Question

A rectangular loop with a sliding connector of length $10cm$ is situated in uniform magnetic field perpendicular to plane of loop. The magnetic induction is $0.1$ tesla and resistance of connector $\left(R\right)$ is $1$ ohm. The sides $AB$ and $CD$ have resistances $2$ ohm and $3$ ohm respectively. Find the current in the connector during its motion with constant velocity $1m/s$.

• A. $\frac{1}{110}\text{A}$
• B. $\frac{1}{220}\text{A}$
• C. $\frac{1}{55}\text{A}$
• D. $\frac{1}{440}\text{A}$

Answer 1

The correct option is B $\frac{1}{220}\text{A}$



$vBl=1\times 0.1\times 0.1$
$={10}^{-2}\text{volt}$
$\frac{x}{2}+\frac{x}{3}+\frac{x-{10}^{-2}}{1}=0$
$\Rightarrow 3x+2x+6x-6\times {10}^{-2}=0$
$11x=6\times {10}^{-2}$
$x=\frac{6\times {10}^{-2}}{11}$
$i=\frac{\frac{6\times {10}^{-2}}{11}-{10}^{-2}}{1}$
$i=-\frac{5\times {10}^{-2}}{11}=-\frac{1}{220}\text{amp}$