A circular coil of 70 turns and radius 5 cm carrying a current of 8 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.5 T. The field lines make an angle of `30^(@)` with the normal of the coil then the magnitude of the counter torque that must be applied to prevent the coil from turning is

Options

A. 33 Nm
B. 3.3 Nm
C. `3.3 xx 10^(-2)Nm`
D. `3.3 xx 10^(-4)Nm`

Explanation

To find the magnitude of the counter torque that must be applied to prevent the coil from turning, we can follow these steps:

### Step 1: Identify the given values
- Number of turns in the coil, \( n = 70 \)
- Radius of the coil, \( r = 5 \, \text{cm} = 0.05 \, \text{m} \)
- Current in the coil, \( I = 8 \, \text{A} \)
- Magnetic field strength, \( B = 1.5 \, \text{T} \)
- Angle with the normal, \( \theta = 30^\circ \)

### Step 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:
\[ A = \pi (0.05)^2 = \pi (0.0025) \approx 0.007854 \, \text{m}^2 \]

### Step 3: Use the torque formula
The torque \( \tau \) on the coil is given by:
\[ \tau = n I B A \sin \theta \]
Substituting the known values:
\[ \tau = 70 \times 8 \times 1.5 \times 0.007854 \times \sin(30^\circ) \]
Since \( \sin(30^\circ) = \frac{1}{2} \):
\[ \tau = 70 \times 8 \times 1.5 \times 0.007854 \times \frac{1}{2} \]

### Step 4: Calculate the torque
Now, calculate the torque step by step:
1. Calculate \( 70 \times 8 = 560 \)
2. Calculate \( 560 \times 1.5 = 840 \)
3. Calculate \( 840 \times 0.007854 \approx 6.596 \)
4. Finally, multiply by \( \frac{1}{2} \):
\[ \tau = 6.596 \times \frac{1}{2} \approx 3.298 \, \text{N m} \]

### Step 5: Final result
The magnitude of the counter torque that must be applied to prevent the coil from turning is approximately:
\[ \tau \approx 3.3 \, \text{N m} \]