# The circuit shown in adjocent figure lied in a uniform magnetic field B coming out of the plane.Initially a capacitor C is uncharfed and the switch is open. A conducting slider of mass m and the lenght l can move freely over parallel tracks. Find the velocity of the slider as soon as the Switch S is closed.

Eshan
5 years ago
Dear student,

As soon as switch is closed, current will start flowing through the circuit and hence through the slider. Therefore, a force will start to act on the rod due to presence of magnetic field.

However at the instant the switch is closed, the force does not have enough time to change the velocity of slider, and so it remains at rest.
Amaan Ullah Siddiqui
24 Points
3 years ago
let the force on the slider due to the current in the circuit be F, therefore
F=BiL.............(1)
and also F=ma..........(2) (’a’ is acceleration)
from (1) and (2)
a=(BiL)/m
now
v=u+at
here u i.e., initial velocity is zero
and t=t
also i=q/t
now
v=at
v=(BiL)/m × t
v=(BqL)/m
q=CE
v=BLCE/M