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

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} * \overline {B}`

For each segment:

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

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

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

`F=\int\limits^((0,a))_((a,0)) {I} \, (√(2) )/(2) (\overline {x}+\overline {y})dx * 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 * 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`