Question

One conducting $U$ tube can slide inside another one, as shown in the figure, while maintaining the electrical contacts. The magnetic field $B$ is perpendicular to the plane of the figure and directed inwards. If each tube moves towards the other at a constant speed $v$, then the induced emf in the circuit is -

Take the width of each tube to be $l$.

• A. $2Blv$
• B. $Blv$
• C. $3Blv$
• D. Zero

Answer 1

The correct option is A $2Blv$

Since, the two $U-$ tubes are in motion, emf will be induced across each of them.

Emf induced across $U-$ tube $ad$,

$E_1=Bvl$

Similarly, across $U-$ tube $bc$,

$E_2=Bvl$

The equivalent circuit diagram is shown below.

The net induced emf in the circuit is,

$E_{net}=E_1+E_2=Bvl+Bvl=2Bvl$

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