A short bar magnet placed with its axis at `30^@` with an external field of 800 G experiences a torque of 0.016 Nm. The magnetic moment of the magnet i

Options

A. `4Am^2`
B. `0.5Am^2`
C. `2Am^2`
D. `0.40Am^2`

Answer

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

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

Where:
- \( \tau \) is the torque (in Nm),
- \( m \) is the magnetic moment (in Am²),
- \( B \) is the magnetic field strength (in T),
- \( \theta \) is the angle between the magnetic moment and the magnetic field (in degrees).

Given:
- \( \tau = 0.016 \, \text{Nm} \)
- \( B = 800 \, \text{G} = 800 \times 10^{-4} \, \text{T} = 0.08 \, \text{T} \) (since 1 G = \( 10^{-4} \) T)
- \( \theta = 30^\circ \)

### Step 1: Convert the angle to radians (if necessary)
For this calculation, we can directly use degrees since the sine function can take degrees.

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

### Step 3: Substitute the known values into the torque formula
\[ 0.016 = m \cdot 0.08 \cdot \sin(30^\circ) \]
\[ 0.016 = m \cdot 0.08 \cdot \frac{1}{2} \]

### Step 4: Simplify the equation
\[ 0.016 = m \cdot 0.04 \]

### Step 5: Solve for \( m \)
\[ m = \frac{0.016}{0.04} \]
\[ m = 0.4 \, \text{Am}^2 \]

### Conclusion
The magnetic moment of the magnet is \( 0.4 \, \text{Am}^2 \).

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