The pole strength of a bar magnet is 48 ampere-metre and the distance between its poles is 25 cm. The amount of the coupled force by which it can be placed at an angle `30^@` with the uniform magnetic intensity of flux density 0.15 newton/ampere-metre will be

Options

A. 0.9 newton x metre
B. 12 newton x metre
C. 12 newton x metre
D. None of these

Explanation

To solve the problem step by step, we will use the given data and the relevant formulas.

### Step 1: Identify the given values
- Pole strength (m) = 48 ampere-metre
- Distance between the poles (2L) = 25 cm = 0.25 m (converted to meters)
- Magnetic flux density (B) = 0.15 newton/ampere-metre
- Angle (θ) = 30 degrees

### Step 2: Calculate the length (L)
Since the distance between the poles is given as 2L, we can find L:
\[ L = \frac{2L}{2} = \frac{0.25 \, \text{m}}{2} = 0.125 \, \text{m} \]

### Step 3: Calculate the magnetic moment (M)
The magnetic moment (M) of the bar magnet can be calculated using the formula:
\[ M = m \cdot 2L \]
Substituting the values:
\[ M = 48 \, \text{A-m} \cdot 0.25 \, \text{m} = 12 \, \text{A-m}^2 \]

### Step 4: Calculate the torque (τ)
The torque (τ) acting on the magnet in a magnetic field is given by the formula:
\[ \tau = M \cdot B \cdot \sin(\theta) \]
Substituting the values:
- M = 12 A-m²
- B = 0.15 N/A-m
- θ = 30 degrees (sin(30°) = 0.5)

Now, substituting these values into the torque formula:
\[ \tau = 12 \, \text{A-m}^2 \cdot 0.15 \, \text{N/A-m} \cdot \sin(30^\circ) \]
\[ \tau = 12 \cdot 0.15 \cdot 0.5 \]
\[ \tau = 0.9 \, \text{Nm} \]

### Final Answer
The amount of the coupled force by which the magnet can be placed at an angle of 30 degrees with the uniform magnetic intensity is **0.9 Nm**.

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