Moving Charged Particles for AP Physics B & C

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By — McGraw-Hill Professional
Updated on Feb 12, 2011

Practice problems for these concepts can be found at:

Magnetism Practice Problems for AP Physics B & C

The whole point of defining a magnetic field is to determine the forces produced on an object by the field. You are familiar with the forces produced by bar magnets—like poles repel, opposite poles attract. We don't have any formulas for the amount of force produced in this case, but that's okay, because this kind of force is irrelevant to the AP exam.

Instead, we must focus on the forces produced by magnetic fields on charged particles, including both isolated charges and current-carrying wires. (After all, current is just the movement of positive charges.)

A magnetic field exerts a force on a charged particle if that particle is moving perpendicular to the magnetic field. A magnetic field does not exert a force on a stationary charged particle, nor on a particle that is moving parallel to the magnetic field.

The magnitude of the force exerted on the particle equals the charge on the particle, q, multiplied by the velocity of the particle, v, multiplied by the magnitude of the magnetic field.

This equation is sometimes written as F = qvB(sin θ). The θ refers to the angle formed between the velocity vector of your particle and the direction of the magnetic field. So, if a particle moves in the same direction as the magnetic field lines, θ = 0°, sin 0° = 0, and that particle experiences no magnetic force!

Nine times out of ten, you will not need to worry about this "sin θ" term, because the angle will either be zero or 90°. However, if a problem explicitly tells you that your particle is not traveling perpendicular to the magnetic field, then you will need to throw in this extra "sin θ" term. This is far more likely to occur on the Physics C exam than on the Physics B exam.

The key to this right-hand rule is to remember the sign of your particle. This next problem illustrates how important sign can be.

This is one of those problems where you're told that the particle is not moving perpendicular to the magnetic field. So the formula we use to find the magnitude of the force acting on the particle is

    F = qvB(sin θ)
    F = (1.6 × 10–19 C)(5 × 106 m/s)(0.4 T)(sin 30°)
    F = 1.6 × 10–13 N

Now we solve for acceleration:

    Fnet = ma
    a = 1.8 × 1017 m/s2

Wow, you say … a bigger acceleration than anything we've ever dealt with. Is this unreasonable? After all, in less than a second the particle would be moving faster than the speed of light, right? The answer is still reasonable. In this case, the acceleration is perpendicular to the velocity. This means the acceleration is centripetal, and the particle must move in a circle at constant speed. But even if the particle were speeding up at this rate, either the acceleration wouldn't act for very long, or relativistic effects would prevent the particle from traveling faster than light. (Notice that we did not include the negative sign on the charge of the electron. We do this because the magnitude of anything is always a positive number.)

Finally, we solve for direction using the right-hand rule. We point our hand in the direction that the particle is traveling—to the right. Next, we curl our fingers upward, so that they point in the same direction as the magnetic field. Our thumb points out of the page. BUT WAIT!!! We're dealing with an electron, which has a negative charge. So the force acting on our particle, and therefore the particle's acceleration, points in the opposite direction. The particle is accelerating into the page.

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