A short bar magnet pleaced with its axis at `30^(@)` with a uniform external magnetic field of `0.16` Tesla expriences a torque of magnitude `0.032` Joule. The magnetic moment of the bar magnet will be

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:

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

Where:
- \( \tau \) is the torque (in Joules),
- \( m \) is the magnetic moment (in Joule per Tesla),
- \( B \) is the magnetic field strength (in Tesla),
- \( \theta \) is the angle between the magnetic moment and the magnetic field (in degrees).

### Step 1: Identify the given values
From the problem, we have:
- Torque (\( \tau \)) = 0.032 Joules
- Magnetic field (\( B \)) = 0.16 Tesla
- Angle (\( \theta \)) = 30 degrees

### Step 2: Convert the angle to radians (if necessary)
In this case, we can use the sine of the angle directly in degrees, so we don't need to convert it to radians.

### Step 3: Calculate \( \sin(30^\circ) \)
We know that:
\[ \sin(30^\circ) = \frac{1}{2} \]

### Step 4: Substitute the values into the torque equation
Now we can rearrange the torque formula to solve for the magnetic moment \( m \):
\[ m = \frac{\tau}{B \cdot \sin(\theta)} \]

Substituting the known values:
\[ m = \frac{0.032}{0.16 \cdot \sin(30^\circ)} \]
\[ m = \frac{0.032}{0.16 \cdot \frac{1}{2}} \]
\[ m = \frac{0.032}{0.08} \]

### Step 5: Calculate the magnetic moment
Now, we perform the division:
\[ m = 0.4 \, \text{Joule per Tesla} \]

### Final Answer
The magnetic moment of the bar magnet is \( 0.4 \, \text{Joule per Tesla} \).

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