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_(0)NI)/(2(b-a))In(b/a)`
B. `(mu_(0)NI)/(2(b-a))In((b+a)/(b-a))`
C. `(mu_(0)NI)/(2b)In (b/a)`
D. `(mu_(0)NI)/(2b) In ((b+a)/(b-a))`

Answer

A

Explanation

If we take a small strip of 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 centre of the coil will be `dB=(mu_(0)(dN)I)/(2r)=(mu_(0)NI)/((b-a))(dr)/r`
`:.B=int_(r-a)^(r=b)dB=(mu_(0)NI)/(2(b-a)) In (b/a)`