Question

A metal rod $AB$ of length $50cm$ is moving at a velocity $8m{s}^{-1}$ in a magnetic field of $2T$. If the field is at $60^{\circ }$ with the plane of motion as shown in the figure, then the potentials $V_A$ and $V_B$ are related by

• A. $V_A-V_B=8V$
• B. $V_A-V_B=4V$
• C. $V_B-V_A=8V$
• D. $V_B-V_A=4V$

Answer 1

Answer: (B)

A metal rod $AB$ shown in the figure moves with speed $8m{s}^{-1}$

As the emf induced in the metal rod,
$e_{AB}=V_A-V_B=l\cdot (v\times B)$
But, $V_A-V_B=lvB\sin \left(90^{\circ }-60^{\circ }\right)$
$V_A-V_B=50\times 10^{-2}\times 8\times 2\times \sin 30^{\circ }$
$=8\times \frac{1}{2}=4V$