Question

A rod $PQ$ of mass $m$ and resistance $r$ is moving on two fixed, resistanceless, smooth conducting rails, closed by resistances $R_1$ and $R_2$ as shown in the figure. Find the current in the rod at the instant when its velocity is $v$.

• A. $\frac{Blv}{r+\frac{R_1{R}_2}{R_1+R_2}}$
• B. $\frac{Blv}{2r+\frac{R_1{R}_2}{R_1+R_2}}$
• C. $\frac{Blv}{r+(R_1+R_2)}$
• D. $\frac{Blv}{r+2(R_1+R_2)}$

Answer 1

The correct option is A $\frac{Blv}{r+\frac{R_1{R}_2}{R_1+R_2}}$

Due to the movement of the rod, an emf is induced in the rod.

The emf is given by,

$\epsilon =Blv$

The equivalent circuit is shown below.


The equivalent external resistance of the above circuit is,

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\Rightarrow R_{eq}=\frac{R_1{R}_2}{R_1+R_2}$


The current $i$ is given by,

$i=\frac{Blv}{r+R_{eq}}=\frac{Blv}{r+\frac{R_1{R}_2}{R_1+R_2}}$

Hence, option $\left(A\right)$ is the correct answer.