A 100 turn closely wound circular coil of radius 10 cm carries a current of 3.2 A.
The coil is placed in a vertical plane and is free to rotate. A uniform magnetic field of 2T, rotates the coil through an angle of `90^@`. What is the magnitude of torque acting on the coil?

Options

A. 10 Nm
B. 20 Nm
C. 50 Nm
D. 25 Nm

Explanation

To find the magnitude of torque acting on a circular coil placed in a magnetic field, we can follow these steps:

### Step 1: Identify the given values
- Number of turns (N) = 100
- Current (I) = 3.2 A
- Radius of the coil (R) = 10 cm = 0.1 m (conversion from cm to m)
- Magnetic field (B) = 2 T
- Angle (θ) = 90° (the angle through which the coil is rotated)

### Step 2: Calculate the area (A) of the coil
The area of a circular coil is given by the formula:
\[ A = \pi R^2 \]
Substituting the radius:
\[ A = \pi (0.1)^2 = \pi (0.01) = 0.01\pi \, \text{m}^2 \]

### Step 3: Calculate the magnetic moment (μ)
The magnetic moment (μ) of the coil is given by the formula:
\[ \mu = NIA \]
Substituting the values:
\[ \mu = 100 \times 3.2 \times (0.01\pi) \]
\[ \mu = 100 \times 3.2 \times 0.01 \times 3.14 \]
\[ \mu = 3.2 \times 3.14 = 10.048 \, \text{A m}^2 \]

### Step 4: Calculate the torque (τ)
The torque (τ) acting on the coil in a magnetic field is given by:
\[ \tau = \mu B \sin \theta \]
Substituting the values:
\[ \tau = 10.048 \times 2 \times \sin(90^\circ) \]
Since \(\sin(90^\circ) = 1\):
\[ \tau = 10.048 \times 2 \]
\[ \tau = 20.096 \, \text{N m} \]

### Step 5: Round the answer
We can approximate the torque to:
\[ \tau \approx 20 \, \text{N m} \]

### Final Answer
The magnitude of the torque acting on the coil is approximately **20 N m**.

---