Question

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

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

Answer 1

Answer: (A)

Initial magnetic flux linked with the loop
${\varphi }_1=B_1{A}_{1}cos\varphi$
$=0.1\times {\left(10\times 10^{-2}\right)}^{2}cos45^{\circ }$
$=\frac{0.1\times \left(10^{-2}\times 1\right)}{\sqrt{2}}=\frac{10^{-3}}{\sqrt{2}}$
Final magnetic flux linked with the loop ${\varphi }_2=0$
Now, induced emf in the loop $e=\frac{-d\varphi }{dt}$
$=\frac{-\left[\frac{10^{-3}}{\sqrt{2}}\right]}{0.7}=10^{-3}V$
$\therefore$ Induced current $=\frac{e}{R}=\frac{10^{-3}}{1}=1mA$