A rectangular coil `20cmxx20cm` has `100` turns and carries a current of `1 A`. It is placed in a uniform magnetic field `B=0.5 T` with the direction of magnetic field parallel to the plane of the coil. The magnitude of the torque required to hold this coil in this position is

Options

A. Zero
B. 200 Nm
C. 2 Nm
D. 10 Nm

Answer

To solve the problem of finding the torque required to hold a rectangular coil in a magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Length of the coil (l) = 20 cm = 0.2 m
- Width of the coil (w) = 20 cm = 0.2 m
- Number of turns (N) = 100
- Current (I) = 1 A
- Magnetic field (B) = 0.5 T

2. **Calculate the Area of the Coil:**
\[ \text{Area (A)} = l \times w = 0.2 \, \text{m} \times 0.2 \, \text{m} = 0.04 \, \text{m}^2 \]

3. **Calculate the Magnetic Moment (m):**
The magnetic moment (m) of the coil can be calculated using the formula:
\[ m = N \times I \times A \]
Substituting the values:
\[ m = 100 \times 1 \, \text{A} \times 0.04 \, \text{m}^2 = 4 \, \text{A m}^2 \]

4. **Determine the Angle (θ):**
The angle (θ) between the magnetic moment and the magnetic field is 90 degrees since the magnetic field is parallel to the plane of the coil. Thus, \(\sin(90^\circ) = 1\).

5. **Calculate the Torque (τ):**
The torque (τ) can be calculated using the formula:
\[ \tau = m \times B \times \sin(\theta) \]
Substituting the values:
\[ \tau = 4 \, \text{A m}^2 \times 0.5 \, \text{T} \times 1 = 2 \, \text{N m} \]

6. **Conclusion:**
The magnitude of the torque required to hold the coil in position is \(2 \, \text{N m}\).

### Final Answer:
The torque required to hold the coil in this position is **2 N m**.

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