Question

A square loop of side $22 cm$ is converted into circular loop in 0.4 sec. A uniform magnetic field of $0.2 T$ is directed normal to the loop then the emf induced in the loop is:

• A. $-6.6 \times 10^{-3} V$
• B. $-6.6 \times 10^{-5} V$
• C. $6.6 \times 10^{-3} V$
• D. $6.6 \times 10^{-5} V$

Answer 1

Answer: (A)

$e =-\frac{ d \phi}{ dt }=-\frac{ BdA }{ dt }=-0.2 \frac{( d A )}{0.4}$
$e =-\frac{ dA }{2}$
$d A =\left[\pi r _{1}^{2}- r _{2}^{2}\right]$
For circle $r _{1}=\frac{44}{\pi} \times 10^{-2} m,$
For square $r _{2}=22 \times 10^{-2} m$
$d A =\left[\pi\left(\frac{44}{\pi}\right)^{2} \times 10^{-4}-\left(22 \times 10^{-2}\right)^{2}\right]$
$=484 \times 10^{-4}\left[\frac{4}{\pi}-1\right]$
$=484 \times 10^{-4}[1.27-1]$
$=130.68 \times 10^{-4}$
$\therefore e =-\frac{ d A }{2}=-\frac{130.68 \times 10^{-4}}{2}$
$=-65.34 \times 10^{-4} V .$