devarshi-dt-logo

Question:

A wire having mass m and length l can freely slide on a pair of parallel smooth horizontal rails placed in a 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=Fm+CB2l2

a=C2B2l−Fm

a=Fm+CBl

a=FC2B2lm

Solution:

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=dq/dt=CBl(dv/dt)=CBla
This same current passes through the wire.
Hence the magnetic force on it=F′=Bil=CB2l2a
Hence, ma=F−F′ ⇒ma=F−CB2l2a ⇒a=F/(m+CB2l2)