Question

A conducting circular loop is placed in a uniform magnetic field of induction B tesla with its plane normal to the field. Now, the radius of the loop starts shrinking at the rate $ \left( \frac{dr}{dt} \right) $ . Then, the induced emf at the instant when the radius is r, is

• A. $ \pi rB\left( \frac{dr}{dt} \right) $
• B. $ 2\pi rB\left( \frac{dr}{dt} \right) $
• C. $ \pi {{r}^{2}}\left( \frac{dr}{dt} \right) $
• D. $ {{\left( \frac{\pi {{r}^{2}}}{2} \right)}^{2}}B\left( \frac{dr}{dt} \right) $

Answer 1

Answer: (B)

Induced emf is given by $ e=-\frac{d\phi }{dt} $ If the radius of loop is r at a time t, then the instantaneous magnetic flux is given by
$ \phi =\pi {{r}^{2}}B $
$ \therefore $ $ e=-\frac{d}{dt}(\pi {{r}^{2}}B) $
$ e=-\pi B\left( \frac{2r\,\,dr}{dt} \right) $
$ e=-2\pi Br\frac{dr}{dt} $
Numerically, $ e=2\pi Br\left( \frac{dr}{dt} \right) $