In fig, a square loop consistaing of an inductor of inductance L and resistor of resistance R is placed between two long parallel wires. The two long straight wires have time - varing current of magnitude `I=(I_0) cos omega t` but the direction of current in them are opposite

Options

A. `(2 mu_(0)a 1n 2 I_(0)omega)/(pisqrt(R^(2(+omega^(2)L^(2))) sin (omega t - phi)`
B. `(2 mu_(0)a 1n 2 I_(0)omega)/(pisqrt(R^(2(+omega^(2)L^(2))) sin (omega t + phi)`
C. `(2 mu_(0)a 1n 2 I_(0)omega)/(pisqrt(R^(2(+omega^(2)L^(2))) sin (omega t)`
D. `( mu_(0)a 1n 2 I_(0)omega)/(pisqrt(R^(2)+omega^(2)L^(2))) sin (omega t - phi)`

Answer

D

Explanation

`d phi = BdA`
`d phi = [(mu_(0)I)/(2 pi x)+(mu_(0)I)/(2 pi (3a-x))]adx`

`phi=(mu_(0)I)/(2 pi)[int_(a)^(2a) (dx)/(x)+int_(a)^(2a)(dx)/((3a-x))]a, phi=(mu_(0)Ia)/(pi) 1n 2`.
Meagnitude of emf in this circuit:
`epsilon = |(d phi)/(dt)| = (mu_(0)a(1n 2))/(pi)|(dI)/(dt)|`
`(mu_(0)a 1n 2 I_(0)omega)/(pi) sin omega t` .
ac current , `I=(mu_(0)a1n2I_(0)omega)/(pisqrt(R^(2)+omega^(2)L^(2)))sin(omegat-phi)`.