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In the figure shown, a uniform magnetic field $| B | = 0.5 T$ is perpendicular to the plane of circuit. The sliding rod of length $l = 0.25 m$ moves uniformly with constant speed $v = 4 m / s$ . If the resistance of the sliding rod is $2 Ω$ , then the current flowing through the sliding rod is

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0.08 A

0.1 A

0.17 A

0.03 A

answer is A.

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Detailed Solution

$e = B v l = 0.5 \times 4 \times 0.25 = 0.5 V$

$12 Ω$ and $4 Ω$ are parallel. Hence, their net resistance $R = 3 Ω$ .

$i = \frac{e}{R + r} = \frac{0.5}{3 + 2} = 0.1 A$ .

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