An equilateral triangle is made by uniform wires, AB, BC, CA. A current I enters at A and leaves from the mid point of BC. If the lengths of each side of the triangle is L, the magnetic field B at the centroid O of the triangle is

Options

A. `(mu_(0))/(4pi) ((4I)/(L))`
B. `(mu_(0))/(2pi) ((4I)/(L))`
C. `(mu_(0))/(4pi) ((2I)/(L))`
D. zero

Answer

D

Explanation

For AB, magnetic field B is along `o.`,
`B= (mu_(0)i)/(4pi r)(sin 60^(@) + sin 60^(@))`
For AC, B is inside `Ox, B= (mu_(0)i)/(4pi r)(sin 60^(@) + sin 60^(@))`.
For BD, is along `o.,B= (mu_(0)i)/(4pi r) sin 60^(@)`,
For DC, B is inside `Ox, B= (mu_(0)i)/(4pi r) sin 60^(@)`,
Since they are causing electric field in two exactly opposite directions eventually, they will cancel out each other effect. So, net magnetic field is zero, i.e., `B_("net")=0`