Question

A conductor of mass$\frac{1}{4}$kg and length 2 m can move without friction along two metallic parallel tracks in a horizontal plane and connected across a capacitor C = 1000
$\mu$ F. The whole system is in a magnetic field of magnetic inductance B = 2 tesla directed outward to the plane. A constant force F = 1.33 N is applied to the middle of conductor perpendicular to it and parallel to the tracks. Find the acceleration of conductor neglecting all resistances. Assume that the conductor started from rest.

Answer 1

Suppose the velocity of wire is 'v' at time 't'
The induced emf will be 'vBl'
As there is no resistor so charge in capacitor will be 'CBlv'
Current in the wire will be derivative of charge i.e. $\frac{d}{dt}$ CBlv = CBla (where 'a' is the acceleration)
Now magnetic force = iBL = C$B^2$$l^2$a
If the applied force is 'F' then free body diagram of conductor gives the equation, F - C$B^2$$l^2$a = ma
Substituting all the values as given in the question we get,
a = 5 $\frac{m}{se{c}^2}$