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A coil rotates in a uniform magnetic field $B$ with angular speed $\omega$. Its plane is parallel to $B$ at $t = 0$. The peak EMF induced equals (let area = $A$, turns = $N$):
A$\tfrac{1}{2}NAB\omega^2$
B$N A B/(2\pi\omega)$
C$NAB$
D$NAB\omega$
Answer & Solution
Correct answer: D. $NAB\omega$
$\phi(t) = NAB\cos(\omega t)$ (when plane is initially parallel, normal is perpendicular to $B$; $\phi$ oscillates as $\sin(\omega t)$). $e = -d\phi/dt = NAB\omega\sin(\omega t)$, peak = $NAB\omega$.
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