Two concentric coils of 10 turns each are placed in the same plane. Their radii are 20 cm and 40 cm and carry 0.2 A and 0.3 A current respectively in opposite directions. The magnetic induction (in tesla) at the centre is

Options

A. `3/4mu_(0)`
B. `5/4mu_(0)`
C. `7/4mu_(0)`
D. `9/4mu_(0)`

Explanation

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

### Step 1: Identify the parameters
- Number of turns in each coil (N1 = N2) = 10 turns
- Radius of the first coil (R1) = 20 cm = 0.2 m
- Radius of the second coil (R2) = 40 cm = 0.4 m
- Current in the first coil (I1) = 0.2 A
- Current in the second coil (I2) = 0.3 A

### Step 2: Formula for magnetic induction
The magnetic induction (B) at the center of a circular coil is given by the formula:
\[ B = \frac{\mu_0 N I}{2R} \]
where:
- \( \mu_0 \) = permeability of free space = \( 4\pi \times 10^{-7} \, \text{T m/A} \)
- N = number of turns
- I = current
- R = radius of the coil

### Step 3: Calculate magnetic induction for each coil
**For the first coil (B1):**
\[ B_1 = \frac{\mu_0 N_1 I_1}{2 R_1} = \frac{\mu_0 \cdot 10 \cdot 0.2}{2 \cdot 0.2} \]
\[ B_1 = \frac{\mu_0 \cdot 10 \cdot 0.2}{0.4} = \frac{\mu_0 \cdot 10 \cdot 0.5}{1} = 5 \mu_0 \]

**For the second coil (B2):**
\[ B_2 = \frac{\mu_0 N_2 I_2}{2 R_2} = \frac{\mu_0 \cdot 10 \cdot 0.3}{2 \cdot 0.4} \]
\[ B_2 = \frac{\mu_0 \cdot 10 \cdot 0.3}{0.8} = \frac{\mu_0 \cdot 10 \cdot 0.375}{1} = 3.75 \mu_0 \]

### Step 4: Determine the direction of magnetic fields
- The magnetic field due to the first coil (B1) is directed into the plane of the paper (let's consider this as positive).
- The magnetic field due to the second coil (B2) is directed out of the plane of the paper (let's consider this as negative).

### Step 5: Calculate net magnetic induction
Since the currents are in opposite directions, we subtract the two magnetic fields:
\[ B_{net} = B_1 - B_2 = 5 \mu_0 - 3.75 \mu_0 = 1.25 \mu_0 \]

### Step 6: Substitute the value of \( \mu_0 \)
\[ B_{net} = 1.25 \cdot (4\pi \times 10^{-7}) \, \text{T} \]
\[ B_{net} = 5\pi \times 10^{-7} \, \text{T} \]

### Final Answer
The magnetic induction at the center due to the two coils is:
\[ B_{net} \approx 1.57 \times 10^{-6} \, \text{T} \quad \text{(approximately)} \]