A wire of mass m and length l is placed on a smooth incline making an angle `theta` with the horizontal, whose front view is shown in Fig. When a finite amount of charge is passed through it in an infinitesimal time, the wire immediately acquires some velocity and then ascends the incline by a distance s. For this small duration, we can neglect the gravitational force because the current can be considered very large due to small time duration. The amount of charge passed through the wire is

Options

A. `(msqrt(2gssin theta))/(Bl)`
B. `(mv)/(Bl)`
C. `(msqrt(2gssin theta))/(Bl cos theta)`
D. information insufficient

Answer

(a,b)

Explanation

When a charge is passed through a wire, a current flows in the wire (for a very small time). As a result of this, the wire experiences a magnetic force for a very small duration, i.e., during the time for which charge has been passed through the wire, due to which it acquires some velocity, then onwards, it moves under gravity. For the wire to move up the incline, the positive charge has to pass through the wire in a direction coinciding with into the plane of paper.
`F_m=IBl=(dq)/(dt)xxBl=ma`
[For this small duration dt, we can neglect gravitational force because I would be very large due to small passage time of charge.]
`implies m(dv)/(dt)=(dq)/(dt)xxBl implies mdv=dqxxBl`
`implies mv=qxxBl`
`implies v=(qxxBl)/m implies q=(mv)/(Bl)`
where q is the charge passed through the wire and v is the velocity aquired by the wire just after charge had been passed through it.
From kinamatics, `v^2=2g sin theta s`
`q=(msqrt(2gssin theta))/(Bl)`