A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0T. The field lines make an angle of `60^@` with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.

Options

A. 3.13 Nm
B. 1.13 Nm
C. 2.13 Nm
D. 4.13Nm

Explanation

To solve the problem, we need to calculate the torque acting on the circular coil in the magnetic field and then determine the counter torque required to prevent the coil from turning.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Number of turns (N) = 30
- Radius of the coil (r) = 8.0 cm = 0.08 m (convert cm to m)
- Current (I) = 6.0 A
- Magnetic field (B) = 1.0 T
- Angle (θ) = 60° (angle between the magnetic field and the normal to the coil)

2. **Calculate the Area of the Coil:**
The area (A) of a circular coil is given by the formula:
\[ A = \pi r^2 \]
Substituting the radius:
\[ A = \pi (0.08)^2 = \pi \times 0.0064 \approx 0.0201 \, \text{m}^2 \]

3. **Calculate the Torque (τ) on the Coil:**
The torque (τ) experienced by the coil in a magnetic field is given by:
\[ \tau = N \cdot B \cdot I \cdot A \cdot \sin(\theta) \]
Substituting the values:
\[ \tau = 30 \cdot 1.0 \cdot 6.0 \cdot 0.0201 \cdot \sin(60^\circ) \]
We know that \(\sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866\):
\[ \tau = 30 \cdot 1.0 \cdot 6.0 \cdot 0.0201 \cdot 0.866 \]
\[ \tau = 30 \cdot 1.0 \cdot 6.0 \cdot 0.0201 \cdot 0.866 \approx 3.13 \, \text{N m} \]

4. **Determine the Counter Torque:**
To prevent the coil from turning, the counter torque must be equal in magnitude to the torque calculated:
\[ \text{Counter Torque} = 3.13 \, \text{N m} \]

### Final Answer:
The magnitude of the counter torque that must be applied to prevent the coil from turning is **3.13 N m**.

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