There are two current-carrying planar coil made each from identical wires of length L. C1 is circular (radius R) and C2 is square (side a). They are so constructed, that they have the same frequency of oscillation when they are placed in the same uniform B and carry the same current. Find a in terms of R.

Hint: Use the formula of frequency of oscillation of a bar magnet.
Step 1: C1 = circular coil of radius R, length L,
number of turns per unit length, n1=L2πR

C2 = square of side a and perimeter L, number of turns per unit length, n2=L4a
 Magnetic moment of C1m1=n1IA1 Magnetic moment of C2
m1=LIR2                         ...i
m2=LIa4                         ...ii
Step 2:
 Moment of inertia of C1I1=MR22                         ...iii Momont of inertia of C2I2=Ma212                         ...iv Frequrncy of C1f1=12πm1BI1
 Frequency of C2f2=12πm2BI2 According to question, f1=f2
                         12πm1BI1=12πm2BI2I1m1=I2m2 or m2m1=I2I1
Step 3:
Plugging the values by Eqs. (i), (ii), (iii) and (iv);
Thus, the value of a is 3R.