How Do Moving Charges Interact with Magnetic Fields? (2024)

Magnetism Related Questions, Please Help!

There are three questions that I need help with, all of which are focused on the main topic of Magnetism (The Magnetic Force on Moving Charges). Before I begin, I would just like to thank anyone that will be able to provide me with some assistance in answering the following questions.

1. A negatively charged ion moves due north with a speed of 1.5 x 10^6 m/s at the Earth's Equator. What is the Magnetic force exerted on this ion?

2. An electron moving with a speed of 9.1 x 10^5 m/s in the positive x direction experiences zero magnetic force. When it moves in the positive y direction it experiences a force of 2.0 x 10^-13 N that points in the negative z direction. What is the direction and magnitude of the magnetic field?

3. A 5.60 x 10^-6 C particle moves through a region of space where an electric field of magnitude 1250 N/C points in the positive x direction, and a magnitude 1.02T points in the positive z direction. If the net force acting on the particle is 6.23 x 10^-3 N in the positive x direction, find the magnitude and direction of the particle velocity.

Just to reiterate, I would prefer if someone were to explain to me how I would go about answering each of the questions; I would also like to thank you all one again for any assistance provided in advance.

1. To solve this question, we need to use the formula F = qvBsinθ, where F is the magnetic force, q is the charge of the ion, v is its velocity, B is the magnetic field, and θ is the angle between the velocity and the magnetic field. In this case, the angle between the velocity and the magnetic field is 90 degrees, since the ion is moving due north and the Earth's magnetic field is directed towards the Earth's center. We also know that the charge of an electron is -1.6 x 10^-19 C, so we can substitute these values into the formula: F = (-1.6 x 10^-19 C)(1.5 x 10^6 m/s)(5 x 10^-5 T)(sin 90) = -1.2 x 10^-13 N. Therefore, the magnetic force exerted on the ion is -1.2 x 10^-13 N, directed towards the Earth's center.

2. In this question, we are given the force and the velocity of the electron in the y direction, and we need to find the direction and magnitude of the magnetic field. We can use the same formula as in the previous question, but this time we are solving for B. We also know that the angle between the velocity and the magnetic field is 90 degrees, since the electron is moving in the positive y direction and the force is in the negative z direction. Substituting the given values, we get: B = (2.0 x 10^-13 N)/[(9.1 x 10^5 m/s)(1.6 x 10^-19 C)(sin 90)] = 1.23 T. Therefore, the magnitude of the magnetic field is 1.23 T, and its direction is in the negative z direction.

3. For this question, we need to use the formula F = qE + qvBsinθ, where F is the net force, q is the charge of the particle, E is the electric field, v is the velocity, B is the magnetic field, and θ is the angle between the velocity and the magnetic field. We know that the net force acting on the particle is in the positive x direction, so we can set up the equation as: 6.23 x 10^-3 N = (5.60 x 10^-