Question

In figure, the wires $P_1{Q}_1$ and $P_2{Q}_2$ are made to slide on the rails with the same speed of $5cm{s}^{-1}$. In this region, a magnetic field of $1T$ exists. The electric current in the $2\Omega$ resistance when both the wires are moving towards it is

• A. $0.2mA$
• B. $0.1mA$
• C. $2mA$
• D. Zero

Answer 1

The correct option is B $0.1mA$

Potential difference developed across a rod of length $l$ moving with velocity $v$ through a perpendicular magnetic field $B$ is
$ϵ=Bvl$
$\therefore ϵ=1\times 5\times {10}^{-2}\times 4\times {10}^{-2}$
$=20\times {10}^{-4}=2\times {10}^{-3}V$

So the rods can be replaced with batteries of equivalent emf in the circuit diagram given below.


Applying KVL across the loop $P_1{R}_1{R}_2{Q}_1$,
$-i\times 2+2\times {10}^{-3}-2i\times 9=0$
$\Rightarrow 20i=2\times {10}^{-3}$
$\Rightarrow i={10}^{-4}A=0.1mA$