A long metal rod XY having length $1 m$ completes the circuit as shown in figure. The plane of the circuit is perpendicular to the magnetic field of flux density 0.15 T. If the resistance of the circuit is 5 Ω, the force needed to move the rod in direction, as indicated, with a constant speed of 4 m s-1 will be _______

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$25 \times 10^{- 3} N$

$18 \times 10^{- 3} N$

$12 \times 10^{- 3} N$

$15 \times 10^{- 3} N$

answer is D.

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Detailed Solution

the force on the rod is given by
$F = i l B \Rightarrow F = \frac{B v l}{R} \times l B \Rightarrow F = \frac{B^{2} l^{2} v}{R}$ …...(i)
The data given is $R = 5 Ω$
$l = 1 m$
$v = 4 m s^{- 1}$
B = 0.15 T
Substituting the values in equation (i)
$F = \frac{{(0.15)}^{2} \times {(1)}^{2} \times 4}{5} = 18 \times 10^{- 3} N$

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