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)/(2(b-a))ln(b/a)`
B. `(mu_0NI)/(2(b-a))ln ((b+a)/(b-a))`
C. `(mu_0NI)/(2b)ln(b/a)`
D. `(mu_0NI)/(2b)((b+a)/(b-a))`

Answer

A

Explanation

a. If we take a small strip of `dr` at distance `r` from centre, then number of turns in this strip would be
`dN=(N/(b-a))dr`
Magnetic field due to this element at the cenntre of the coil will be
`dB=(mu_0)(dN)I/(2r)=(mu_0NI)/((b-a)) (dr)/(2r)`
`:. Bint_(r=a)^(r=b)dB=(mu_0NI)/(2(b-a))ln(b/a)`
`:. current answer is a.