Question

A conduting rod $PQ$ of length $l$ intially at rest is dragged with a constant force $F$ along two long smooth parallel rails separated by a distance $l$ as shown in the figure. Then choose the correct statement(s).

• A. Terminal velocity of the rod, $V_T=\frac{2FR}{B^2{l}^2}$
• B. Terminal velocity of the rod, $V_T=\frac{FR}{B^2{l}^2}$
• C. Maximum charge on the capacitor, $q_{max}=\frac{FCR}{Bl}$
• D. Maximum charge on the capacitor, $q_{max}=\frac{2FCR}{Bl}$

Answer 1

Answer: (B), (C)

The correct options are
B Terminal velocity of the rod, $V_T=\frac{FR}{B^2{l}^2}$
C Maximum charge on the capacitor, $q_{max}=\frac{FCR}{Bl}$


At any instant ${}^{\prime }{t}^{\prime }$, the motional $\text{emf}$ in rod = $BVl$

Charge on capacitor is $q=C\left(BlV\right)$
Current $I$ in the rod =$\frac{dq}{dt}+\frac{BVl}{R}$

FBD of rod,

When terminal velocity is achieved
$F=IBl$
$F=(\frac{dq}{dt}+\frac{BVl}{R})Bl$
$\Rightarrow F=[CBl\frac{dV}{dt}+\frac{BVl}{R}]Bl$
But as rod moves with terminal velocity,
$a=\frac{dV}{dt}=0\Rightarrow F=\frac{B^{2}V{l}^2}{R}$
$\Rightarrow V_t=\frac{FR}{B^2{l}^2}$ $....\left(b\right)$
Maximum charge on capacitor
$q_{max}=C\left[B{V}_{t}l\right]=CBl\times \left[\frac{FR}{B^2{l}^2}\right]=\frac{FCR}{Bl}....\left(c\right)$