Question

A metal rod $A B$ of length $50\, cm$ is moving at a velocity $8 \,ms ^{-1}$ in a magnetic field of $2 \,T$. 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 = 8 V $
• B. $V_A - V_B = 4 V $
• C. $V_B - V_A = 8 V $
• D. $V_B - V_A = 4 V $

Answer 1

Answer: (B)

A metal rod $AB$ shown in the figure moves with speed $8 \,ms^{-1}$

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