A magnetic needle is pivoted through its centre of mass and is free to rotate in a plane containing uniform magnetic field `200 xx 10^(-4)T`. When it is displaced slightly from the equilibrium it makes 20 oscillations per second. If the moment of inertia of the needle about the axis of oscillation is `0.75 xx10^(-5) kg m^(2)`, find the magnetic moment of the needle.

Explanation

To solve the problem, we need to find the magnetic moment of the needle given the following information:

- Magnetic field strength (B) = \(200 \times 10^{-4} \, T\)
- Frequency of oscillation (f) = \(20 \, Hz\)
- Moment of inertia (I) = \(0.75 \times 10^{-5} \, kg \, m^2\)

### Step-by-Step Solution:

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