Sure, here's an AP Physics 2 exam question and its solution about magnetic fields:

Question: A straight wire of length 0.5 m carries a current of 2 A. The wire is placed perpendicular to a uniform magnetic field of 0.4 T. Calculate the magnitude and direction of the magnetic force acting on the wire.

Answer: The magnetic force acting on a current-carrying wire in a magnetic field is given by the equation F = I * L * B * sin(θ), where I is the current, L is the length of the wire, B is the magnetic field, and θ is the angle between the wire and the magnetic field.

Given: I = 2 A (current) L = 0.5 m (length of the wire) B = 0.4 T (magnetic field)

First, let's find the magnitude of the magnetic force: F = I * L * B * sin(90°) F = 2 A * 0.5 m * 0.4 T * 1 (since sin(90°) = 1) F = 0.4 N

Now, let's determine the direction of the magnetic force: By the right-hand rule, if you point your thumb in the direction of the current and your fingers in the direction of the magnetic field, the palm will point in the direction of the force. In this case, using the right-hand rule, the force will be directed into the page.

So, the magnitude of the magnetic force acting on the wire is 0.4 N, and the direction of the force is into the page.