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)`

Explanation

To find the magnetic field at the center of two concentric circular coils carrying currents in opposite directions, we can follow these steps:

### Step 1: Identify the parameters
- 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 Coil 1
The magnetic field at the center of a circular coil is given by the formula:
\[ B = \frac{\mu_0 n I}{2r} \]
Where:
- \( \mu_0 \) is the permeability of free space, \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)
- \( n \) is the number of turns
- \( I \) is the current
- \( r \) is the radius of the coil

For Coil 1:
\[ B_1 = \frac{\mu_0 n_1 I_1}{2 r_1} = \frac{4\pi \times 10^{-7} \times 10 \times 0.2}{2 \times 0.2} \]
\[ B_1 = \frac{4\pi \times 10^{-7} \times 10 \times 0.2}{0.4} = \frac{4\pi \times 10^{-7} \times 10 \times 0.5}{1} = 2\pi \times 10^{-6} \, \text{T} \]

### Step 3: Calculate the magnetic field due to Coil 2
For Coil 2:
\[ B_2 = \frac{\mu_0 n_2 I_2}{2 r_2} = \frac{4\pi \times 10^{-7} \times 10 \times 0.3}{2 \times 0.4} \]
\[ B_2 = \frac{4\pi \times 10^{-7} \times 10 \times 0.3}{0.8} = \frac{4\pi \times 10^{-7} \times 10 \times 0.375}{1} = 1.5\pi \times 10^{-6} \, \text{T} \]

### Step 4: Determine the net magnetic field
Since the currents are in opposite directions, the magnetic fields will also be in opposite directions. Therefore, the net magnetic field \( B \) at the center will be:
\[ B = B_1 - B_2 \]
Substituting the values:
\[ B = 2\pi \times 10^{-6} - 1.5\pi \times 10^{-6} = 0.5\pi \times 10^{-6} \, \text{T} \]

### Step 5: Convert to Wb/m³
To express the magnetic field in Wb/m³ (Tesla), we can write:
\[ B = 0.5\pi \times 10^{-6} \, \text{T} \approx 1.57 \times 10^{-6} \, \text{T} \]

### Final Answer
Thus, the magnetic field at the center is approximately:
\[ B \approx 1.57 \times 10^{-6} \, \text{Wb/m}^2 \]