Question

An elastic conducting material is stretched into a circular loop of 12.4 cm radius. It is placed with its plane perpendicular to a uniform 0.979 T magnetic field. When released, the radius of the loop starts to shrink at an instantaneous rate of 84.8 cm/s. What emf is induced in volts in the loop at that instant?

Answer 1

Magnetic flux through an area A due to a magnetic field B is given by

$\phi=\overrightarrow{B}.\overrightarrow{A}=BA\cos\theta$

$\theta$ is the angle between the magnetic field and the area vector.

The plane of the loop is perpendicular to the magnetic field i.e., the magnetic field vector is parallel to the area vector

$\theta=0^{\circ}$

Hence, flux becomes

$\phi=BA=\pi r^2B$

r is the radius of the loop

Emf induced is given by the rate of change of flux i.e,

$e=\frac{\mathrm{d\phi} }{\mathrm{d} t}=\frac{d}{dt}\left ( \pi r^2B \right )=\pi B\frac{d}{dt}r^2=2\pi rB\frac{dr}{dt}$

dr/dt denotes the rate of change of radius of the loop

Putting the values,

$e=2\pi\times\frac{12.4}{100}\times0.979\times\frac{84.8}{100}\ V=0.6468\ V$

This is the emf induced.