First slide

Magnetic dipole in a uniform magnetic field

Question

A bar magnet takes $π$ /10 seconds to complete one oscillation in uniform magnetic field. The moment of inertia of the magnet about the axis of rotation is 1.2 × 10^–4kgm² and the earth’s horizontal magnetic field is 30 $μ$ T. The magnetic moment of the magnet is...........Am²

Moderate

Solution

$\begin{array}{rcl} T & = & 2 π \sqrt{\frac{I}{MB}} \\ \frac{π}{10} & = & 2 π \sqrt{\frac{1.2 x 10^{- 4}}{M x 30 x 10^{- 6}}}, \\ M & = & 1600 {Am}^{2} \end{array}$

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