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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). |
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Answer» Solution :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 taccording 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) ` |
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