A rectangular coil of 20 turns, each of area `20 cm^(2) `is suspended freely in a uniform magnetic field of induction `6 xx 10^(-2)Wb//m^(2)`,with its plane inclined at 60◦ with the field. What is the magnitude of the torque acting on the coil, if a current of 100mA is passed through it?

Options

A. `4 xx 10^(-5)N-m`
B. `8 xx 10^(-5)N-m`
C. `1.2 xx 10^(4)N-m`
D. `2.4 xx 10^(-4)N-m`

Explanation

To find the magnitude of the torque acting on the rectangular coil, we can use the formula for torque (\( \tau \)) acting on a current-carrying coil in a magnetic field:

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

where:
- \( n \) = number of turns in the coil
- \( B \) = magnetic field induction (in Wb/m²)
- \( I \) = current (in Amperes)
- \( A \) = area of the coil (in m²)
- \( \theta \) = angle between the magnetic field and the normal to the plane of the coil

### Step 1: Convert the area from cm² to m²
The area given is \( 20 \, \text{cm}^2 \). To convert it to square meters:

\[ A = 20 \, \text{cm}^2 = 20 \times 10^{-4} \, \text{m}^2 = 0.002 \, \text{m}^2 \]

### Step 2: Convert the current from mA to A
The current given is \( 100 \, \text{mA} \). To convert it to Amperes:

\[ I = 100 \, \text{mA} = 100 \times 10^{-3} \, \text{A} = 0.1 \, \text{A} \]

### Step 3: Identify the values for \( n \), \( B \), and \( \theta \)
From the problem statement:
- \( n = 20 \) turns
- \( B = 6 \times 10^{-2} \, \text{Wb/m}^2 \)
- \( \theta = 60^\circ \)

### Step 4: Calculate \( \sin(\theta) \)
We need to find \( \sin(60^\circ) \):

\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]

### Step 5: Substitute the values into the torque formula
Now we can substitute all the values into the torque formula:

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

Substituting the values:

\[ \tau = 20 \cdot (6 \times 10^{-2}) \cdot (0.1) \cdot (0.002) \cdot \left(\frac{\sqrt{3}}{2}\right) \]

### Step 6: Calculate the torque
Calculating step by step:

1. \( 20 \cdot 6 \times 10^{-2} = 1.2 \)
2. \( 1.2 \cdot 0.1 = 0.12 \)
3. \( 0.12 \cdot 0.002 = 0.00024 \)
4. \( 0.00024 \cdot \frac{\sqrt{3}}{2} \approx 0.00024 \cdot 0.866 = 0.00020784 \)

Thus, the torque \( \tau \) is approximately:

\[ \tau \approx 0.00020784 \, \text{N m} \approx 2.08 \times 10^{-4} \, \text{N m} \]

### Final Answer:
The magnitude of the torque acting on the coil is approximately \( 2.08 \times 10^{-4} \, \text{N m} \).