The final torque on a coil having magnetic moment 25 A m2 in a 5 T uniform external magnetic field, if the coil rotates through an angle of `60^(@)` under the influence of the magnetic field is

Options

A. 216.5Nm
B. 108.25Nm
C. 102.5Nm
D. 258.1Nm

Explanation

To find the final torque on a coil with a given magnetic moment when placed in a magnetic field and rotated through a certain angle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Magnetic moment (M) = 25 A·m²
- Magnetic field (B) = 5 T
- Angle of rotation (θ) = 60°

2. **Understand the Torque Formula:**
The torque (τ) experienced by a magnetic moment in a magnetic field is given by the formula:
\[ \tau = M \cdot B \cdot \sin(\theta) \]
where:
- τ is the torque,
- M is the magnetic moment,
- B is the magnetic field strength,
- θ is the angle between the magnetic moment and the magnetic field.

3. **Convert the Angle to Radians (if necessary):**
Since the sine function can work with degrees, we can directly use θ = 60°.

4. **Calculate the Sine of the Angle:**
\[ \sin(60°) = \frac{\sqrt{3}}{2} \]

5. **Substitute the Values into the Torque Formula:**
\[ \tau = 25 \, \text{A·m²} \times 5 \, \text{T} \times \sin(60°) \]
\[ \tau = 25 \times 5 \times \frac{\sqrt{3}}{2} \]

6. **Perform the Multiplication:**
\[ \tau = 125 \times \frac{\sqrt{3}}{2} \]

7. **Calculate the Final Value:**
\[ \tau = \frac{125\sqrt{3}}{2} \approx 108.25 \, \text{N·m} \]

### Final Answer:
The final torque acting on the coil is approximately **108.25 N·m**.

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