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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
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Verified Answer
The correct answer is:
$V_A-V_B=4 \mathrm{~V}$
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).
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).
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