A proton has a charge of 1.6 × 10-19 C and a mass of 1.67 × 10-27 kg. A proton moves with velocity 1.8 × 105 m/s in the z-direction. A magnetic field with B = 2.1 × 10-1 T points in the y-direction. Find the magnitude and direction of the magnetic force on the proton.What is the angle between the proton’s velocity and the magnetic field?
45°
90°
180°
50°

I'll report you if you don't actually help. I'd like an actual explanation, please.

Answers

Answer 1
Answer:

Hey

So first we need to know what the direction of the force is, using your right hand rule point your right hand in the direction of the velocity. You're saying its the z direction, not telling me whether it's into the page or out? Since its a positive z im assuming its coming out. The magnetic field is pushing it upwards, so the force is going in the negative x direction.

The force of a magnetic field is

F = Qv X B

What's weird is that you don't need mass in this equation. Actually you don't even need the formula, its telling you that they're all going in perpendicar directions. the answer is 90 degrees.

Now if you want to know the F just multiply the charge, velocity and magnetic field .

F = GVB

F = 6.048 E -15

Answer : 90 degrees, sin(90) = 1

Answer 2
Answer:

Final answer:

To find the magnitude and direction of the magnetic force on a proton moving in a magnetic field, you can use the equation F = qvBsinθ, where F is the force, q is the charge, v is the velocity, B is the magnetic field, and θ is the angle between the velocity and the magnetic field. The magnitude of the magnetic force can be calculated using the equation, and its direction can be determined using the right-hand rule. In this case, the angle between the proton's velocity and the magnetic field is 90°.

Explanation:

To determine the magnitude of the magnetic force on the proton, we need to use the equation F = qvBsinθ, where F is the force, q is the charge, v is the velocity, B is the magnetic field, and θ is the angle between the velocity and the magnetic field.

Plugging in the values, we have F = (1.6 × 10-19 C)(1.8 × 105 m/s)(2.1 × 10-1 T)sinθ.

To find the angle θ, we can use the fact that the force is perpendicular to both the velocity and the magnetic field, which means that sinθ = 1.

Therefore, the magnitude of the magnetic force on the proton is F = (1.6 × 10-19 C)(1.8 × 105 m/s)(2.1 × 10-1 T) = 6.048 × 10-14 N. The direction of the magnetic force is given by the right-hand rule, which shows that the force is perpendicular to both the velocity and the magnetic field, pointing in the positive x-direction.

The angle between the proton's velocity and the magnetic field is 90°.

Learn more about magnetic force on a charged particle here:

brainly.com/question/19525738

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