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

**(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 elementdy, `"d"phi` = BdA

Where,

dA = Area of element dy = a dy

B = Magnetic field at distance y

= `(mu_0"I")/(2pi"y")`

I = Current in the wire

`mu_0` = Permeability of free space = 4π × 10^{−7} T m A^{−1}

∴ `"d"phi = (mu_0"Ia")/(2pi)("dy")/"y"`

`phi = (mu_0"Ia")/(2pi)int("dy")/"y"`

y tends from x to a + x

∴ `phi = (mu_0"Ia")/(2pi) int_"x"^("a" + "x")("dy")/"y"`

= `(mu_0"Ia")/(2pi) [log_"e""y"]_"x"^("a" + "x")`

= `(mu_0"Ia")/(2pi) log_"e"(("a" + "x")/"x")`

for mutual inductunce M, the flux is given as :

`phi = "MI"`

∴ 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")/(2pi"x"))"av"`

Given,

I = 50 A

x = 0.2 m

a = 0.1 m

v = 10 m/s

`"e" = (4pi xx 10^-7 xx 50 xx 0.1 xx 10)/(2pi xx 0.2)`

`"e" = 5 xx 10^-5` V