Question

A conducting rod $P Q$ of length $l=2\, m$ is moving at a speed of $2\, m s ^{-1}$ making an angle of $30^{\circ}$ with its length. A uniform magnetic field $B=2\, T$ exists in a direction perpendicular to the plane of motion. Then

• A. $V_{P}-V_{Q}=8\, V$
• B. $V_{P}-V_{Q}=4\, V$
• C. $V_{Q}-V_{P}=8\, V$
• D. $V_{Q}-V_{P}=4\, V$

Answer 1

Answer: (B)

Emf induced across the rod $A B$ is
$e=\vec{B} \cdot(\vec{\ell} \times \vec{v})=B \ell v \sin \theta$
$=2 \times 2 \times 2 \times \sin 30$
$e=4\, V$
Free electrons of the rod shift towards right due to force
$q(\vec{v} \times \vec{B})$
Thus, end $P$ is at higher potential or $V_{P}-V_{Q}=4 V$.