A thin insulated wire form a spiral of N=100 turns carrying a current of i=8mA . The inner and outer radii are equal to a=5cm and b=10cm. Find the magnetic field at the centre of the coil.

Explanation

Let n = no.of turns per unit length along the radial of spiral. Consider a ring of radii x and x + dx .
No. of turns in the ring = ndx. `n=N/((b-a))`
Magnetic field at the centre due to the ring is
`dB=(mu_(0)(ndx)i)/(2x)`
So net field `B=intdB=int_(a)^(b)(mu_(0)nidx)/(2x)=(mu_(0)ni)/2int_(a)^(b)(dx)/x`
or `B=(mu_(0)ni)/2"ln"b/a` or `B=(mu_(0)Ni)/(2(b-a))"ln"b/a`
`=(4pixx10^(-7)xx100x8xx10^(-3))/(2(10-5)xx10^(-2))"ln"10/5B=6.9xx10^(-6)T`