A circular coil of 16 turns and radius 10cm carrying a current of 0.75A rests with its plane normal to an external field of magnitude `5*0xx10^-2T`. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of `2*0s^-1`. What is the moment of inertia of the coil about its axis of rotation?

Explanation

To find the moment of inertia of a circular coil about its axis of rotation, we will follow these steps:

### Step 1: Identify the Given Values
- Number of turns, \( n = 16 \)
- Radius of the coil, \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Current, \( I = 0.75 \, \text{A} \)
- External magnetic field, \( B = 5 \times 10^{-2} \, \text{T} \)
- Frequency of oscillation, \( \nu = 2 \, \text{s}^{-1} \)
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