Question

A wire having mass m and length I can freely slide on a pair of parallel smooth horizontal rails placed in vertical 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. Then, the acceleration of the wire can be given as

• A. $a=\frac{C^2{B}^{2}l-F}{m}$
• B. $a=\frac{F}{m+CBl}$
• C. $a=\frac{F{C}^2{B}^{2}l}{m}$
• D. $a=\frac{F}{m+C{B}^2{l}^2}$

Answer 1

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

Due to movement of rail with instantaneous velocity $v$, the emf produced across the capacitor $=V=Bvl$

Charge stored in the capacitor $=q=CV$ $=CBvl$

Therefore the current through the capacitor $=i=\frac{dq}{dt}$ $=CBl\frac{dv}{dt}$ $=CBla$

This same current passes through the wire.

Hence the magnetic force on it=$F^{\prime }=Bil=C{B}^2{l}^{2}a$

Hence, $ma=F-F^{\prime }$
$⟹ma=F-C{B}^2{l}^{2}a$

$⟹a=\frac{F}{m+C{B}^2{l}^2}$