A bar magnet of moment of inertia `I` is vibrated in a magnetic field of inducton is `0.4xx10^(-4)T`. The time period period of vibration is `12` sec. The magnetic moment of the magnet is `120Am^(2)`. The moment of inertia of the magnet is ("in"kgm^(2))` approximately

Options

A. `172.8xx10^(-4)`
B. `2.1xx10^(-2)`
C. `1.57xx10^(2)`
D. `1728xx10^(-2)`

Explanation

To find the moment of inertia \( I \) of the bar magnet, we can use the formula for the time period \( T \) of a vibrating bar magnet in a magnetic field:

\[ T = 2\pi \sqrt{\frac{I}{MB}} \]

Where:
- \( T \) is the time period of vibration,
- \( I \) is the moment of inertia,
- \( M \) is the magnetic moment,
- \( B \) is the magnetic field induction.

We want to rearrange this formula to solve for the moment of inertia \( I \):

1. **Square both sides** to eliminate the square root:
\[ T^2 = (2\pi)^2 \frac{I}{MB} \]

2. **Rearranging the equation** to isolate \( I \):
\[ I = \frac{MB T^2}{(2\pi)^2} \]

3. **Substituting the known values**:
- \( M = 120 \, \text{Am}^2 \)
- \( B = 0.4 \times 10^{-4} \, \text{T} \)
- \( T = 12 \, \text{s} \)

Now substituting these values into the equation:
\[ I = \frac{(120)(0.4 \times 10^{-4})(12^2)}{(2\pi)^2} \]

4. **Calculating \( 12^2 \)**:
\[ 12^2 = 144 \]

5. **Calculating \( (2\pi)^2 \)**:
\[ (2\pi)^2 = 4\pi^2 \approx 39.478 \]

6. **Substituting the values back into the equation**:
\[ I = \frac{(120)(0.4 \times 10^{-4})(144)}{39.478} \]

7. **Calculating the numerator**:
\[ 120 \times 0.4 \times 10^{-4} \times 144 = 6912 \times 10^{-4} \]

8. **Now dividing by \( 39.478 \)**:
\[ I \approx \frac{6912 \times 10^{-4}}{39.478} \approx 175.5 \times 10^{-4} \, \text{kg m}^2 \]

9. **Converting to a more standard form**:
\[ I \approx 0.01755 \, \text{kg m}^2 \text{ or } 172.8 \times 10^{-4} \, \text{kg m}^2 \]

Thus, the moment of inertia \( I \) of the magnet is approximately \( 172.8 \times 10^{-4} \, \text{kg m}^2 \).