Question

A square loop of wire, side length $ 10\,cm $ is placed at angle of $ 45^{\circ} $ with a magnetic field that changes uniformly from $ 0.1 \,T $ to zero in $ 0.7\, s $ . The induced current in the loopa (its resistance is $ 1\,\Omega $ ) is

• A. $ 1.0 \,mA $
• B. $ 2.5 \,mA $
• C. $ 3.5 \,mA $
• D. $ 4.0 \,mA $

Answer 1

Answer: (A)

Initial magnetic flux linked with the loop
$\phi_{1}=B_{1}A_{1}\,cos\, \phi $
$=0.1\times\left(10\times10^{-2}\right)^{2} cos\,45^{\circ}$
$=\frac{0.1\times\left(10^{-2}\times1\right)}{\sqrt{2}}=\frac{10^{-3}}{\sqrt{2}}$
Final magnetic flux linked with the loop $\phi_{2}=0$
Now, induced emf in the loop $e=\frac{-d\phi}{dt}$
$=\frac{-\left[\frac{10^{-3}}{\sqrt{2}}\right]}{0.7}=10^{-3}V$
$\therefore $ Induced current $=\frac{e}{R}=\frac{10^{-3}}{1}=1\,mA$