There are two current carrying planar coils made each from identical wires of length L. `C_1` is the circular (radius R) and `C_2` is square (side a). They are so constructed that they have same frequency of oscillation when they are placed in the same uniform `vecB` and carry the same current i. Find a in terms of R.

Explanation

To solve the problem, we need to find the side length \( a \) of the square coil \( C_2 \) in terms of the radius \( R \) of the circular coil \( C_1 \). Both coils are made from identical wires of length \( L \) and carry the same current \( I \) while oscillating in the same uniform magnetic field \( \vec{B} \).

### Step-by-Step Solution:

1. **Determine the number of turns for the square coil \( C_2 \)**:
The perimeter of the square coil is \( 4a \). Since the total length of the wire is \( L \), the number of turns \( N_1 \) in the square coil is given by:
\[
N_1 = \frac{L}{4a}
...