InterviewSolution
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InterviewSolution
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A magnetic field in a certain region is given by vecB=B_0 cos (omegat) hatkand a coil of radius a with resistance R is placed in the xy-plane with its centre at the origin in the magnetic field (see figure). Find the magnitude and the direction of the current at (a, 0, 0) at t=pi/(2omega), pi/omega and t=(3pi)/(2omega). |
Answer» Solution : Magnetic flux linked with coil, `phi=vecB.vecA=BA cos theta = BA` `phi=B_0 (pia^2)cos omegat[ because A=pia^2]` Induced emf, `epsilon=(dphi)/(dt)=-d/(dt) (B_0 pia^2 cos omegat)` `=-B_0 pia^2 [-omega sin omegat]` `epsilon=B_0 pia^2 omega sin omegat` Induced current `I=epsilon/R=(B_0 pia^2 omega)/R sin omegat` At `t=pi/(2OMEGA)`time, `I=(B_0pia^2omega)/R sin(omegaxxpi/(2omega))` `I=(B_0 pi a^2 omega)/R` Now according to `vecB = B_0 cos (omegat) hatk`magnetic flux came out from coil decreases (cos is decreasing FUNCTION for t = 0 to `t =pi/omega` ). Hence, according to Lenz.s law induced current will try to increase that outward flux. For that induced current will flow in anticlockwise direction so at (a, 0, 0) current is along `+HATJ` direction. At `t=pi/omega` time `I=(B_0 pi a^2 omega)/R sin (omegaxx pi/omega)=0` At `t=(3pi)/(2omega)` time `I=(B_0 pia^2 omega)/R sin (omegaxx(3pi)/(2omega))` `=(B_0 pi a^2 omega)/R sin ((3pi)/2)` `I=-(B_0 pia^2 omega)/R` Negative SIGN appearing here indicates that induced current at (a, 0, 0) is in `-hatj` direction. [According to CASE (i)] |
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