A rectangular coil 6 cm long and 2 cm wide is placed in a magnetic field of 0.02 T. If the loop contains 200 turns and carries a current of 50 mA, the torque on the coil (with loop face parallel to the field) is

Options

A. A) `2.4 xx 10^(-4) Nm`
B. B) `2.4 Nm`
C. C) `0.24 Nm`
D. D) `0.024 Nm`

Answer

To find the torque on the rectangular coil placed in a magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Length of the coil (l) = 6 cm = 0.06 m
- Width of the coil (w) = 2 cm = 0.02 m
- Magnetic field strength (B) = 0.02 T
- Number of turns (n) = 200
- Current (I) = 50 mA = 50 × 10^(-3) A = 0.050 A

2. **Calculate the Area of the Rectangular Coil:**
\[ \text{Area (A)} = \text{Length} \times \text{Width} = l \times w = 0.06 \, \text{m} \times 0.02 \, \text{m} = 0.0012 \, \text{m}^2 \]

3. **Use the Formula for Torque (τ):**
The torque (τ) on a rectangular coil in a magnetic field is given by the formula:
\[ \tau = n \times B \times I \times A \]
where:
- \( n \) = number of turns
- \( B \) = magnetic field strength
- \( I \) = current
- \( A \) = area of the coil

4. **Substitute the Values into the Formula:**
\[ \tau = 200 \times 0.02 \, \text{T} \times 0.050 \, \text{A} \times 0.0012 \, \text{m}^2 \]

5. **Perform the Calculation:**
\[ \tau = 200 \times 0.02 \times 0.050 \times 0.0012 \]
\[ = 200 \times 0.0000012 = 0.00024 \, \text{N m} = 2.4 \times 10^{-4} \, \text{N m} \]

6. **Conclusion:**
The torque on the coil is \( \tau = 2.4 \times 10^{-4} \, \text{N m} \).