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A metallic coil of N turns of radius resistance R, and inductance L is held fixed with its axis along a spatial uniform magnetic field bar(B) whose magnitude is given by B_(0) sin(omegat). Write the equation for the current i in the coil. (b) Assuming that in the steady state oscillates with the same frequency as the magnetic field, obtain the expression for i. (c )Obtain the force per unit length. Further obtain its oscillatory part and the time-averaged compressional part. (d) Calculate the time-averaged compressional force per unit length given that B_(0)=1.00"tesla", N=10 a=10cm, omega=1000.0 rad s^(-1), R=10.0Omega, L-100,0mH. |
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Answer» Solution :`(a)iR+L(DI)/(DT)=-Npia^(2)B_(0)omegacos omegat` `(B)i=(Npi a^(2)B_(0)omega (R cos omegat+omegaL sin omegat))/(R^(2)+omega^(2)L^(2))` `(c)(dF)/(dt)=-(NB_(0)pia^(2)omega)^(2)/(R^(2)omega^(2)L^(2))(R sin omegat cos omegat+omegaL sin^(2)omegat)` `(dF)/(DL)|_(osc)=-(NB_(0)^(2)pia^2omega^(2)L)/(2(R^(2)+omega^(2)L^(2)))` `(dF)/(dl)|_(osc)=-(NB_(0)^(2)pia^2omega^(2))/(2(R^(2)+omega^(2)L^(2)))=(R sin 2 omegat-omegaL cos 2 omegat)` `(dF)/(dl)|_(osc)=1.55N.m^(-1)` |
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