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 magnetic field at the centre of the spiral is

Options

A. `(mu_(0)NI)/b`
B. `(2mu_(0)NI)/a`
C. `(mu_(0)NI)/(2(b-a)) ln(b/a)`
D. `(mu_(0)I^(N))/(2(b-a))ln(b/a)`

Answer

C

Explanation

Number of turns per unit width `=N/(b-a)` Consider an elemental ring of radius `x` and with thickness `dx` Number of turns in the ring `=dN=(Ndx)/(b-a)` Magnetic field at the centre due to the ring element

`dB=(mu_(0)(dN)i)/(2x) =(mu_(0)i)/2. (Ndx)/((b-a)).1/x`
`:.` Field at the centre
`=intdB=(mu_(0)Ni)/(2(b-a))int_(a)^(b)(dx)/(x)`
`=(mu_(0)Ni)/(2(b-a)) ln(b/a)`.