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A magnetic field in a certain region is ...

A magnetic field in a certain region is given by `vecB=B_0 cos (omegat) hatk` and 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)`.

Text Solution

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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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