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

`C_(1)=` circular coil of radius L, number of turns per unit length
`n_(1)=(L)/(2piR)`
`C_(2)=` square of side a and perimeter L, number of turns per unit length `n_(2)=(L)/(4a)`

Magnetic moment of `C_(1)`
`rArr m_(1)=n_(1)IA_(1) `
Magnetic moment of `C_(2)`
`rArr m_(2)=n_(2)IA_(2)`
`m_(1)=(L.I.piR^(2))/(2piR)`
`m_(2)=(L)/(4a) I.a^(2)`
`m_(1)=(LIR)/(2)`
`m_(2)=(LIa)/(4)`
Moment of inertia of `C_(1)rArrI_(1)=(MR^(2))/(2)`
Moment of inertia of `C_(2)rArrI_(2)=(Ma^(2))/(12)`
Frequency of `C_(1)rArr=2pisqrt((I_(1))/(m_(1)B))`
Frequency of `C_(2)rArr=2pisqrt((I_(2))/(m_(2)B))`
According to question, `f_(1)=f_(2)`
`2pisqrt((I_(1))/(m_(1)B))=2pisqrt((I_(2))/(M_(2)B))`
`(I_(1))/(m_(1))=(I_(2))` or `(m_(2))/(m_(1))=(I_(2))/(I_(1))`
Plugging the values by Eqs. (i),(ii) and (iii) and (iv)
`(LIa.2)/(4xxLIR)=(Ma^(2).2)/(12.MR^(2))`
`(a)/(2R)=(a^(2))/(6R^(2))`
3R=a
Thus, the value of a is 3R.