A solenoid of length 10 cm and radius 1 cm contains 200 turns and carries a current of 10 A. Find the magnetic field at a point on the axis at a distance of 10cm from the centre.

Explanation

The dipole moment of each turn is `mu = iA = (10 A) (pi cm^2)`
`= pi xx 10^(-3) A m^2`.
if each current loop is replaced by a dipole having pole strength `m` and separation between the poles `d`, we have
`mu =md`.
As there are 200 turns,
`200 d = 10cm`
or, `d = 5 xx 10^(-4) m`
Thus,
`m = (mu)/d = (pi xx 10^(-3) A m^2)/(5 xx 10^(-4) m) = 2pi A m`.
We can replace the solenoid by a south pole and a north pole of equal pole strength `2pi A m`, separated by 10cm. The equivalent picture is shown in figure.
The megnetic field at `P` due to the north pole is
`B_N = (mu_o)/(4pi) (2pi A m)/(5 cm)^2) = 2.5 xx 10^(-4) T`.
The magnetic field at `P` due to the south pole is
`B_S = (mu_0)/(4pi) (2pi A m)/(15 cm)^2 = 0.3 xx 10^(-4) T`.
The field `B_N` is away from the poles and `B_S` is towards the poles. The resultant field at `P` is
B=B_N - B_S`
`= 2.2 xx 10^(-4) T`
away from the solenoid.