A conducting rod PQ of length $l$ is dragged with a constant force F along two smooth parallel rails separated by a distance $l$ in uniform magnetic field ‘B’ perpendicular to the plane of the coil (into) as shown in the adjacent figure. Then choose the correct statement(s)

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terminal velocity of the rod, $v_{t} = \frac{2 F R}{B^{2} l^{2}}$

maximum charge on the capacitor, $q_{\max} = \frac{F C R}{B l}$

terminal velocity of the rod, $v_{t} = \frac{F R}{B^{2} l^{2}}$

maximum charge on the capacitor, $q_{\max} = \frac{2 F C R}{B l}$

answer is B, C.

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Detailed Solution

In steady state.
Question Image
$B l V_{t e r \min a l} = \frac{q_{\max}}{C} a l s o i B l = F & i = \frac{B l v_{t e r \min a l}}{R} \Rightarrow \frac{B l v_{T}}{R} . B l = F \Rightarrow v_{T} = \frac{F R}{B^{2} l^{2}} a n d q_{\max} = B l C (\frac{F R}{B^{2} l^{2}}) \Rightarrow i_{\max} = \frac{v_{i n}}{Z} \Rightarrow v_{o u t} = i \times R v_{o u t} = \frac{v_{i n}}{Z} \times R \frac{v_{i n}}{v_{o u t}} = \frac{Z}{R} = \frac{\sqrt{{(\frac{1}{ω C})}^{2} + R^{2}}}{R}$