A short bar magnet placed with its axis at `30^(@)` with a uniform external magnetic field of 0.35 T experiences a torque of magnitude equal to `4.5xx10^(-2)Nm`. The magnitude of magnetic moment of the given magnet is

Options

A. `0.26JT^(-1)`
B. `2.6JT^(-1)`
C. `0.26JT^(-1)`
D. `0.026JT^(-1)`

Explanation

To find the magnitude of the magnetic moment of the bar magnet, we will use the formula for torque experienced by a magnetic dipole in a magnetic field:

\[ \tau = mB \sin \theta \]

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:
- Torque, \(\tau = 4.5 \times 10^{-2} \, \text{Nm}\)
- Magnetic field strength, \(B = 0.35 \, \text{T}\)
- Angle, \(\theta = 30^\circ\)

### Step 2: Rearrange the formula to solve for magnetic moment \(m\)
We need to isolate \(m\) in the torque equation:

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

### Step 3: Calculate \(\sin \theta\)
We need to calculate \(\sin 30^\circ\):

\[ \sin 30^\circ = \frac{1}{2} \]

### Step 4: Substitute the values into the equation
Now we can substitute the known values into the equation for \(m\):

\[ m = \frac{4.5 \times 10^{-2}}{0.35 \times \sin 30^\circ} \]

Substituting \(\sin 30^\circ = \frac{1}{2}\):

\[ m = \frac{4.5 \times 10^{-2}}{0.35 \times \frac{1}{2}} \]

### Step 5: Simplify the equation
Now simplify the denominator:

\[ m = \frac{4.5 \times 10^{-2}}{0.35 \times 0.5} = \frac{4.5 \times 10^{-2}}{0.175} \]

### Step 6: Perform the division
Now calculate the value:

\[ m = \frac{4.5 \times 10^{-2}}{0.175} \approx 0.257 \, \text{J/T} \]

### Step 7: Final answer
Thus, the magnitude of the magnetic moment of the given magnet is approximately:

\[ m \approx 0.257 \, \text{J/T} \]