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

Answer

C

Explanation

`"For coil" C_(1)`
`n_(1)=(L)/(2piR),m_(1)=n_(1)iA_(1)=((L)/(2piR))(i)(piR^(2))=(LiR)/(2)`
`I_(1)` (moment of inertia of the coil `C_(1)` about an axis through its diameter) `=(1)/(2)MR^(2)`
`"Also", omega_(1)=sqrt((m_(1)B)/(I_(1))) ("as" T_(1)=2pi sqrt(I_(1))/(mB))`
For the coil `C_(2)`,
As, `n_(2)=(L)/(4a), m_(2)=n_(2)i A_(2)=((L)/(4a))(i)(a^(2))=(Lia)/(4)`
`I_(2)` (momenty of inertia of coil `C_(2)` about an axis through its center and parallel to one of its sides) `=(1)/(12)Ma^(2)`
`"Also", omega_(2)=sqrt(m_(2)B)/(I_(2))`
`"Since", omega_(1)=omega_(2),(m_(1))/(l_(1))=(omega_(2))/(I_(2))`
`or (LiR//2)/((1)/(2)MR^(2))=(LiA//4)/((1)/(12)Ma^(2)) or (Li)/(MR)=(3Li)/(Ma)`
Here, a=3R