The pole strength of a bar magnet is 48A-m and the distance between its poles is 25 cm . The torque by which it can be held at an angle of `30^(@)` with the uniform magnetic field of strength 0.15 `Wb//m^(2)` , will be

Options

A. 0.90 N-m
B. 1.80 N-m
C. 2.70 N-m
D. 3.6 N-m

Answer

To solve the problem, we need to calculate the torque acting on a bar magnet when it is held at an angle of \(30^\circ\) in a uniform magnetic field. We will use the formula for torque (\(\tau\)) in a magnetic field, which is given by:

\[ \tau = M B \sin \theta \]

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

### Step 1: Calculate the Magnetic Moment (M)

The magnetic moment \(M\) can be calculated using the formula:

\[ M = P \cdot L \]

where:
- \(P\) is the pole strength,
- \(L\) is the distance between the poles.

Given:
- Pole strength \(P = 48 \, \text{A-m}\)
- Distance between the poles \(L = 25 \, \text{cm} = 0.25 \, \text{m}\) (convert cm to m)

Now, substituting the values:

\[ M = 48 \, \text{A-m} \times 0.25 \, \text{m} = 12 \, \text{A-m}^2 \]

### Step 2: Identify the Magnetic Field Strength (B)

The magnetic field strength is given as:

\[ B = 0.15 \, \text{Wb/m}^2 \]

### Step 3: Calculate \(\sin \theta\)

For \(\theta = 30^\circ\):

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

### Step 4: Substitute Values into the Torque Formula

Now we can substitute \(M\), \(B\), and \(\sin \theta\) into the torque formula:

\[ \tau = M B \sin \theta \]

Substituting the known values:

\[ \tau = 12 \, \text{A-m}^2 \times 0.15 \, \text{Wb/m}^2 \times \frac{1}{2} \]

### Step 5: Calculate the Torque

Now, calculate the torque:

\[ \tau = 12 \times 0.15 \times \frac{1}{2} = 12 \times 0.075 = 0.9 \, \text{N-m} \]

### Final Answer

The torque by which the bar magnet can be held at an angle of \(30^\circ\) with the uniform magnetic field is:

\[ \tau = 0.9 \, \text{N-m} \]