A uniform wire of mass m and length l slides freely along parallel rails on application of force F which starts acting at t = 0. Resistance of circuit is zero. System is placed in uniform magnetic field B, perpendicular to plane of figure. What is current through capacitor at any time t. (Ignore any inductance)

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$\frac{BICF}{m + B^{2} I^{2} C}$

$\frac{BICF}{m} t$

$\frac{{BICFt}^{2}}{m}$

$\frac{{BICFt}^{2}}{m + B^{2} I^{2} C}$

answer is A.

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Detailed Solution

$\frac{q}{C} = vBl \Rightarrow \Rightarrow \Rightarrow (1) n e t f o r c e = F - i l B = m a \Rightarrow \Rightarrow (2) f r o m (1) \frac{d q}{d t} \frac{1}{C} = \frac{d v}{d t} l B i \frac{1}{C} = a B l a = \frac{i}{C B l} s u b s t i t u t i n g i n (2) F - i l B = m \frac{i}{C B l} F = i (\frac{m}{C B l} + l B) i = \frac{F}{\frac{m}{C B l} + l B} = \frac{F C B l}{m + C B^{2} l^{2}}$