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.

Answer

A, B, D

Explanation

Flux inside the strip =`B.l.dr` ,`=B_(0)(1+ar^(2)).Idr`
Total flux in the loop =`underset(0)overset(y)intB_(0)(1+ar^(2))Idt phi=B_(0)L(y+(ay^(3))/3)`
Induced `emf epsilon=(-dphi)/(dt)=-B_(0)l(1+(3ay^(3))/3)(dy)/(dt)=-B_(0)l(1+ay^(2))(dy)/(dt)`
Given `v=beta t`
`(dy)/(dt)=betat` `underset(0)overset(y)intdy-underset(0)overset(y)int betatdt rArry=l^(2)`
`therefore epsilon=-B_(0)l(1+(alphaxxbeta^(2)t^(5))/4)betat rArrepsilon=-B_(0)ibeta(t-(alphaxxbeta^(2)t^(5))/4)`