**Solve the following problem.**

A magnetic pole of a bar magnet with a pole strength of 100 A m is 20 cm away from the centre of a bar magnet. The bar magnet has a pole strength of 200 A m and has a length of 5 cm. If the magnetic pole is on the axis of the bar magnet, find the force on the magnetic pole.

#### Solution

Given that, (q_{m})_{1} = 200 Am and

(2l) = 5 cm = 5 × 10^{–2} m

∴ m = 200 × 5 × 10^{–2} = 10 Am^{2}

For a bar magnet, magnetic dipole moment is,

m = q_{m} (2l)

For a point on the axis of a bar magnet at distance, r = 20 cm = 0.2 m,

B_{a} = `μ_0/(4π)xx(2"m")/"r"^3`

= `10^-7xx(2xx10)/(0.2)^3`

= 0.25 × 10^{−3}

= 2.5 × 10^{−4} Wb/m^{2}

The force acting on the pole will be given by,

F = q_{m} B_{a} = 100 × 2.5 × 10^{–4}

= 2.5 × 10^{–2} N

The force acting on the magnetic pole due to the bar magnet is **2.5 × 10 ^{–2} N**.