A proton traveling at 4.10 km/s suddenly enters a uniform magnetic field of 0.80

Question

A proton traveling at 4.10 km/s suddenly enters a uniform magnetic field of 0.800T traveling at an angle of 55 degrees with the field lines shown:

Part A

Find the direction of the force this magnetic field exerts on the proton

( into or out of the page?)

Part B

Find the magnitude of the force this magnetic field exerts on the proton.

Part C

If you can vary the direction of the proton's velocity, find the magnitude of the maximum and minimum forces you could achieve.

Part D

How the velocity should be oriented to achieve the forces in part (C).

Part E

What would the answers to part (A) be if the proton were replaced by an electron traveling in the same way as the proton?

(into or out of the page?)

Part F

What would the answers to part (B) be if the proton were replaced by an electron traveling in the same way as the proton?

Explanation / Answer

part A : into the page. part B: the magnitude is F = qvB sin ? so substuting the values we get, 3.99x 10^-16 N part C& E:maximum force occurs when the particle is perpendicular to the field so, F= qvB= 5.248 x 10^-16N minimum force is zero if the velocity is in the direction of magnetic field part E Out of the page. part F : the magnitude of force remains the same, only the direection of force is reversed.

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