Question

A rectangular loop with a sliding connector of length $10\text{cm}$ 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 $1\text{m/s}$.

• 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 $\frac{1}{220}\text{A}$


Motional emf generated is
$vBl=1\times 0.1\times 0.1={10}^{-2}\text{volt}$
Let potential at point $X$ be $x\text{V}$ and potential at other end of connector R is $0\text{V}$
Now using Kirchoff's junction rule
$\Rightarrow \frac{x}{2}+\frac{x}{3}+\frac{x-{10}^{-2}}{1}=0$
$\Rightarrow 3x+2x+6x-6\times {10}^{-2}=0$
$\Rightarrow 11x=6\times {10}^{-2}$
$\Rightarrow x=\frac{6\times {10}^{-2}}{11}$
Current through connector $R$ is
$i=\frac{\frac{-6\times {10}^{-2}}{11}+{10}^{-2}}{1}$
$i=\frac{5\times {10}^{-2}}{11}=\frac{1}{220}\text{amp}$