A U-shaped conducting frame is placed in a magnetic field B in such a way that the plane of the frame is perpendicular to the field lines. A conducting rod is supported on the parallel arms of the frame, perpendicular to them and is given a velocity `v_0` at time t = 0. Prove that the velocity of the rod at time t will be given by `v_t=v_0 "exp" ((-B^2l^2)/(mR)t)`.

Explanation

emf induced in a rod at time t is ,
`epsilon=-Bv_t l`
`therefore IR=-Bv_t l`
`therefore I=- (Bv_t l)/R to` current at time t
The force acting on the rod at time t according to Lenz.s law is ,
`F=BI l sin 90^@`
`therefore ` F=BI l [ `because sin 90^@ =1]`
`therefore F=-(B^2l^2)/R. v_t`...(1)
`therefore m.(dv_t)/(dt) =(B^2l^2)/R v_t [because F=ma=m . (dv_r)/(dt)]`
`therefore (dv_t)/(v_t)=-(B^2l^2)/(mR). dt`
Integrating on both sides
`int_(v_o)^(v_t) 1/v_t . dv_t = -(B^2l^2)/(mR) int_(t=0)^t dt`
`therefore [ln v_t]_0^t =-(B^2l^2)/(mR) [t]_0^t` ...(2)
`therefore ln v_t - ln v_0 = - (B^2l^2)/(mR)t`
`therefore ln ((vt)/v_0)=-(B^2l^2)/(mR)t`
`therefore v_t=v_0 e^(-B^2l^2)/(mR)t` OR `v_t=v_0 "exp" (-(B^2l^2)/(mR) t) `