A long insulated copper wire is closely wound as a spiral of `N` turns. The spiral has inner radius a and outer radius `b`. The spiral lies in the `xy`-plane and a steady current I flows through the wire. The`z`-component of the magetic field at the centre of the spiral is

Options

A. `(mu_0NI)/b`
B. `(2mu_0NI)/a`
C. `(mu_0NI)/(2(b-a)) In b/a`
D. `(mu_0IN)/(2(b-a)) In b/a`

Answer

C

Explanation

(c) Let us consider a thickness dx of wire.
Let it be at a distance x form the centre
O.
Number of turns per unit length =`N/(b-a)`
Number of turns in thickness
`dx=N/(b-a)dx`
Small amount of magnetic field produced at O due to thick-
ness dx of the wire
`dB=(mu_0)/2 (NI)/((b-a)) (dx)/x`
On integrating , we get
`B= int _a^b (mu_0)/2 (NI)/((b-a)) (dx)/x=(mu_0)/2 (NI)/((b-a)) int_a^b (dx)/x`
`=(mu_0)/2 (NI)/((b-a))[log_ex]_a^b=(mu_0)/2 (NI)/((b-a)) log_e(b/a)`
Correct option is (c).