Correct Answer - D
Take an element of thickness dr at a distance r from the centre of spiral coil as shown in figure.
No. of turns per unit thickness of spiral coil
`=(N)/((b-a))`
No. of turns in the element of thickness dr is
`n=(N)/((b-a))dr`
Magnetic field induction at the centre of the spiral coil due to current in element of thickness `dr` is
`dB=(mu_0)/(4pi)(2pinI)/(r)=(mu_0)/(2)((Ndr)/(b-a))I/r=(mu_0NI)/(2(b-a))(dr)/(r)`
Total magnetic field induction at the centre of spiral wire due to current through the wire is
`B=int_(a)^(b)(mu_0NI)/(2(b-a))(dr)/(r)=(mu_0NI)/(2(b-a))int_(a)^(b)(dr)/(r)`
`=(mu_0NI)/(2(b-a))1nb/a`