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

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 = int dB = int_a^b (mu_0 "nidx")/(2x) =(mu_0 "ni")/(2) int_a^b (dx)/(x)`
or `B = (mu_0 "ni")/(2) ln b/a " or " B = (mu_0 Ni)/(2(b-a) ) ln b/a`
` = (4pi xx 10^(-7) xx 100 xx 8 xx 10^(-3) )/(2(10 - 5) xx 10^(-2) ) ln 10/5 , B = 6.9 xx 10^(-6) T`