Two circular coils of wire each having a radius of `4cm` and `10 turns` have a common axis and are `3cm` apart If a current of `1A` passes through each coil in the opposite direction find the magnetic induction
(i) At the center of either coil
(ii) At a point on the axis, midway between them .

Answer

To solve the problem, we will calculate the magnetic induction at two different points: (i) at the center of either coil and (ii) at a point on the axis, midway between the two coils.

### Given Data:
- Radius of coils, \( r = 4 \, \text{cm} = 0.04 \, \text{m} \)
- Number of turns in each coil, \( N = 10 \)
- Current in each coil, \( I = 1 \, \text{A} \)
- Distance between the coils, \( d = 3 \, \text{cm} = 0.03 \, \text{m} \)

### (i) Magnetic Induction at the Center of Either Coil

1. **Magnetic Field due to a Circular Coil at its Center**:
The formula for the magnetic field \( B \) at the center of a circular coil is given by:
\[ B = \frac{\mu_0 N I}{2r} \]
where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) is the permeability of free space.

2. **Calculate Magnetic Field for Coil 1**:
\[ B_1 = \frac{(4\pi \times 10^{-7}) \cdot 10 \cdot 1}{2 \cdot 0.04} \]
\[ B_1 = \frac{4\pi \times 10^{-6}}{0.08} = \frac{4\pi \times 10^{-6}}{8 \times 10^{-2}} = \frac{\pi \times 10^{-4}}{2} \approx 1.57 \times 10^{-4} \, \text{T} \]

3. **Magnetic Field for Coil 2**:
Since the current in coil 2 is in the opposite direction, the magnetic field at the center of coil 2 will be:
\[ B_2 = \frac{(4\pi \times 10^{-7}) \cdot 10 \cdot 1}{2 \cdot 0.04} = 1.57 \times 10^{-4} \, \text{T} \]

4. **Net Magnetic Field at the Center**:
Since the fields are in opposite directions:
\[ B_{\text{net}} = B_1 - B_2 = 1.57 \times 10^{-4} - 1.57 \times 10^{-4} = 0 \, \text{T} \]

### (ii) Magnetic Induction at the Midpoint Between the Coils

1. **Distance from Each Coil to Midpoint**:
The distance from each coil to the midpoint is:
\[ x = \frac{d}{2} = \frac{0.03}{2} = 0.015 \, \text{m} \]

2. **Magnetic Field at Midpoint due to Coil 1**:
The formula for the magnetic field at a point on the axis of a coil is given by:
\[ B = \frac{\mu_0 N I}{2} \cdot \frac{r^2}{(r^2 + x^2)^{3/2}} \]
For Coil 1:
\[ B_1' = \frac{(4\pi \times 10^{-7}) \cdot 10 \cdot 1}{2} \cdot \frac{(0.04)^2}{((0.04)^2 + (0.015)^2)^{3/2}} \]

Calculate \( (0.04)^2 + (0.015)^2 \):
\[ (0.04)^2 + (0.015)^2 = 0.0016 + 0.000225 = 0.001825 \]
Now calculate \( (0.001825)^{3/2} \):
\[ (0.001825)^{3/2} \approx 0.000078 \]
Now substituting back:
\[ B_1' \approx \frac{(4\pi \times 10^{-7}) \cdot 10 \cdot 1}{2} \cdot \frac{0.0016}{0.000078} \]

3. **Magnetic Field at Midpoint due to Coil 2**:
The magnetic field \( B_2' \) at the midpoint due to coil 2 will be the same magnitude but in the opposite direction:
\[ B_2' = B_1' \]

4. **Net Magnetic Field at Midpoint**:
Since both magnetic fields are equal in magnitude but opposite in direction:
\[ B_{\text{net}} = B_1' - B_2' = 0 \]

### Final Answers:
(i) The magnetic induction at the center of either coil is \( 0 \, \text{T} \).

(ii) The magnetic induction at the midpoint between the coils is also \( 0 \, \text{T} \).