Question

A conducting rod $\text{PQ}$ of length $l=2\text{m}$ is moving at a velocity $v=8{\text{ms}}^{-1}$ making an angle ${30}^{\circ }$ with its length. A uniform magnetic field $B=3\text{T}$ exits in a direction perpendicular to the plane of motion. Then

• A. $V_P-V_Q=41.6\text{V}$
• B. $V_P-V_Q=24\text{V}$
• C. $V_Q-V_P=41.6\text{V}$
• D. $V_Q-V_P=24\text{V}$

Answer 1

The correct option is B $V_P-V_Q=24\text{V}$

The induced emf in the rod is,

$E=B(l\times V)=3\times 2\times 8\sin 30$
$E=24\text{V}$

From Fleming's Left-hand Rule, positive charges experiences force towards end $\text{P}$ and the negative charges experiences force towards end $\text{Q}.$ Thus, $\text{P}$ is at a higher potential.

$\therefore V_P-V_Q=24\text{V}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.