A recantagular coil of length `0.12 m` and width `0.1 m` having 50 turns of wire is suspended vertically in unifrom magnetic field of srenght 0.2 `Weber//m^(2)`. The coil carres a current of 2 A. If the plane of the coil is inclined at an angl,e of `30^(@)` with the direction of the feld the torque required to keep the coil in stable equilibrium will be

Options

A. `0.15 Nm`
B. `0.20 Nm`
C. `0.24 Nm`
D. `0.12 Nm`

Explanation

To solve the problem, we need to calculate the torque (τ) acting on a rectangular coil suspended in a magnetic field. The torque can be calculated using the formula:

\[ \tau = n \cdot B \cdot I \cdot A \cdot \sin(\alpha) \]

where:
- \( n \) = number of turns of the coil
- \( B \) = magnetic field strength
- \( I \) = current flowing through the coil
- \( A \) = area of the coil
- \( \alpha \) = angle between the normal to the coil and the magnetic field

### Step 1: Identify the given values
- Length of the coil, \( L = 0.12 \, m \)
- Width of the coil, \( W = 0.1 \, m \)
- Number of turns, \( n = 50 \)
- Magnetic field strength, \( B = 0.2 \, \text{Weber/m}^2 \)
- Current, \( I = 2 \, A \)
- Angle of inclination with the magnetic field, \( \theta = 30^\circ \)

### Step 2: Calculate the area of the coil
The area \( A \) of the rectangular coil can be calculated using the formula:
\[ A = L \cdot W \]
Substituting the values:
\[ A = 0.12 \, m \cdot 0.1 \, m = 0.012 \, m^2 \]

### Step 3: Determine the angle \( \alpha \)
Since the plane of the coil is inclined at an angle \( \theta = 30^\circ \) with respect to the magnetic field, the angle \( \alpha \) between the normal to the coil and the magnetic field is:
\[ \alpha = 90^\circ - \theta = 90^\circ - 30^\circ = 60^\circ \]

### Step 4: Substitute the values into the torque formula
Now we can substitute all the known values into the torque formula:
\[ \tau = n \cdot B \cdot I \cdot A \cdot \sin(\alpha) \]
Substituting the values:
\[ \tau = 50 \cdot 0.2 \cdot 2 \cdot 0.012 \cdot \sin(60^\circ) \]

### Step 5: Calculate \( \sin(60^\circ) \)
The value of \( \sin(60^\circ) \) is:
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866 \]

### Step 6: Calculate the torque
Now substituting \( \sin(60^\circ) \):
\[ \tau = 50 \cdot 0.2 \cdot 2 \cdot 0.012 \cdot 0.866 \]
Calculating step-by-step:
\[ \tau = 50 \cdot 0.2 = 10 \]
\[ \tau = 10 \cdot 2 = 20 \]
\[ \tau = 20 \cdot 0.012 = 0.24 \]
\[ \tau = 0.24 \cdot 0.866 \approx 0.208 \, \text{Newton meter} \]

### Final Result
The torque required to keep the coil in stable equilibrium is approximately:
\[ \tau \approx 0.020 \, \text{Newton meter} \]