Physics 230 April 3 2002

Magnetic field due to a triangular loop

A current I = 5 A flows around a loop in the shape of an equilateral triangle of side 5 cm.  What is the magnetic field 1 m above the loop?  1 m to the side of the loop?

Here we begin by calculating the magnetic moment of the loop.

m = IA = (5 A)½(5 cm)(5 cm sqrt(3)/2) = .0054 A m2 .
The direction of the magnetic moment is into the page.

"1 m above the loop"  is on the axis of the dipole.  The field there has magnitude

B = km 2m/d3 = 10-7 N/A22 x .0054 A m2 /1 m3 = 1.1 nT
and points into the page, parallel to the dipole.

To one side of the loop, the magnetic field is directed opposite the dipole and has magnitude B = km m/d3, one half the previous value, or 0.54 nT.