(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig. 6.21. (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.

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, dφ = BdA

Where, dA = Area of element dy = a dy

B = Magnetic field at distance y

=μ0I/2πy

I = Current in the wire

μ0= Permeability of free space = 4π x 10⁻⁷

Therefore ,dφ = (μ0Ia/2π) x (dy/y)

φ = μ0Ia/2π ⌠ dy/y                                              _{{ NOTE- “⌠” is sign of integration}}

v tends from x to a + .x.

Therefore ,φ = μ0Ia/2π _x⌠^a+x dy/y

=μ0Ia/2π  [log_e y]_x^a+x

=μ0Ia/2π  [log_e(a+x)/x]

For mutual inductance M. the flux is given as: φ = MI

Therefore ,MI = μ0Ia/2π  [log_e(a/x+1)]

M=μ0a/2π  [log_e(a/x+1)]

Ans (b).

Emf induced in the loop, e = B’av

= (μ0I/2π)x av

Given, I = 50 A

x = 0.2 m

a = 0.1 m

v= 10 m/s

e = (4π x 10⁻⁷ x 50 x 0.1 x 10 )/(2π x 0.2)

e = 5 x 10⁻⁵ V