Two identical wires `A and B` , each of length 'l', carry the same current `I`. Wire A is bent into a circle of radius `R and wire B` is bent to form a square of side 'a' . If ` B_(A) and B_(B)` are the values of magnetic field at the centres of the circle and square respectively , then the ratio `(B_(A))/(B_(B))` is :

Options

A. `(pi^(2)//8)`
B. `(pi^(2)//8sqrt(2))`
C. `(pi^(2)//16)`
D. `(pi^(2)//16sqrt(2))`

Answer

D

Explanation

`B_(1)=(mu_(0))/(4pi)xx(2piI)/R=(mu_(0))/(4R)xx(2piIxx2pi)/L.....(1)`
(`:' L=2piR`, for circular loop)
`B_(2)=(mu_(0))/(4pi)xx1/((a//2))[sin45^(@)+sin45^(@)]xx4`
where `a=(L//4)`
`:. B_(2)=(mu_(0)I)/(4piL)xx8xx4xx[1/(sqrt(2))+1/(sqrt(2))]`
`=(mu_(0)I)/(4piL)xx64/(sqrt(2))`
`:. (B_(1))/(B_(2))=((mu_(0))/(4pi))(4pi^(2)I)/L//(mu_(0))/(4piL)xx641/(sqrt(2))`
or `(B_(1))/(B_(2))=(pi^(2))/(8sqrt(2))`