A circular loop of area `0.02 m^2` carrying a current of 10A, is held with its plane perpendicular to a magnetic field induction 0.2T. The torque acting on the loop is

Options

A. 0.01 m
B. 0.001 Nm
C. zero
D. 0.8 Nm

Answer

To find the torque acting on a circular loop placed in a magnetic field, we can use the formula for torque (\( \tau \)) acting on a current-carrying loop in a magnetic field:

\[ \tau = \vec{m} \times \vec{B} \]

Where:
- \( \tau \) is the torque,
- \( \vec{m} \) is the magnetic moment of the loop,
- \( \vec{B} \) is the magnetic field induction.

### Step-by-Step Solution:

1. **Identify the given values:**
- Area of the loop (\( A \)) = \( 0.02 \, m^2 \)
- Current (\( I \)) = \( 10 \, A \)
- Magnetic field induction (\( B \)) = \( 0.2 \, T \)

2. **Calculate the magnetic moment (\( m \)):**
The magnetic moment (\( m \)) for a circular loop is given by the formula:
\[ m = I \cdot A \]
Substituting the values:
\[ m = 10 \, A \times 0.02 \, m^2 = 0.2 \, A \cdot m^2 \]

3. **Determine the angle between the magnetic moment and the magnetic field:**
Since the loop is held with its plane perpendicular to the magnetic field, the angle (\( \theta \)) between the magnetic moment (\( m \)) and the magnetic field (\( B \)) is \( 90^\circ \).

4. **Calculate the torque (\( \tau \)):**
The torque can be calculated using the formula:
\[ \tau = m \cdot B \cdot \sin(\theta) \]
Since \( \theta = 90^\circ \), we have \( \sin(90^\circ) = 1 \):
\[ \tau = 0.2 \, A \cdot m^2 \times 0.2 \, T \times 1 = 0.04 \, N \cdot m \]

5. **Final Result:**
The torque acting on the loop is:
\[ \tau = 0.04 \, N \cdot m \]