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.