Question

A constant current I circulates counterclockwise around a wire loop in the shape of a triangle, lying in x-y plane, with vertices at the points (0, 0, 0), (a, 0, 0), and (0,a,0). Show that the total force on the loop is zero in the presence of a constant magnetic field B = Bo(z).

Answer 1

Answer:

Considering a triangle like the one of the figures, you can obtain the total magnetic force applied on it like the addition of the forces applied to each of the 3 sides.

Been the magnetic force formula:

$F=\int\limits^a_b {I} \, \overline {dx} \times \overline {B}$

For each segment:

Segment 1 (from [0,0,0] to [a,0,0]):

$F=\int\limits^a_0 {I} \, \overline {x}dx \times B \overline {z}=-IBa\overline {y}$

Segment 2 (from [a,0,0] to [0,a,0]):

$F=\int\limits^{(0,a)}_{(a,0)} {I} \, \frac{\sqrt{2} }{2} (\overline {x}+\overline {y})dx \times B \overline {z}=IB2a(\overline {x}+\overline {y})$

Segment 3 (From [0,a,0] to [0,0,0]):

$F=\int\limits^0_a {I} \, \overline {y}dx \times B \overline {z}=-IBa\overline {x}$

Total force on the wire loop:

$F_{T} =F_{1} +F_{2} +F_{3} =-IBa\overline {y}+IB2a(\overline {x}+\overline {y})-IBa\overline {x}=0N$