A bar magnet of moment of inertia `9xx10^(-5) kgm^(2)` placed in a vibration magnetometer and oscillating in a uniform magnetic field `16pi^(2)xx10^(-5)T` makes 20 oscillations in 15 s . The magnetic moment of the bar magnet is

Options

A. `3Am^(2)`
B. `1Am^(2)`
C. `5Am^(2)`
D. `4Am^(2)`

Answer

To find the magnetic moment of the bar magnet, we will follow these steps:

### Step 1: Determine the time period of oscillation
The time period \( T \) can be calculated from the number of oscillations and the total time taken. Given that the bar magnet makes 20 oscillations in 15 seconds, we can find \( T \) as follows:

\[ T = \frac{\text{Total time}}{\text{Number of oscillations}} = \frac{15 \, \text{s}}{20} = 0.75 \, \text{s} \]

### Step 2: Use the formula for the time period of oscillation
The time period \( T \) for a bar magnet oscillating in a magnetic field is given by the formula:

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

Where:
- \( I \) is the moment of inertia of the bar magnet,
- \( m \) is the magnetic moment,
- \( B \) is the magnetic field strength.

### Step 3: Rearranging the formula to find the magnetic moment
We can rearrange the formula to solve for the magnetic moment \( m \):

\[ m = \frac{4\pi^2 I}{T^2 B} \]

### Step 4: Substitute the known values
Now, we will substitute the known values into the equation:
- Moment of inertia \( I = 9 \times 10^{-5} \, \text{kg m}^2 \)
- Magnetic field \( B = 16\pi^2 \times 10^{-5} \, \text{T} \)
- Time period \( T = 0.75 \, \text{s} \)

Substituting these values into the equation:

\[ m = \frac{4\pi^2 \times (9 \times 10^{-5})}{(0.75)^2 \times (16\pi^2 \times 10^{-5})} \]

### Step 5: Simplifying the equation
Now, we simplify the equation:

1. The \( \pi^2 \) terms cancel out.
2. The \( 10^{-5} \) terms also cancel out.

Thus, we have:

\[ m = \frac{4 \times 9 \times 10^{-5}}{(0.75)^2 \times 16} \]

Calculating \( (0.75)^2 = 0.5625 \):

\[ m = \frac{36}{0.5625 \times 16} \]

Calculating \( 0.5625 \times 16 = 9 \):

\[ m = \frac{36}{9} = 4 \, \text{A m}^2 \]

### Final Answer
The magnetic moment of the bar magnet is \( 4 \, \text{A m}^2 \).

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