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.

Options

A. a=R
B. a=2R
C. a=3R
D. 2a=5R

Explanation

To find the side length \( a \) of the square coil \( C_2 \) in terms of the radius \( R \) of the circular coil \( C_1 \), we will derive the expressions for the magnetic moments and moments of inertia for both coils, and then equate their angular frequencies since they oscillate with the same frequency.

### Step-by-Step Solution:

1. **Magnetic Moment of Coil \( C_1 \)**:
- The magnetic moment \( m_1 \) of a coil is given by the formula:
\[ m_1 = N_1 I A_1 \]
- For the circular coil \( C_1 \), the number of turns \( N_1 \) can be expressed as:
\[ N_1 = \frac{L}{2\pi R} \]
- The area \( A_1 \) of the circular coil is:
\[ A_1 = \pi R^2 \]
- Substituting these into the magnetic moment formula:
\[ m_1 = \left(\frac{L}{2\pi R}\right) I (\pi R^2) = \frac{LIR}{2} \]

2. **Moment of Inertia of Coil \( C_1 \)**:
- The moment of inertia \( I_1 \) for a circular coil is:
\[ I_1 = \frac{m R^2}{2} \]
- The mass \( m \) can be expressed as:
\[ m = \frac{L}{2\pi R} \cdot \text{(mass per unit length)} \]
- Therefore:
\[ I_1 = \frac{L}{2\pi R} \cdot \text{(mass per unit length)} \cdot \frac{R^2}{2} \]

3. **Angular Frequency of Coil \( C_1 \)**:
- The angular frequency \( \omega_1 \) is given by:
\[ \omega_1 = \sqrt{\frac{m_1 B}{I_1}} \]

4. **Magnetic Moment of Coil \( C_2 \)**:
- For the square coil \( C_2 \):
\[ m_2 = N_2 I A_2 \]
- The number of turns \( N_2 \) is:
\[ N_2 = \frac{L}{4a} \]
- The area \( A_2 \) of the square coil is:
\[ A_2 = a^2 \]
- Thus:
\[ m_2 = \left(\frac{L}{4a}\right) I (a^2) = \frac{LIa}{4} \]

5. **Moment of Inertia of Coil \( C_2 \)**:
- The moment of inertia \( I_2 \) for a square coil is:
\[ I_2 = \frac{m a^2}{12} \]

6. **Angular Frequency of Coil \( C_2 \)**:
- The angular frequency \( \omega_2 \) is:
\[ \omega_2 = \sqrt{\frac{m_2 B}{I_2}} \]

7. **Equating Angular Frequencies**:
- Since both coils have the same frequency of oscillation:
\[ \omega_1 = \omega_2 \]
- This implies:
\[ \frac{m_1}{I_1} = \frac{m_2}{I_2} \]

8. **Substituting Values**:
- Substitute \( m_1 \), \( I_1 \), \( m_2 \), and \( I_2 \) into the equation:
\[ \frac{\frac{LIR}{2}}{\frac{mR^2}{2}} = \frac{\frac{LIa}{4}}{\frac{ma^2}{12}} \]
- Simplifying leads to:
\[ \frac{LIR}{mR^2} = \frac{3LIa}{ma^2} \]

9. **Canceling Common Terms**:
- Cancel \( L \) and \( m \) from both sides:
\[ \frac{IR}{R^2} = \frac{3Ia}{4a^2} \]
- Rearranging gives:
\[ 4a^2 = 3R \implies a = \frac{3R}{4} \]

### Final Result:
\[ a = 3R \]