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

For circular coil `C_1`, `n_1=(L)/(2piR)`, Magnetic moment, `M_1=n_1iA_1=(L)/(2piR)xxixxpiR^2=(LiR)/(2)`
For square coil `C_2`, `n_2=(L)/(4a)`, Magnetic moment, `M_2=n_2iA_2=(L)/(4a)xxixxa^2=(Lia)/(4)`
Moment of inertia of circular coil about the diameter as axis, `I_1=(massxx (radius)^2)/(2)=(mR^2)/(2)`
Moment of inertia of square coil about an axis passing through its centre parallel to breadth
`I_2=(ma^2)/(12)`.
Time period of oscillation of the magnet in magnetic field is given by
`T=2pisqrt(I)/(MB)` and `omega=(2pi)/(T)=sqrt((MB)/(I))`
`:.` `omega_1^2=(M_1B)/(I_1)` and `omega_2^2=(M_2B)/(I_2)`
Given, `omega_1^2=omega_2^2`, so `(M_1)/(I_1)=(M_2)/(I_2)` or `(LIR//2)/(mR^2//2)=(LIa//4)/(ma^2//12)`
On solving, `a=3R`