InterviewSolution
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InterviewSolution
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A line charge lambda per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Fig. 6.08). A uniform magnetic field extends over a circular region within the rim. It is given by, vecB = - B_(0)hatk (r le a, aItR) = 0 (otherwise) What is the angular velocity of the wheel after the field is suddenly switched off ? |
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Answer» Solution :If the MAGNETIC field changes with TIME, then, Induced emf `varepsilon=-(dphi)/dt` This implies the presence of an electric field `vceE` tangential to all points of the periphery of the circular region denoted by a. If we move a test charge `qvarepsilon` round this periphery a work qe will be done once round it. The electric force on the test charge is q E and the work done by this force round this periphery is `(qE)(2pia)`. EQUATING the two values of work done, we have `qvarepsilon = (qE)(2pia) implies E = varepsilon/(2pia)` Substituting the value of `varepsilon`, we get `E = - 1/(2pia)(dphi_(B))/dt=-a/2 (dB)/dt [therefore phi_(B) = BA = pia^(2)B]` In the given problem, the total charge on the rim` q = lambda .2pia` Therefore, force on this charge `F = q E = -(lambda.2pia)(a/2(dB)/dt)`. According to Newton.s second law of motion, force F = M(acceleration) `= M (DV)/dt,` `-(lambda2pia)(a/2(dB)/dt)=M.d/dt(Romega)` `therefore -(lambda2pia^(2))/2.(dB)/dt=MR(domega)/dt impliesdomega=-(lambdapia^(2))/(MR)dB` On integrating, we get `omega=-(lambdapia^(2)B)/(MR) or `vec omega=-(Bpia^(2)lambda)/(MR)hatk` |
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