A metal $\operatorname{rod} A B$ of length $50 \mathrm{~cm}$ is moving at a velocity $8 \mathrm{~ms}^{-1}$ in a magnetic field of $2 \mathrm{~T}$. 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

Question

Aicharya · Accepted Answer

A metal rod $A B$ shown in the figure moves with speed $8 \mathrm{~ms}^{-1}$.

As the emf induced in the metal rod, $ \begin{aligned} e_{A B} & =V_A-V_B=l \cdot(\mathbf{V} \times \mathbf{B}) \\ \text { 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 \mathrm{~V} \end{aligned} $ Hence, the correct option is (b)

A metal $\operatorname{rod} A B$ of length $50 \mathrm{~cm}$ is moving at a velocity $8 \mathrm{~ms}^{-1}$…

Solution