Question

A loop of wire carrying a current of 2.0A is in the shape of a right triangle with two equal sides, each 15 cm long. A 0.7T uniform magnetic field is in the plane of the triangle and is perpendicular to the hypotenuse. The magnetic force on either of the two equal sides has a magnitude of:

A. zero
B. 0.105N
C. 0.15N
D. 0.21N
E. 0.25N

Answer 1

The magnitude of the magnetic force on either of the two equal sides of the right triangle loop is 0.21N.

Step-by-step explanation:

The magnetic force on each of the two equal sides of the right triangle loop can be calculated using the formula F = I * B * L * sin(theta), where F is the force, I is the current, B is the magnetic field strength, L is the length of the wire, and theta is the angle between the wire and the magnetic field.

In this case, the force on each side of the loop can be calculated as:

1. Length 1: F1 = 2.0A * 0.7T * 0.15m * sin(90°) = 0.21N
2. Length 2: F2 = 2.0A * 0.7T * 0.15m * sin(90°) = 0.21N

Therefore, the magnitude of the magnetic force on either of the two equal sides is 0.21N (option D).