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)/(16 sqrt2)`
C. `pi^(2)/16`
D. `pi^(2)/(8sqrt2)`

Answer

b

Explanation

Let r be the radius of circle formed then we can write ` l = 2 pi r rArr r = l//2 pi`
Magnetic field at the centre of circle
`B_(A) = (mu_(0)I)/(2 r) = (mu_(0)I)/(2 xx l/(2 pi)) = (mu_(0) pi I)/l `

There are four sides hence magnetic field due to one side is multiplied to four.
`B_(B) = 4 xx (mu_(0)I)/(2 pi l/8) ( sin 45 + sin 45 ) = (16 mu_(0) I sqrt2)/(pi l)`
`B_(A)/B_(B) = ((mu_(0)pi I)/l)/((16 mu_(0)I sqrt2)/(pi l)) = pi^(2)/(16 sqrt2) `
Thus, option (b) is correct.