(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) Take a small element dy in the loop at a distance y from the long straight wire (as shown in the given figure).

Magnetic flux associated with element `dy, dphi= Bda`
Where
dA = Area of element dy = a dy
B= magnetic field at distance y
`=(mu_(0)I)/(2piy)`
I= current in the wire
`mu_(0)` =Permeability of free space `=4pixx10^(-7) T m A^(-1)`
`therefore dphi=(mu_(0)Ia)/(2pi) (dy)/(y)`
`phi=(mu_(0)Ia)/(2pi)int(dy)/(y)`
y tends from x to `a+x`
`therefore phi=(mu_(0)Ia)/(2pi)underset(x)overset(a+x)int(dy)/(y)`
`=(mu_(0)Ia)/(2pi)["log"_(e)y]_(x)^(a+x)`
`=(mu_(0)Ia)/(2pi)"log"_(e)((a+x)/(x))`
For mutyual inductance M, the flux is given as :
`phi=MI`
`therefore MI=(mu_(0)Ia)/(2pi)"log"_(e)((a)/(x)+1)`
`M=(mu_(0)a)/(2pi)"log"_(e)((a)/(x)+1)`
(b) Emf induced in the loop, `e=B'av=((mu_(0)I)/(2pix))av`
Given
I=50 A
x=0.2 m
a=0.1 m
v=10 m/s
`e=(4pixx10^(-7)xx50xx0.1xx10)/(2pixx0.2)`
`e=5xx10^(-5)V`