A loop of wire carrying a current of 2.0 A is in the shape of a right triangle with two equal sides, each 15 cm long.A 0.7 T uniform magnetic field is parallel to the hypotenuse.The total magnetic force on the two equal sides has a magnitude of:
A)0 N
B)0.21 N
C)0.30 N
D)0.41 N
E)0.51 N

Question

Quizplus · Accepted Answer

The magnetic force on a current-carrying wire is given by $F = ILB \sin(\theta)$, where $I$ is the current, $L$ is the length of the wire, $B$ is the magnetic field strength, and $\theta$ is the angle between the wire and the magnetic field. In this case, the magnetic field is parallel to the hypotenuse, meaning it is perpendicular to the two equal sides of the triangle. However, since the forces on the two equal sides are equal in magnitude but opposite in direction (due to the symmetry of the situation and the direction of the current), they cancel each other out, resulting in a net force of 0 N.

A Loop of Wire Carrying a Current of 2