A symmetric star shaped conducting wire loop is carrying a steady state current I as shown in the figure. The distance between the diametrically opposite vertices of the star is 4a. The magnitude of the magnetic field at the center of the loop is

Options

A. `(mu_(0)I)/(4pi a) 3[2-sqrt3]`
B. `(mu_(0)I)/(4pi a)6[sqrt3-1]`
C. `(mu_(0)I)/(4pi a)3[sqrt3-I]`
D. `(mu_(0)I)/(4pi a)6 [sqrt3+1]`

Answer

B

Explanation

Let us first calculate the magnetic field due to element AB. Perpendicular distance of centre O from element AB, OP= a
Angle subtended by centre at element are `theta_(1)= 30^(@), theta_(2)= 60^(@)` as shown in figure.
`:. vec(B) =(mu_(0)I)/(4pi a) (cos 30^(@) - cos 60^(@)) o.`
`=(mu_(0)I)/(4pi a) ((sqrt3)/(2)-(1)/(2)) o. = (mu_(0)I)/(8pi a) (sqrt3-1) o.`
Since the direction of magnetic field due to each element side is out of the paper.
`:.` Net magnetic field at `O= 12 xx (mu_(0)I)/(8pi a) (sqrt3-1)`
`=(mu_(0)I)/(4pi a)6 (sqrt3-1)`