A bar magnet of magnetic moment `9.85 A-m^2` and moment of inertia `I = 10^-6 kg-m^2` makes 10 oscillation in 5secs inuniform magnetic field. Find intensity of magnetic field (take `pi ^2 = 9.85`)

Options

A. `20 mu T`
B. `25 mu T`
C. `16 mu T`
D. `10 mu T`

Answer

To solve the problem, we need to find the intensity of the magnetic field given the magnetic moment and the moment of inertia of the bar magnet.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Magnetic moment \( m = 9.85 \, \text{A-m}^2 \)
- Moment of inertia \( I = 10^{-6} \, \text{kg-m}^2 \)
- Number of oscillations = 10
- Time for oscillations = 5 seconds

2. **Calculate the Time Period (T):**
The time period \( T \) for one oscillation can be calculated as:
\[ T = \frac{\text{Total Time}}{\text{Number of Oscillations}} = \frac{5 \, \text{s}}{10} = 0.5 \, \text{s} \]

3. **Use the Formula for Time Period of a Bar Magnet in a Magnetic Field:**
The formula for the time period \( T \) of a bar magnet in a magnetic field is given by:
\[ T = 2\pi \sqrt{\frac{I}{mB}} \]
where \( B \) is the magnetic field intensity.

4. **Square Both Sides:**
Squaring both sides of the time period equation gives:
\[ T^2 = 4\pi^2 \frac{I}{mB} \]

5. **Rearranging the Equation to Solve for B:**
Rearranging the equation to find \( B \):
\[ B = \frac{4\pi^2 I}{m T^2} \]

6. **Substituting the Known Values:**
Now, substitute \( I = 10^{-6} \, \text{kg-m}^2 \), \( m = 9.85 \, \text{A-m}^2 \), and \( T = 0.5 \, \text{s} \):
\[ B = \frac{4 \cdot 9.85 \cdot 10^{-6}}{9.85 \cdot (0.5)^2} \]

7. **Simplifying the Expression:**
The \( 9.85 \) in the numerator and denominator cancels out:
\[ B = \frac{4 \cdot 10^{-6}}{(0.5)^2} = \frac{4 \cdot 10^{-6}}{0.25} = 16 \cdot 10^{-6} \, \text{T} \]

8. **Final Result:**
Therefore, the intensity of the magnetic field \( B \) is:
\[ B = 16 \, \mu\text{T} \]