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A wire of mass m and length l is placed on a smooth incline making an angle theta with the horizontal, whose fornt 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 |
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Answer» `(msqrt(2gssin theta))/(Bl)` `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)` 
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