A circular loop of area `0.01 m^(2)` carrying a current of `10 A`, is held perpendicular to a magnetic field of intensity `0.1 T`. The torque acting on the loop is

Options

A. (a)0.01 Nm
B. (b)0.001 Nm
C. (c)zero
D. (d)0.8 Nm

Explanation

To solve the problem of finding the torque acting on a circular loop in a magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Area of the loop, \( A = 0.01 \, m^2 \)
- Current flowing through the loop, \( I = 10 \, A \)
- Magnetic field intensity, \( B = 0.1 \, T \)

2. **Calculate the magnetic moment (\( \mu \)) of the loop:**
The magnetic moment (\( \mu \)) for a current-carrying loop is given by the formula:
\[ \mu = I \times A \]
Substituting the values:
\[ \mu = 10 \, A \times 0.01 \, m^2 = 0.1 \, A \cdot m^2 \]

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

4. **Use the formula for torque (\( \tau \)):**
The torque (\( \tau \)) acting on the loop in a magnetic field is given by:
\[ \tau = \mu \times B \times \sin(\theta) \]
Substituting the values we have:
\[ \tau = 0.1 \, A \cdot m^2 \times 0.1 \, T \times \sin(90^\circ) \]

5. **Calculate \( \sin(90^\circ) \):**
We know that \( \sin(90^\circ) = 1 \).

6. **Calculate the torque:**
\[ \tau = 0.1 \, A \cdot m^2 \times 0.1 \, T \times 1 = 0.01 \, N \cdot m \]

### Final Answer:
The torque acting on the loop is \( 0.01 \, N \cdot m \).

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