Figure 3.188 shows two parallel and coaxial loops. The smaller loop (radius r) is above the larger loop (radius R), by distance `x gtgt R`. The magnetic field due to current `i` in the larger loop is nearly constant throughout the smaller loop. Suppose that `x` is increasing at a constant rate of `dx//dt = v`.

The induced emf in the smaller loop is

Options

A. (a) `(mu_(0)piiR^(2)r^(2))/(x^(4))v`
B. (b) `(mu_(0)pi iR^(2)r^(2))/(2x^(4))v`
C. ( c) `(3)/(2)(mu_(0)piiR^(2)r^(2))/(x^(4))v`
D. (d) `(mu_(0)pi iR^(2)r^(2))/(3x^(4))v`

Answer

C

Explanation

(c )Magnetic field on the axis of a circular coil is given by
`B = (mu_(0)iR^(2))/(2(R^(2) + x^(2))^(2)`
Since `xgtgt R`, therefore, nagnetic field at the centre of the smaller loop is
`B~~(mu_(0)piiR^(2)r^(2))/(2x^(3))` ltbr. Flux linked with coil is
`phi = B(pir^(2)) = (mu_(0)piiR^(2)r^(2))/(2x^(3))`
From Faraday's law we have
`E = -(dphi)/(dt) = (3)/(2)(mu_(0)piiR^(2)r^(2))/(x^(4))v`