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,
$A=10cm\times 10cm$
$=100cm^{2}=10^{- 2}m^{2}$
Initial magnetic flux linked with the loop
$\phi_{1}=B_{1}Acos\phi$
$=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
$\left(\phi\right)_{2}=0Wb\left(\because B_{2} = 0\right)$
The induced emf in the loop is
$e=-\frac{d \phi}{dt}=-\frac{\left(\left(\phi\right)_{2} - \left(\phi\right)_{1}\right)}{t}$
$=-\frac{\left(0 - \frac{\left(10\right)^{- 3}}{\sqrt{2}}\right)}{0 . 7}=\frac{\left(10\right)^{- 3}}{0 . 7 \times \sqrt{2}}=\left(10\right)^{- 3}volt$
The induced current in the loop is
$I=\frac{e}{R}=\frac{10^{- 3} V}{1 \Omega}=10^{- 3}A=1.0mA$