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

Answer

D

Explanation

For A, `l=2piR` or `R=l//2pi`
`B_A=(mu_0I)/(2R)=(mu_0I)/(2xxl//2pi)=(mu_0piI)/(l)`

`l=4a` or `a=l//4`
`OT=a//2=l//8`
Magnetic field at O due to current through a square is
`B_B=4xx(mu_0)/(4pi)(I)/((OT))[sin 45^@+sin 45^@]`
`=4xx(mu_0)/(4pi)(I)/(l//8)[1/sqrt2+1/sqrt2]=(16mu_0I)/(pisqrt2l)`
`:. B_A/B_B=((mu_0piI//l))/((16mu_0I//sqrt2pil))=(sqrtpi^2)/(16)=(pi^2)/(8sqrt2)`