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 a circular coil (coil 1) and a quarter disc coil (coil 2) given that the magnetic field induction at their centers is equal and both coils have equal areas.

### Step-by-Step Solution:

1. **Identify the Areas of the Coils:**
- For coil 1 (circular coil), the area \( A_1 \) is given by:
\[ A_1 = \pi R_1^2 \]
- For coil 2 (quarter disc), the area \( A_2 \) is:
\[ A_2 = \frac{1}{4} \pi R_2^2 \]

2. **Set the Areas Equal:**
Since both coils have equal areas:
\[ A_1 = A_2 \implies \pi R_1^2 = \frac{1}{4} \pi R_2^2 \]
Dividing both sides by \( \pi \):
\[ R_1^2 = \frac{1}{4} R_2^2 \]
Taking the square root:
\[ R_1 = \frac{1}{2} R_2 \implies \frac{R_2}{R_1} = 2 \]

3. **Magnetic Field Induction for Coil 1:**
The magnetic field induction \( B_1 \) at the center of a circular coil is given by:
\[ B_1 = \frac{\mu_0 I_1}{2 R_1} \]

4. **Magnetic Field Induction for Coil 2:**
The magnetic field induction \( B_2 \) at the center of a quarter disc is given by:
\[ B_2 = \frac{\mu_0 I_2}{8 R_2} \]

5. **Set the Magnetic Fields Equal:**
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:
\[ \frac{I_1}{2 R_1} = \frac{I_2}{8 R_2} \]

6. **Substituting \( R_2 \) in terms of \( R_1 \):**
From the previous step, we know \( R_2 = 2 R_1 \):
\[ \frac{I_1}{2 R_1} = \frac{I_2}{8 \cdot 2 R_1} \]
Simplifying:
\[ \frac{I_1}{2 R_1} = \frac{I_2}{16 R_1} \]
Canceling \( R_1 \) from both sides:
\[ \frac{I_1}{2} = \frac{I_2}{16} \]

7. **Finding the Ratio of Currents:**
Rearranging gives:
\[ I_2 = 8 I_1 \]
Thus, 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:
\[ \frac{I_2}{I_1} = 8:1 \]