A certain elastic conducting material is stretched into a circular loop. It is placed with its plane perpendicular to a uniform magnetic field $$B = 0.8$$ T. When released the radius of the loop starts shrinking at a constant rate of $$2$$ cm s$$^{-1}$$. The induced emf in the loop at an instant when the radius of the loop is $$10$$ cm will be ______ mV.

Solution :

Magnetic flux through the circular loop is :

$$\phi = BA$$

$$= B\pi r^2$$

Induced emf :

$$\mathcal{E} = \left|\frac{d\phi}{dt}\right|$$

$$= \left|\frac{d}{dt}(B\pi r^2)\right|$$

Since \(B\) is constant,

$$\mathcal{E} = 2\pi Br\left|\frac{dr}{dt}\right|$$

Given :

$$B = 0.8\text{ T}$$

$$r = 10\text{ cm} = 0.1\text{ m}$$

$$\left|\frac{dr}{dt}\right| = 2\text{ cm s}^{-1} = 0.02\text{ m s}^{-1}$$

Substituting values :

$$\mathcal{E} = 2\pi \times 0.8 \times 0.1 \times 0.02$$

$$= 0.0032\pi\text{ V}$$

$$\approx 0.01\text{ V}$$

$$= 10\text{ mV}$$

Final Answer :

$$10$$