Question

Discuss with theory the method of inducing emf in a coil by changing its orientation with respect to the direction of the magnetic field.

Answer 1

Emf induced by changing the orientation of the coil :

$PQRS$ is a rectangular coil of $N$ turns and area $A$ placed in a uniform magnetic field $B$ (figure). The coil is rotated with an angular velocity $\omega$ in the clockwise direction about an axis perpendicular to the direction of the magnetic field. Suppose, initially the coil is in vertical position, so that the angle between normal to the plane of the coil and magnetic field is zero. After a time $t$, let $q$ be the angle through which the coil is rotated. If $\phi$ is the flux linked with the coil at this instant, then $\phi =NBA\cos \theta$ (since, $\theta =\omega t$)
The induced emf is, $e=-\frac{d\varphi }{dt}=-NBA\frac{d}{dt}\cos \left(\omega t\right)$
$\therefore e=NBA\omega \sin \omega t$ ....(1)
The maximum value of the induced emf is, $E_o=NAB\omega$
Hence, the induced emf can be represented as $e=E_o\sin \omega t$
The induced emf $e$ varies sinusoidally with time $t$ and the frequency being $v$ cycles per second $(v=\frac{\omega }{2\pi })$.
(i) When $\omega t=\theta$, the plane of the coil is perpendicular to the field $B$ and hence $e=0$.
(ii) When $\omega t=\pi /2$, the plane of the coil is parallel to $B$ and hence $e=E_o$.
(iii) When $\omega t=\pi$, the plane of the coil is at right angle to $B$ and hence $e=0$.
(iv) When $\omega t=3\pi /2$, the plane of the coil is again parallle to $B$ and the induced emf is $e=-E_o$.
(v) When $\omega t=2\pi$, the plane of the coil is again perpendicular to $B$ and hence $e=0$.
If the ends of the coil are connected to an external circuit through a resistance $R$, current flows through the circuit, which is also sinusoidal in nature.