A wire shaped to a regular hexagon of side x carries a current I ampere. Calculate the strength of the magnetic field at the centre of the hexagon.

Options

A. `(sqrt3 mu_(0)I)/(pi x)`
B. `(3mu_(0)I)/(pi x)`
C. `(mu_(0)I)/(3pi x)`
D. `(mu_(0)I)/(sqrt3 pi x)`

Answer

A

Explanation

Let ABCDEF be a regular hexagon of side x m with its centre at P (figure). The magnetic field at P due to the current I (ampere) in the arm AB is given by
`B=(mu_(0))/(4pi). (I)/(a) (sin theta_(1) + sin theta_(2))`
where a= OP
Here, `theta_(1)= theta_(2)= 30^(@)` (from the geometry of the figure), PB= AB=x and a = OP = PB `cos 30^(@)= x.(sqrt3)/(2)`
`:. B= (mu_(0))/(4pi) .(I)/(x.(sqrt3)/(2)) (sin 30^(@) + sin 30^(@)) = (mu_(0))/(4pi) .(2I)/(sqrt3x) ((1)/(2) + (1)/(2)`
`=(mu_(0))/(2pi).(I)/(sqrt3x)` tesla.
The direction of B at point P is perpendicular to the plane of the page directed downward. Similarly, the magnetic fields at P due to the other arms of the hexagon, i.e, BC, CD, DE, EF and FA are same and act in the same direction. Therefore, the total magneitc field at `P= 6B = 6 xx (mu_(0))/(2pi).(I)/(sqrt3x)= (sqrt3 mu_(0)I)/(pi x)` tesla
acting in a direction perpendicular to the plane of the page direction downward.