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A square frame made of thin copper wire is made to move in a uniform magnetic field of induction 1.5 Tesla with a velocity of 10 m/s as shown in the figure. If length of side of the square is 1m, potential difference between

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A and C is $15 \sqrt{2}$ volt

A and C is $8 \sqrt{2}$ volt

B and D is $5 \sqrt{2}$ volt

B and D is zero

answer is A.

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Detailed Solution

Length of diagonal $= \sqrt{1^{2} + 1^{2}} m = \sqrt{2} m$ .

The diagonal BD is parallel to the direction of motion,

So $V_{B} - V_{D} = 0$ .

The diagonal AC is perpendicular to the direction of motion, So $V_{A} - V_{C} = B l_{e} V = 1.5 \times \sqrt{2} \times 10$ volt

$\Rightarrow V_{A} - V_{C} = 15 \sqrt{2}$ volt.

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