Two concentric circular coils of ten turns each are situated in the same plane. Their radii are `20` and `40 cm` and they carry respectively `0.2` and `0.3` ampere current in opposite direction. The magnetic field in `Wb//m^(3)` at the centre is

Options

A. `(35//4)mu_(0)`
B. `(mu_(0)//80)`
C. `(7//80)mu_(0)`
D. `(5/4)mu_(0)`

Answer

To find the magnetic field at the center of two concentric circular coils carrying currents in opposite directions, we can use the formula for the magnetic field due to a circular coil at its center:

\[ B = \frac{\mu_0 n I}{2R} \]

where:
- \( B \) is the magnetic field,
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( n \) is the number of turns,
- \( I \) is the current,
- \( R \) is the radius of the coil.

### Step 1: Identify the parameters for both coils
- For coil 1:
- Number of turns \( n_1 = 10 \)
- Radius \( R_1 = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Current \( I_1 = 0.2 \, \text{A} \)

- For coil 2:
- Number of turns \( n_2 = 10 \)
- Radius \( R_2 = 40 \, \text{cm} = 0.4 \, \text{m} \)
- Current \( I_2 = 0.3 \, \text{A} \)

### Step 2: Calculate the magnetic field due to each coil
Using the formula for the magnetic field:

1. **Magnetic field due to coil 1**:
\[ B_1 = \frac{\mu_0 n_1 I_1}{2R_1} = \frac{4\pi \times 10^{-7} \times 10 \times 0.2}{2 \times 0.2} \]
Simplifying this:
\[ B_1 = \frac{4\pi \times 10^{-7} \times 10 \times 0.2}{0.4} = \frac{4\pi \times 10^{-7} \times 10 \times 0.5}{1} = 2\pi \times 10^{-6} \, \text{T} \]

2. **Magnetic field due to coil 2**:
\[ B_2 = \frac{\mu_0 n_2 I_2}{2R_2} = \frac{4\pi \times 10^{-7} \times 10 \times 0.3}{2 \times 0.4} \]
Simplifying this:
\[ B_2 = \frac{4\pi \times 10^{-7} \times 10 \times 0.3}{0.8} = \frac{4\pi \times 10^{-7} \times 10 \times 0.375}{1} = 1.5\pi \times 10^{-6} \, \text{T} \]

### Step 3: Determine the net magnetic field at the center
Since the currents are in opposite directions, the net magnetic field \( B \) at the center will be the difference between \( B_1 \) and \( B_2 \):
\[ B = B_1 - B_2 = 2\pi \times 10^{-6} - 1.5\pi \times 10^{-6} = 0.5\pi \times 10^{-6} \, \text{T} \]

### Step 4: Convert to Wb/m³
Since \( 1 \, \text{T} = 1 \, \text{Wb/m}^2 \), we can express the magnetic field in Wb/m³ as:
\[ B = 0.5\pi \times 10^{-6} \, \text{Wb/m}^2 \]

### Final Answer
\[ B \approx 1.57 \times 10^{-6} \, \text{Wb/m}^2 \]