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A conducting wire XY of mass m and neglibile resistance slides smoothly on two parallel conducting wires as shown in figure. The closed circuit has a resistance R due to AC. AB and CD are perfect conductors. There is a magnetic field vecB=B(t)hatk (i) Write down equation for the acceleration of the wire XY. (ii) If vecB is independent of time, obtain v(t), assuming v (0) =u_0. (iii) For (ii), show that the decrease in kinetic energy of XY equals the heat lost in. |
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Answer» Solution :Suppose two parallel wire are at y = 0 and y = l Suppose at t = 0 time wire xy is at x = 0 and at t = t time it is at x = x place. (i) Magnetic flux linked with closed loop, `phi=vecB.vecA=BA cos 0=BA` =[B(t)](lx) Induced emf, `EPSILON=-(dphi)/(dt)` `epsilon=-d/(dt)[B(t)lx]` `=-l[B(t)(dx)/(dt)+x d/(dt)B(t)]` `epsilon=-lB(t)v-lx(DB(t))/(dt)`....(1) (Motional emf) (emf DUE to variation in magnetic field) Magnetic force on wire , F=IlB(t) `=epsilon/R lB(t) [ because I=epsilon/R]` `=[-lB(t)v-lx(dB(t))/(dt)]l/R B(t)` `ma=-l^2/R B(t) [vB(t)+x(dB(t))/(dt)]` Acceleration of wire, `a=-(l^2B(t))/(mR)[vB(t)+x(dB(t))/(dt)]` ...(2) (II) If magnetic field does not change with time `(dB(t))/(dt)=0` `therefore a=(-l^2B^2)/(mR)v` `therefore (dv)/(dt)=(-l^2B^2)/(mR)v` `therefore 1/v dv=(-B^2l^2)/(mR)dt` Taking integration and at t=0, v=`u_0` `int_(u_0)^v 1/v dv = int_0^t (-B^2l^2)/(mR)dt` `[ln v]_(u_0)^v =(-B^2l^2)/(mR)[t]_0^t` `ln v- m u_0=(-B^2l^2t)/(mR)` `ln v/u_0=e^(-(B^2l^2)/(mR)t)` `v=u_0 e^(-(B^2l^2)/(mR)t)` `v(t)=u_0 exp (-(B^2l^2t)/(mR))`...(3) (iii) Change in kinetic energy of wire, `DeltaK=1/2m u_0^2 -1/2 mv^2` `=1/2m u_0^2-1/2 m u_0^2 e^(-(2B^2l^2t)/(mR))` `DeltaK=1/2 m u_0^2[1-e^(-(2B^2l^2t)/(mR))]`...(4) Power dissipated in the terms of heat, `P=I^2R` `(dH)/(dt)=I^2R = epsilon^2/R` `dH=epsilon^2/R dt` `H=int_0^t epsilon^2/R dt`...(5) If B is constant then from equation (1), `epsilon=-B vl` `H=int_0^t (B^2v^2l^2)/R dt` `H=(B^2l^2)/R int_0^t v^2dt` `=(B^2l^2)/R int_0^t u_0^2 e^(-(2B^2l^2t)/(mR))dt` `=(B^2l^2)/R u_0^2([e^(-(2B^2l^2t)/(mR))]_0^t)/(((-2B^2l^2)/(mR)))` `=-m/2u_0^2[e^(-(2B^2l^2t)/(mR)-1)]` `H=1/2 m u_0^2 [1-e^(-(2B^2l^2t)/(mR))]`...(6) From equation (5) and (6) it is clear that change in kinetic energy is equal to produced heat energy. |
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