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 acting on a magnetic dipole in a magnetic field. The torque (\( \tau \)) experienced by a magnetic dipole is given by the equation:

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

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 magnet is perpendicular to the magnetic field)

### Step 2: Substitute the known values into the torque formula
Since \( \theta = 90^\circ \), we know that \( \sin(90^\circ) = 1 \). Therefore, the formula simplifies to:

\[ \tau = m \cdot B \]

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

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

### Step 4: Substitute the values of \( \tau \) and \( B \)
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 5: Calculate the magnetic moment \( m \)
Now, performing the calculation:

\[ m = \frac{4.5 \times 10^{-4}}{0.5 \times 10^{-4}} = \frac{4.5}{0.5} = 9 \, \text{J/T} \]

### Conclusion
Thus, the magnetic moment of the magnet is:

\[ \boxed{9 \, \text{J/T}} \]