An arrangement of smooth pair of long conducting horizontal rails which are resistance less are joined to a charged capacitor. The rails are connected by a rod of mass $m,$ length $l$ and resistance $R .$ Vertical magnetic field $B$ is uniform. The rod is initially at rest. It starts moving due to the influence of magnetic field. What should be the mass of the rod (in mg) so that its kinetic energy in steady state is maximum ?
$[Take : l = 1 m, B = 2 T, C = 1 μ F, R = 1 Ω]$

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answer is 4.

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Detailed Solution

$F = B i l m \frac{d v}{d t} = - B \frac{d q}{d t} l m \int_{0}^{v} d v = - B l \int_{Q}^{q} d q m v = B l (Q - q)______(1)$

n steady state, $\frac{q}{c} = B l v$

$C B l v = q____(2) l B Q - C l^{2} B^{2} v = m v v = \frac{B l Q}{m + C B^{2} l^{2}} k = \frac{1}{2} m v^{2} = \frac{1}{2} m {(\frac{B l Q}{m + C B^{2} l^{2}})}^{2} \frac{d k}{d m} = 0 \Rightarrow \frac{1}{2} {(\frac{B l Q}{m + C B^{2} l^{2}})}^{2} - \frac{2}{2} \frac{m {(B l Q)}^{2}}{{(m + C B^{2} l^{2})}^{3}} \times 1 = 0 \Rightarrow \frac{1}{2} - \frac{m}{m + C B^{2} l^{2}} = 0 C B^{2} l^{2} = m \Rightarrow m = 4 \times 10^{- 6} k g = 4 m g$