Your Question:

3. (Problem 22-4) A proton moving at 4.00 x 106 m/s through a magnetic field of magnitude 1.70 T experiences a magnetic force of magnitude 8.20 x 10-13 N. What is the angle between the proton's velocity and the field?

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Step By Step Answers with Explanation

Step-by-step explanation of Problem 22-4

Step 2: Write down the relevant equation

The magnetic force on a moving charged particle is given by the following equation:

• v is the velocity of the particle

• B is the magnetic field

• v = 4.00e6 m/s

• B = 1.70 T

Dividing both sides of the equation by 1.60217662e-19 C * 4.00e6 m/s * 1.70 T, we get:

sin(theta) = 8.20e-13 N / (1.60217662e-19 C * 4.00e6 m/s * 1.70 T)

Step 5: Interpret the answer

The angle between the proton's velocity and the magnetic field is 48.82 degrees. This means that the proton is not moving directly perpendicular to the magnetic field. The magnetic force will cause the proton to move in a circular path, with the radius of the circle determined by the magnitude of the magnetic force, the proton's velocity, and the angle between the velocity and the magnetic field.

where:

• F is the magnetic force

r = mv / qB

where:

• B is the magnetic field

If the particle is not moving perpendicular to the magnetic field, then the magnetic force will cause the particle to move in a helical path. A helical path is a three-dimensional spiral that is formed by combining a circular path with a forward motion.