A thin copper rod ABC is bent at point B as shown in figure. The bent rod is rotating with angular velocity $ω = 10 r a d / s$ about an axis passing through A and perpendicular to the plane of the bent rod. There is a uniform magnetic field of induction B = 0.5 T perpendicular to the plane of the page then induced potential difference between end C and point B is

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20 volt

60 volt

125 volt

40 volt

answer is A.

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Detailed Solution

$l_{e} =$ Distance between A and C $= \sqrt{3^{2} + 4^{2}} m = 5 m$

$∴ V_{C} - V_{A} = \frac{1}{2} B ω l_{e}^{2} a n d V_{B} - V_{A} = \frac{1}{2} B ω {(A B)}^{2}$

Now $(V_{B} - V_{A}) + (V_{C} - V_{B}) = (V_{C} - V_{A})$

$\Rightarrow (V_{C} - V_{B}) = \frac{1}{2} B ω [l_{e}^{2} - {(A B)}^{2}] = \frac{1}{2} \times 0.5 \times 10 \times (5^{2} - 3^{2}) V$

$\Rightarrow V_{C} - V_{B} = 40 volt$

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