Question

A wire of mass $m$ and length $l$ can freely slide on a pair of parallel, smooth, horizontal rails placed in a vertical uniform magnetic field $B$. The rails are connected by a capacitor of capacitance $C$. The electric resistance of the rails and the wire is zero. If a constant force $F$ acts on the wire as shown in the figure, find the acceleration of the wire.

• A. $\frac{4F}{2m+C{B}^2{l}^2}$
• B. $\frac{F}{m+C{B}^2{l}^2}$
• C. $\frac{F}{4(m+C{B}^2{l}^2)}$
• D. $\frac{Fl}{m+C{B}^2{l}^2}$

Answer 1

The correct option is B $\frac{F}{m+C{B}^2{l}^2}$

Suppose the velocity of the wire is $v$ at time $t$.

The induced emf, $E=Blv$.

As there is no resistance anywhere, the charge on the capacitor will be,

$q=CE=CBlv$

At time $t$, the current in the circuit will be,

$I=\frac{dq}{dt}=CBl\frac{dv}{dt}=CBla$

$a$ is the acceleration of the wire.

Magnetic force acting on the wire is,

$F^{\prime }=BIl=Bl\left(CBla\right)=C{B}^2{l}^{2}a$

The net force on the wire is $F-F^{\prime }$.

From Newton's second law,

$F-F^{\prime }=ma$

or, $F-C{B}^2{l}^{2}a=ma$

or, $a=\frac{F}{m+C{B}^2{l}^2}$

Hence, option $\left(B\right)$ is the correct answer.