = \varepsilon_0/\sqrt{2}$ . Many JEE problems ask for the RMS value directly.

Common Mistake

Students write $\varepsilon = NBA\omega\cos(\omega t)$ instead of $\sin(\omega t)$ . The confusion arises from not being careful about the starting position. If at $t = 0$ the coil is perpendicular to $\vec{B}$ (maximum flux), the flux is $\cos(\omega t)$ and the EMF is $\sin(\omega t)$ . If at $t = 0$ the coil is parallel to $\vec{B}$ (zero flux), the flux is $\sin(\omega t)$ and the EMF is $\cos(\omega t)$ . The standard NCERT convention uses the first case.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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