A square coil of side `10cm` consists of 20 turns and carries a current of 12A. The coil is suspended vertically and normal to the plane of the coil makes an angle of `30^@` with the direction of a uniform horizontal magnetic field of magnitude `0*80T`. What is the magnitude of torque experienced by the coil?

Options

A. 1.6 N m
B. 1.2 N m
C. 1.4 N m
D. 1.8 N m

Explanation

To find the magnitude of the torque experienced by the square coil, we can follow these steps:

### Step 1: Identify the given values
- Side of the square coil, \( a = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Number of turns, \( n = 20 \)
- Current, \( I = 12 \, \text{A} \)
- Magnetic field strength, \( B = 0.8 \, \text{T} \)
- Angle between the normal to the coil and the magnetic field, \( \theta = 30^\circ \)

### Step 2: Calculate the area of the coil
The area \( A \) of the square coil can be calculated using the formula:
\[ A = a^2 \]
Substituting the value of \( a \):
\[ A = (0.1 \, \text{m})^2 = 0.01 \, \text{m}^2 \]

### Step 3: Calculate the magnetic moment \( m \)
The magnetic moment \( m \) of the coil is given by the formula:
\[ m = n \cdot A \cdot I \]
Substituting the values:
\[ m = 20 \cdot 0.01 \, \text{m}^2 \cdot 12 \, \text{A} = 2.4 \, \text{A} \cdot \text{m}^2 \]

### Step 4: Calculate the torque \( \tau \)
The torque \( \tau \) experienced by the coil can be calculated using the formula:
\[ \tau = m \cdot B \cdot \sin(\theta) \]
Substituting the values:
\[ \tau = 2.4 \, \text{A} \cdot \text{m}^2 \cdot 0.8 \, \text{T} \cdot \sin(30^\circ) \]
Since \( \sin(30^\circ) = 0.5 \):
\[ \tau = 2.4 \cdot 0.8 \cdot 0.5 = 0.96 \, \text{N} \cdot \text{m} \]

### Final Result
Thus, the magnitude of the torque experienced by the coil is:
\[ \tau = 0.96 \, \text{N} \cdot \text{m} \]

---