A rectangular metallic loop of length l and width b is placed coplanerly with a long wire carrying a current I . The loop is moved perpendicular to the wire with a speed v in the plane containing the wire and the loop. Calculate the emf induced in the loop when the rear end of hte loop is at a distance a from the wire. Solve by unsing Faraday's law for the flux thrugh the loop and also by replacing different sagments with equivalent batteries.

Explanation

(a) Using Faraday's law`
`Cosider a unit length dx at a distance x `
`B=(mu_0i)/(2 pi x) Area of strip =bdx`
`d phi = (mu_0i)/(2 pi x) bdx`
` implies phi= int^(a+l)_(a) (mu_0 i)/(2 pi x) bdx`
`=(mu_0 i)/(2 pi)b int_(a)^(a+l)((dx)/(x))`
`=(mu_0 i)/(2 pi x) log ((a+l)/(a))`
`Emf= (d phi)/(dt)=d/(dt)[(mu_0 i)/(2 pi) log ((a+l)/(a))]`
`=(mu_0 i)/(2 pi)((a)/(a+l)) ((va-(a+l)v)/(a^2))`
`(where (da)/(dt)=v)`
`= (mu_0ib)/(2 pi) (a)/(a+l) (vl)/(a^2)`
=(mu_0 ibvl)/(2 pi (a+l)a)`
the velocities of AB and CD create e.m/f because the emf 's due to to AD and BC are equal and opposite to each other.
`:. B_(AB)=(mu_0i)/(2 pi a)`
Emf_(AB)=(mu_0i)/(2 pi a)`
`length =b, velocity =v `
`B_(CD)=(mu_0i)/(2 pi(a+l))`
`Emf'_(CD)=(mu_0ibv)/(2 pi(a+l))`
`Net E.m.f.=(mu_0ibv)/(2 pi a))-(mu_0ibv)/(2 pi(a+l))`
`=(mu_0ibvl (a+l)- mu ib va)/(2 pi a(a+l))`
`(mu_0ibvl)/(2 pi(a+l))`.