(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in figure.
(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m/s. Calculate the induced emf in the loop at the instant when x = 0.2 m. Take a = 0.1 m and assume that the loop has a large resistance.

Solution :
Consider area element in the form of rectangular strip with area dA = a dr at distance r from current carrying extremely long straight wire. Magnetic field through this element is perpendicularly INSIDE the plane of figure. Since the width of above strip is extremely small, magnetic field `B=(mu_0I)/(2pir)` can be taken almost constant. 2nr Hence, magnetic flux linked with this strip is,
`d phi=B dA`
`therefore dphi=((mu_0I)/(2pir))(a dr)`
`int_0^Phi dphi=(mu_0Ia)/(2PI) int_x^(x+a) 1/r dr`
`therefore {phi}_0^Phi=(mu_0Ia)/(2pi){ln r}_x^(x+a)`
`therefore {Phi-0}=(mu_0Ia)/(2pi){ln (x+a)-ln(x)}`
`therefore Phi=(mu_0Ia)/(2pi)ln((x+a)/x)`
`therefore Phi=(mu_0Ia)/(2pi)ln(1+a/x)`...(1)
Mutual inductance of given system will be ,
`M=Phi/I=(mu_0a)/(2pi) ln (1+a/x)`
(b) Now, when above loop moves with constant velocity V towards right, magnetic flux linked with it goes on decreasing with time.
When loop is at distance x from the wire at time t, emf induced in it will be ,
`epsilon=-(dPhi)/(dt)`
`therefore epsilon=-((dPhi)/(dx))((dx)/(dt))`
`therefore epsilon=-v (dPhi)/(dx)` ( `because (dx)/(dt)=v=` velocity of loop )
`=-vd/(dx){(mu_0Ia)/(2pi) ln (1+a/x)}`
`=-v(mu_0Ia)/(2pi) 1/(1+a/x)d/(dx)(1+a/x)`
`=-v (mu_0Ia)/(2pi) (x/(x+a)){0+a d/(dx)(1/x)}`
`=-v(mu_0Ia)/(2pi) (x/(x+a)){a(-1/x^2)}`
`therefore epsilon=(mu_0Ia^2v)/(2PIX(x+a))`
`therefore epsilon=((4pixx10^(-7))(50)(0.1)^2(10))/((2pi)(0.2)(0.2+0.1))`
`therefore epsilon=1.667xx10^(-5)` V