Question

A circuit $ABCD$ is held perpendicular to uniform magnetic field of $5\times 10^{-2}$ T extending over the region $PQRS$ and directed into the plane of the paper. The circuit is pulled out of the field at a uniform speed of $0.2m{s}^{-1}$ for $1.5s$. During this time, the current in the $5\Omega$ resistor is

• A. $0.6mA$ from $B$ to $C$
• B. $0.9mA$ from $B$ to $C$
• C. $0.9mA$ from $C$ to $B$
• D. $0.6mA$ from $C$ to $B$

Answer 1

Answer: (A)

$\varphi =Blx$
$|\epsilon |=\frac{d\varphi }{dt}=Bl\left(\frac{dx}{dt}\right)$

$\epsilon =\left(5\times 10^{-2}T\right)\cdot (0.3m)\cdot {\left(0.2\frac{m}{s}\right)}^2=3\times 10^{-3}V$
$i=\frac{\epsilon }{R}=\frac{3\times 10^{-3}V}{5\Omega }=0.6mA$
Applying, Lenz's law current will go clockwise or from $B$ to $C$ through the $5\Omega$ resistor.