AP Physics 2 Exam Question

A long straight wire carries a current of 5 A in the positive x-direction. The wire is placed in a uniform magnetic field of 2 T in the positive y-direction. A square loop of side length 10 cm is also placed in the same magnetic field. The loop is oriented such that one side is parallel to the wire and is located 5 cm away from the wire. The loop is now moved outwards away from the wire with an acceleration of 2 m/s². Determine the magnitude and direction of the net force acting on the loop at this instant.

Answer with Step-by-Step Explanation

The magnetic force on a current-carrying wire is given by the formula:

F = |I| * |B| * L * sin(θ)

where F is the magnitude of the force, |I| is the magnitude of current, |B| is the magnitude of magnetic field, L is the length of the conductor, and θ is the angle between the current direction and the magnetic field.

In this case, we are interested in finding the total force on the square loop. Since the loop is composed of four wires, the net force on the loop will be the sum of forces on individual wires. Let's start by calculating the force on one side of the loop.

The length of one side of the square loop is 10 cm = 0.1 m. Given that the current is 5 A and the magnetic field is 2 T, the force on one side of the loop is given by:

F₁ = |I| * |B| * L * sin(θ) = |5 A| * |2 T| * 0.1 m * sin(90°) (since the side is parallel to the wire) = 10 * 0.1 * 1 * 1 = 1 N

Since the other three sides of the square loop experience the same force due to symmetry, the net force on the loop will be:

Net force = 4 * F₁ = 4 N

Therefore, the magnitude of the net force acting on the loop is 4 N.

To determine the direction of the net force, we use the right-hand rule. If the current is flowing in the positive x-direction and the magnetic field is in the positive y-direction, the force on each side of the loop will be towards the center of the loop (or inwards). The net force will point towards the center of the loop as well.

Hence, the direction of the net force acting on the loop is towards the center of the loop.

Therefore, the magnitude and direction of the net force acting on the loop at this instant are 4 N towards the center of the loop.