(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig.
(b) Evaluate the induced emf in loop if the wire carries a current of 50 A and the loop has an instantaneous velocity `upsilon = 10 ms^(-1)` at the location x = 0.2 m, as shown. Take a = 0.1m and assume that the loop has a large resistance.

Explanation

(a) Let us consider a small strip of thickness dr at a distance r from the straight wire as shown in the figure given below, assuming that the magnetic field is uniform over this slit.
Area of the strip, dA = adr .
`:.` Magnetic field at a distance from the long straight current-carrying wire carrying current I is
`B =(mu_(0) I)/(2pir)`

So magnetic flux linked with this strip
`dphi=B xx dA`
or `dphi=(mu_(0) I)/(2pir) xx adr`
`:.` Total magnetic flux linked with the whole square loop
`phi= int dphi= int_(r=x)^(r=x+a) (mu_(0) Ia)/(2pir)dr`
`=(mu_(0) Ia)/(2pi)int_(r=x)^(r=x+a)(dr)/(r)`
`=(mu_(0) Ia)/(2pi)"In"[(x+a)/(x)]`
`=(mu_(0) Ia)/(2pi)"In"[1+(a)/(x)]`
`phi=MIimplies M=(mu_(0) a)/(2pi)"In"[1+(a)/(x)]`
(b) Given, current in the straight wire, 1 = 50 A,
Velocity of loop, `v = 10 m s^(-1)`,
Distance between the wire and loop, x = 0.2 m,
Side of the square loop, a = 0.1 m
The induced emf is
`epsilon=-(dphi)/(dt)`
`=-(d)/(dt)[(mu_(0)Ia)/(2pi){"In(x + a) - Inx"}]`
`=-(mu_(0)Ia)/(2pir)[(1)/(x+a)-(1)/(x)](dx)/(dt)`
or `epsilon=-(mu_(0)Ia)/(2pi)[(1)/(x+a)-(1)/(x)]v`
`=-(4pi xx 10^(-7) xx 50 xx 0.1)/(2pi) xx [(1)/(0.2+0.1)-(1)/(0.2)] xx 10`
`=-2 xx 10^(-7) xx 5 xx [-(1)/(0.6)] xx 10`
`=1.66 xx 10^(-5) V=1.7 xx 10^(-5) V`