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

(a) From Biot-Savart's law, the magnetic induction due to a circular current carrying wire loop at its centre is given by,
`B_(r) = (mu_(0))/(2r) i`
The plane spiral is made up of concentric circular loops, having different radii, varying from `a` to `b` Therefore, the total magentic induction at the centre,
`B_(0) = int (mu_(0))/(2r) dN` .....(1)
where `(mu_(0))/(2r) i` is the contribution of one turn of radius `r` and `dN` is the number of turns in the interval `(r,r + dr)`
`i.e., dN = (N)/(b - a) dr`
Substibuiting is equaction (1) and intergrating the result over `r` between `a` and `b`, we obtain,
`B_(0) = int_(a)^(b) (mu_(0)i)/(2r) (N)/((b - a)) dr = (mu_(0) i N)/(2 (b-a)) In (b)/(a)`
(b) The magetic moment of a turn of radius `r` is `P_(m) - i pi r^(2)` and of all turns,
`p = int p_(m) dN = int_(a)^(b) i pi r^(2) (N)/(b - a) dr = (pi i N(b^(3) - a^(2)))/(3 (b - a))`