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Question

Two conducting wires are joined together and are moving in a magnetic field perpendicular to the plane of the wires, as shown in the figure. If $V_{P}$ and $V_{Q}$ are the potentials at points $P$ and $Q$ respectively, then -

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Solution

From the figure, we can see that, $V_{O} > V_{P}$ using the right-hand rule and,

$V_{O P} = \frac{B ω l^{2}}{2} = \frac{0.8 \times 20 \times 1^{2}}{2} = 8 V$

Similarly,

$V_{O Q} = \frac{B ω l^{2}}{2} = \frac{0.8 \times 20 \times (1^{2} + 2^{2})}{2} = 40 V$

Now,

$V_{P Q} = V_{P} - V_{Q} = V_{P} - V_{O} - V_{Q} + V_{O}$

$\Rightarrow V_{P Q} = - (V_{O} - V_{P}) + (V_{O} - V_{Q}) = V_{O Q} - V_{O P}$

$\Rightarrow V_{P Q} = 40 - 8 = 32 V > 0 \Rightarrow V_{P} > V_{Q}$

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