Question

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

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

Answer 1

Answer: (A)

$ \phi=B l x$
$| \varepsilon |=\frac{d \phi}{d t}=B l\left(\frac{d x}{d t}\right)$

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