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A square loop ABCD of side l is moving t...

A square loop `ABCD` of side `l` is moving the `xy` plane with velocity `vecv=betat hatj`.There exists a non-uniform magnetic field `vecB=-B_(0)(1+alphay^(2)) hatk (y gt 0)`, where `B_(0)` and `alpha` are positive constants. Initially, the upper wire of the loop is at `y=0`.Find the induced voltage across the resistance `R` as a function of time.Neglect the magnetic force due to induced current.

Text Solution

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The correct Answer is:
`epsi=-B_(0)Ibeta(t+(alphaxxbeta^(2)t^(5))/(4))`
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