A circular loop of radius `12` cm carries a current of `7.2` A. A flat coil of radius `0.82` cm, having `50` turns and a current of `1.3` A, is concentric with the loop. The plane of the loop is perpendicular to the plane of the coil. Assume the loop's magnetic field is uniform across the coil. What is the magnitude of (a) the magnetic field produced by the loop at its centre and (b) the torque on the coil due to the loop ?

Answer

To solve the problem step by step, we will address both parts of the question: (a) finding the magnetic field produced by the loop at its center and (b) calculating the torque on the coil due to the loop.

### Step-by-Step Solution

#### Part (a): Magnetic Field at the Center of the Loop

1. **Identify the Given Values:**
- Radius of the loop, \( R = 12 \, \text{cm} = 0.12 \, \text{m} \)
- Current in the loop, \( I = 7.2 \, \text{A} \)
- Permeability of free space, \( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \)

2. **Use the Formula for the Magnetic Field at the Center of a Circular Loop:**
The magnetic field \( B \) at the center of a circular loop is given by the formula:
\[ B = \frac{\mu_0 I}{2R} \]

3. **Substitute the Values into the Formula:**
\[ B = \frac{(4 \pi \times 10^{-7} \, \text{T m/A}) \times (7.2 \, \text{A})}{2 \times (0.12 \, \text{m})} \]

4. **Calculate the Magnetic Field:**
\[ B = \frac{(4 \pi \times 10^{-7} \times 7.2)}{0.24} \approx 3.77 \times 10^{-5} \, \text{T} \]
Converting to microtesla:
\[ B \approx 37.7 \, \mu\text{T} \]

#### Part (b): Torque on the Coil Due to the Loop

1. **Identify the Given Values for the Coil:**
- Radius of the coil, \( r = 0.82 \, \text{cm} = 0.0082 \, \text{m} \)
- Number of turns in the coil, \( n = 50 \)
- Current in the coil, \( I_c = 1.3 \, \text{A} \)

2. **Calculate the Area of the Coil:**
The area \( A \) of the coil is given by:
\[ A = \pi r^2 = \pi (0.0082)^2 \approx 2.11 \times 10^{-4} \, \text{m}^2 \]

3. **Use the Formula for Torque:**
The torque \( \tau \) on the coil in a magnetic field is given by:
\[ \tau = n I_c A B \sin \theta \]
Here, \( \theta = 90^\circ \) (since the plane of the coil is perpendicular to the magnetic field), so \( \sin 90^\circ = 1 \).

4. **Substitute the Values into the Torque Formula:**
\[ \tau = 50 \times (1.3) \times (2.11 \times 10^{-4}) \times (3.77 \times 10^{-5}) \]

5. **Calculate the Torque:**
\[ \tau \approx 50 \times 1.3 \times 2.11 \times 10^{-4} \times 3.77 \times 10^{-5} \approx 5.2 \times 10^{-7} \, \text{N m} \]

### Final Answers:
- (a) The magnetic field produced by the loop at its center is approximately \( 37.7 \, \mu\text{T} \).
- (b) The torque on the coil due to the loop is approximately \( 5.2 \times 10^{-7} \, \text{N m} \).