The magnetic field induction at the centre of a current - carrying circular coil (coil 1) and a closed coil (coil 2), shaped as a quarter of a disc is found to be equal in magnitude. If both the coils have equal area, then find the ratio of the currents flowing in coil 2 and coil 1.

Answer

To solve the problem, we need to find the ratio of the currents flowing in two coils, given that the magnetic field induction at the center of both coils is equal and that both coils have the same area.

### Step-by-Step Solution:

1. **Understanding the Coils**:
- Coil 1 is a circular coil.
- Coil 2 is a quarter of a circular disc.
- Let's denote the radius of coil 1 as \( r_1 \) and the radius of coil 2 as \( r_2 \).

2. **Area of the Coils**:
- The area of coil 1 (circular) is given by:
\[ A_1 = \pi r_1^2 \]
- The area of coil 2 (quarter disc) is given by:
\[ A_2 = \frac{1}{4} \pi r_2^2 \]
- Since the areas are equal:
\[ \pi r_1^2 = \frac{1}{4} \pi r_2^2 \]
- Canceling \( \pi \) from both sides:
\[ r_1^2 = \frac{1}{4} r_2^2 \]
- Rearranging gives:
\[ r_2^2 = 4 r_1^2 \quad \Rightarrow \quad r_2 = 2 r_1 \]

3. **Magnetic Field Induction**:
- The magnetic field induction at the center of coil 1 (circular) is given by:
\[ B_1 = \frac{\mu_0 I_1}{2 r_1} \]
- The magnetic field induction at the center of coil 2 (quarter disc) can be calculated considering only the arc's contribution:
\[ B_2 = \frac{\mu_0 I_2 \theta}{4 \pi r_2} \]
- For coil 2, since it is a quarter disc, \( \theta = \frac{\pi}{2} \):
\[ B_2 = \frac{\mu_0 I_2 \frac{\pi}{2}}{4 \pi r_2} = \frac{\mu_0 I_2}{8 r_2} \]

4. **Equating the Magnetic Inductions**:
- Since \( B_1 = B_2 \):
\[ \frac{\mu_0 I_1}{2 r_1} = \frac{\mu_0 I_2}{8 r_2} \]
- Canceling \( \mu_0 \) from both sides:
\[ \frac{I_1}{2 r_1} = \frac{I_2}{8 r_2} \]
- Cross-multiplying gives:
\[ 8 I_1 r_2 = 2 I_2 r_1 \]

5. **Substituting \( r_2 \)**:
- Substitute \( r_2 = 2 r_1 \):
\[ 8 I_1 (2 r_1) = 2 I_2 r_1 \]
- Simplifying:
\[ 16 I_1 r_1 = 2 I_2 r_1 \]
- Dividing both sides by \( r_1 \) (assuming \( r_1 \neq 0 \)):
\[ 16 I_1 = 2 I_2 \]
- Rearranging gives:
\[ I_2 = 8 I_1 \]

6. **Finding the Ratio**:
- The ratio of the currents \( \frac{I_2}{I_1} \) is:
\[ \frac{I_2}{I_1} = 8 \]

### Final Answer:
The ratio of the currents flowing in coil 2 and coil 1 is:
\[ \frac{I_2}{I_1} = 8 \]