A conducting rod is bent as a parabola `y=Kx^(2)`, where `K` is a constant and it is placed in a unifrom magnetic field of induction `B`. At `t=0` a conductor of resistance per unit length `lambda` starts sliding up on the parabola with a constant acceleration `a` and the parabolic frame starts rotating with constant angular frequency `omega` about the axis of symmetry, as shown in the figure. Find the instantaneous current induced in the rod, when the frame turns through `pi//4rad`.

Explanation

When the frame has turned through an angle `theta`,
`Phi=BA costheta`
where `A=int_(0)^(y)2xdy=(2)/(sqrt(k))int_(0)^(y)sqrt(y)dy=(4)/(3sqrt(k))y^(3//2)`
Since `y=(1)/(2)at^(2)`
`:. Phi=(B)/(3)sqrt((2)/(K))a^(3//2)t^(3)costheta`
By Faraday's law, `E_("ind")=-(dPhi)/(dt)`
or `E_("ind")=(Ba^(3//2))/(3)sqrt((2)/(K))[t^(3)sintheta((d theta)/(dt))-3t^(2)costheta]`
When the frame thus through `pi//4`,
`t=(theta)/(omega)=(pi)/(4omega)`
`E_("ind")=(Ba^(3//2))/(3)sqrt((2)/(K))t^(2)[omegat sintheta-3costheta]`
or `E_("ind")=(pi^(2)Ba^(3//2))/(48)sqrt((2)/(K))[(pi)/(4)-3](1)/(sqrt(2))`
`E_("ind")=(pi^(2)(pi-12)Ba^(3//2))/(192omega^(2)sqrt(k))`
`I=(E_("ind"))/(R )=(pi^(2)(pi-12)Ba^(3//2))/(192omega^(2)Rsqrt(k))`
`R=lambda2x=lambda.2sqrt((y)/(k))=(2lambda)/(sqrt(k))sqrt((a)/(2))t=sqrt((2a)/(k))(lambda pi)/(4omega)`
`I=(E_("ind"))/(R )=(pi^(2)(pi-12)Ba^(3//2)sqrt(k)4omega)/(192omega^(2)sqrt(k)sqrt(2)pilambda)=(Bapi(pi-12))/(48sqrt(2)omegalambda)`