Question

A rectangle coil ABCD is rotated anticlockwise with a uniform angular velocity about an axis as shown in the figure. The axis of rotation of the coil as well as the magnetic field B are horizontal. The induced emf in the coil would be minimum when

• A. the plane of coil is horizontal.
• B. the plane of coil is vertical.
• C. the plane of coil makes an angle of 45${}^o$ with the direction of the magnetic field.
• D. the plane of coil makes an angle of 30${}^o$ with the field.

Answer 1

Answer: (B)

the plane of coil is vertical.

Let at time $t=0$, the coil ABCD is vertical as shown in figure and at time $t$, plane of coil makes an angle $\theta$ with the vertical, then

$\theta =\omega t$, ($\omega =$ uniform angular velocity),

in this position , magnetic flux linked with coil will be,

$\varphi =BA\cos \theta$, (where, $A=$ area of coil,)

or $\varphi =BA\cos \omega t$,

now, differentiating this equation w.r.t. time, we get

$\frac{d\varphi }{dt}=\frac{d}{dt}\left(BA\cos \omega t\right)$,

or $\frac{d\varphi }{dt}=-BA\omega \sin \omega t$,

if $e$ is the emf induced in coil then by Faraday's law,

$e=-\frac{d\varphi }{dt}=BA\omega \sin \omega t$,

now if, $\sin \omega t=1$ (maximum),

then $e=BA\omega =e_0$ (maximum),

therefore $e=e_0\sin \omega t$

it is clear from equation that $e$ will be minimum when

$\omega t=0$,

or $\theta =0$, the plane of coil is vertical.