Question

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

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

Answer 1

Answer: (A)

Area of a square loop,
$A=10cm\times 10cm$
$=100c{m}^2=10^{-2}{m}^2$
Initial magnetic flux linked with the loop
${\varphi }_1=B_{1}Acos\varphi$
$=0.1\times 10^{-2}\times cos45^{\circ }$
$=\frac{0.1\times 10^{-2}\times 1}{\sqrt{2}}=\frac{10^{-3}}{\sqrt{2}}Wb$
Final magnetic flux linked with loop
${\left(\varphi \right)}_2=0Wb\left(\because B_2=0\right)$
The induced emf in the loop is
$e=-\frac{d\varphi }{dt}=-\frac{\left({\left(\varphi \right)}_2-{\left(\varphi \right)}_1\right)}{t}$
$=-\frac{\left(0-\frac{{\left(10\right)}^{-3}}{\sqrt{2}}\right)}{0.7}=\frac{{\left(10\right)}^{-3}}{0.7\times \sqrt{2}}={\left(10\right)}^{-3}volt$
The induced current in the loop is
$I=\frac{e}{R}=\frac{10^{-3}V}{1\Omega }=10^{-3}A=1.0mA$