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As shown in the figure, a metal rod makes contact and complete the circuit. The circuit is perpendicular to the magnetic field with B = 0.15 T. If the resistance is $3 Ω$ force needed to move the rod as indicated with a constant speed of 2 m/s is

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$3.75 \times 10^{- 2} N$

$3.75 \times 10^{- 3} N$

$3.75 \times 10^{2} N$

$3.75 \times 10^{- 4} N$

answer is A.

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Detailed Solution

Induced current in the circuit, $i = \frac{Bvl}{R}$

Magnetic force acting on the wire,

$F_{m} = Bil = B (\frac{Bvl}{R}) l$

$\Rightarrow F_{m} = \frac{B^{2} {vl}^{2}}{R}$

External force needed to move the rod with constant velocity is

$(F_{m}) = \frac{B^{2} {vl}^{2}}{R} = \frac{(0.15)^{2} \times (2) \times (0.5)^{2}}{3}$

$= 3.75 \times 10^{- 3} N$

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