At a place the hoizontal componet of earth 's field is `0.5 xx 10^(-4)` T . A bar magnet suspended horizontally perpendicular to to earth' field experience a torque of `4.5xx 10^(-4) N - M` at the place . The magnetic moment of the magnet is .

Options

A. `2.25 xx 10^(-8) JT`
B. `1//9 J//T`
C. `2.25 J//T`
D. `9 J//T`

Explanation

To find the magnetic moment of the bar magnet, we can use the formula for torque (\(\tau\)) experienced by a magnetic dipole in a magnetic field. The torque is given by:

\[ \tau = M \cdot B \cdot \sin(\theta) \]

Where:
- \(\tau\) is the torque,
- \(M\) is the magnetic moment,
- \(B\) is the magnetic field strength,
- \(\theta\) is the angle between the magnetic moment and the magnetic field.

### Step 1: Identify the given values
From the problem, we have:
- Horizontal component of Earth's magnetic field, \(B = 0.5 \times 10^{-4} \, \text{T}\)
- Torque experienced by the magnet, \(\tau = 4.5 \times 10^{-4} \, \text{N-m}\)
- The angle \(\theta = 90^\circ\) (since the bar magnet is perpendicular to the magnetic field).

### Step 2: Substitute the values into the torque formula
Since \(\sin(90^\circ) = 1\), we can simplify the torque equation:

\[ \tau = M \cdot B \cdot 1 \]

Thus, we can rewrite it as:

\[ M = \frac{\tau}{B} \]

### Step 3: Calculate the magnetic moment
Now we can substitute the values of \(\tau\) and \(B\):

\[ M = \frac{4.5 \times 10^{-4} \, \text{N-m}}{0.5 \times 10^{-4} \, \text{T}} \]

### Step 4: Perform the calculation
Calculating the right-hand side:

\[ M = \frac{4.5}{0.5} \times 10^{-4} \div 10^{-4} \]

This simplifies to:

\[ M = 9 \, \text{J/T} \]

### Step 5: Write the final answer
Thus, the magnetic moment of the magnet is:

\[ M = 9 \, \text{J/T} \]