A bar has a magnetic moment equal to `5xx10^(-5) weberxxm`. It is suspended in a magnetic field which has a magnetic induction (B) equal to `8pixx10^(-4)` tesla. The magnet vibrates with a period of vibration equal to 15 sec. The moment of intertia of the magnet is

Answer

To solve the problem, we need to find the moment of inertia (I) of a bar that has a magnetic moment (m) and is suspended in a magnetic field (B). The period of vibration (T) is given as 15 seconds.

### Step-by-Step Solution:

1. **Identify the given values:**
- Magnetic moment, \( m = 5 \times 10^{-5} \, \text{Weber meter} \)
- Magnetic induction, \( B = 8 \pi \times 10^{-4} \, \text{Tesla} \)
- Period of vibration, \( T = 15 \, \text{seconds} \)

2. **Use the formula for the period of oscillation:**
The formula for the period of oscillation of a magnetic dipole in a magnetic field is given by:
\[ T = 2\pi \sqrt{\frac{I}{mB}} \]
where:
- \( T \) is the period,
- \( I \) is the moment of inertia,
- \( m \) is the magnetic moment,
- \( B \) is the magnetic field.

3. **Square both sides of the equation:**
To eliminate the square root, we square both sides:
\[ T^2 = (2\pi)^2 \frac{I}{mB} \]
This simplifies to:
\[ T^2 = 4\pi^2 \frac{I}{mB} \]

4. **Rearrange the equation to solve for moment of inertia (I):**
Rearranging gives:
\[ I = \frac{T^2 mB}{4\pi^2} \]

5. **Substitute the known values:**
Now we substitute the values into the equation:
\[ I = \frac{(15)^2 \times (5 \times 10^{-5}) \times (8\pi \times 10^{-4})}{4\pi^2} \]

6. **Calculate the values:**
- Calculate \( T^2 \):
\[ T^2 = 15^2 = 225 \]
- Substitute \( T^2 \) into the equation:
\[ I = \frac{225 \times (5 \times 10^{-5}) \times (8\pi \times 10^{-4})}{4\pi^2} \]

7. **Simplify the expression:**
- The \( \pi \) terms cancel out:
\[ I = \frac{225 \times 5 \times 8 \times 10^{-5} \times 10^{-4}}{4\pi} \]
- Calculate the numerator:
\[ 225 \times 5 = 1125 \]
\[ 1125 \times 8 = 9000 \]
- Thus:
\[ I = \frac{9000 \times 10^{-9}}{4\pi} \]

8. **Final calculation:**
- Approximating \( \pi \approx 3.14 \):
\[ I \approx \frac{9000 \times 10^{-9}}{12.56} \approx 7.16 \times 10^{-7} \, \text{kg m}^2 \]

### Final Answer:
The moment of inertia of the magnet is approximately \( 7.16 \times 10^{-7} \, \text{kg m}^2 \).