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A wire of length 0.1 m moves with a speed of $10 m/s$ perpendicular to a magnetic field of induction ${1 wb/m}^{2}$ then the value of induced emf is

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1 v

2 v

3 v

answer is B.

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Detailed Solution

$ε = BLV [Motional E.M.F] B = 1 {we/m}^{2}, L = 0.1 m , V = 10 m/sec ∴ ε = 1 \times 0.1 \times 10 = 1 volt$

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