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?

Options

A. `1.13xx10^(-1)"kg m"^(2)`
B. `1.13xx10^(-2)"kg m"^(2)`
C. `1.113xx10^(-3)"kg m"^(2)`
D. `1.13xx10^(-4)"kg m"^(2)`

Explanation

To find the moment of inertia of the coil about its axis of rotation, we can use the relationship between the frequency of oscillation, the moment of inertia, and the torque due to the magnetic field. The steps to solve the problem are as follows:

### Step 1: Calculate the Magnetic Moment of the Coil
The magnetic moment \( M \) of a circular coil is given by the formula:
\[ M = n \cdot I \cdot A \]
where:
- \( n \) = number of turns of the coil = 16
- \( I \) = current through the coil = 0.75 A
- \( A \) = area of the coil = \( \pi r^2 \)

First, we need to calculate the area \( A \):
\[ A = \pi (0.1)^2 = \pi \cdot 0.01 \approx 0.0314 \, \text{m}^2 \]

Now, substituting the values into the magnetic moment formula:
\[ M = 16 \cdot 0.75 \cdot 0.0314 \approx 0.375 \, \text{A m}^2 \]

### Step 2: Calculate the Torque on the Coil
The torque \( \tau \) acting on the coil in a magnetic field is given by:
\[ \tau = M \cdot B \]
where:
- \( B \) = magnetic field strength = \( 5 \times 10^{-2} \, \text{T} \)

Substituting the values:
\[ \tau = 0.375 \cdot (5 \times 10^{-2}) = 0.01875 \, \text{N m} \]

### Step 3: Relate Torque to Angular Frequency
The angular frequency \( \omega \) of the oscillation is related to the moment of inertia \( I \) and the torque by:
\[ \tau = I \cdot \alpha \]
where \( \alpha \) is the angular acceleration. For small oscillations, we can relate \( \alpha \) to the angular frequency \( \omega \) as:
\[ \alpha = \omega^2 \]
Thus,
\[ \tau = I \cdot \omega^2 \]

### Step 4: Calculate the Moment of Inertia
We know the frequency \( f \) is given as \( 2 \, \text{s}^{-1} \). The angular frequency \( \omega \) is related to the frequency by:
\[ \omega = 2\pi f = 2\pi \cdot 2 \approx 12.57 \, \text{rad/s} \]

Now substituting \( \tau \) and \( \omega \) into the equation:
\[ 0.01875 = I \cdot (12.57)^2 \]
Calculating \( (12.57)^2 \):
\[ (12.57)^2 \approx 158.0 \]

Now we can solve for \( I \):
\[ I = \frac{0.01875}{158.0} \approx 0.000118 \, \text{kg m}^2 \]

### Final Answer
The moment of inertia of the coil about its axis of rotation is approximately:
\[ I \approx 0.000118 \, \text{kg m}^2 \]