A thin copper rod AB of length 1m is rotating with constant angular velocity $ω$ = 5 rad/s. A magnetic field of induction B = 0.5 T with its direction perpendicular to the rod exists in that region as shown in the figure. Then induced potential difference between mid point C and end B of the rod is

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1.25 V

1.5 V

0.9375 V

0.7528 V

answer is A.

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Detailed Solution

$V_{C}$ = Speed of C = $5 x \frac{1}{2} m / s = \frac{5}{2} m / s$

$V_{B}$ = Speed of B = $5 x 1 m / s = 5 m / s$

Point B is at higher potential.

$∴$ Potential difference between B and C

$= B x (Length BC) x (\frac{V_{B} + V_{C}}{2})$

$= 0.5 x 0.5 \times (\frac{5 + 2.5}{2}) volt$

$= 0.9375 volt$

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