Four identical wires of length L each are bent into four different planar figures: an equilateral triangle, a square, a regular hexagon and a circle. The loops are kept far away from each other and made to carry equal current I. The magnetic field intensity at the centre of the figures are recorded as `B_(T),B_(S),B_(H)` and `B_(C)`. Choose the correct option(s):

Options

A. `B_(T)=(27mu_(0)I)/(2piL)`
B. `B_(H)=(6 sqrt(3)mu_(0)I)/(piL)`
C. `B_(S)=(8sqrt(2)mu_(0)I)/(piL)`
D. `B_(T)gtB_(S)gtB_(H)gtB_(C)`

Answer

A, B, C, D

Explanation

Let the side length of the triangle be a
Magnetic field at the centre due to one side of the triangle,
`B_(1)=(mu_(0)I)/(4pi(a/(2sqrt(3))))(sin 60^(@)+sin 60^(@))=(3mu_(0)I)/(2pia)`
Therefore, net magnetic field at the centre of the triangle,
`B_(T)=3B_(1)=(9mu_(0)I)/(2pia)`. But `a=L/3` therefore `B_(T)=(27mu_(0)I)/(2piL0`
b. Let the side length of the square be a
Magnetic field at the centre due to one side of the square,
`B_(1)=(mu_(0)I)/(4pi(a/2))(sin 45^(@)+sin 45^(@))=(mu_(0)I)/(sqrt(2)pia)`
Therefore, net magnetic field at the centre due to square,
`B_(S)=4B_(1)=(2sqrt(2)mu_(0)I)/(pia)`
But `a=L/4` therefore `B_(S)=(8sqrt(2)mu_(0)I)/(piL)`
(c) Let the side length of the hexagon be a
Magnetic field at the centre due to one side of the hexagon,
`B_(1)=(mu_(0)I)/(4pi((asqrt(3))/2))(sin 30^(@)+sin 30^(@))=(mu_(0)I)/(2sqrt(3)pia)`
Therefore, net magnetic field at the centre due to hexagon,
`B_(H)=6B_(1)=(sqrt(3)mu_(0)I)/(pia)`, But `a=L/6`, therefore `B_(H)=(6sqrt(3)mu_(0)I)/(piL)`
(d) Let the radius of the circle be R , We already know that `B_(C)=(mu_(0)I)/(2R)`
But `R=L/(2pi)` therefoe `B_(C)=(pi mu_(0)I)/L`