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 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 .
`dB = (mu_0(ndx)i)/(2x)`
So net field
`B = int dB = int_(a)^(b)(mu_0nidx)/(2x) = (mu_0ni)/(2) int_a^b (dx)/(x)`
`or B = (mu_0 ni )/(2) ln"" (b)/(a) or B = (mu_0Ni)/(2(b-a)) "ln"(b)/(a)`
` = (4pixx 10^(-7) xx 100 xx 8 xx 10^(-3))/(2(10 -5)xx10^(-2)) "ln"(10)/(5)`
`B = 6.96 xx 10^(-6) T`