A proton traveling at 3.80km/s suddenly enters a uniform magnetic field of 0.780

Question

A proton traveling at 3.80km/s suddenly enters a uniform magnetic field of 0.780T , traveling at an angle of 55.0o with the field lines (see the figure(Figure 1) ).

Part A

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

Part B

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

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

Answer the question in the order indicated. Separate your answers using comma.

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?

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?

A proton traveling at 3.80km/s suddenly enters a uniform magnetic field of 0.780T , traveling at an angle of 55.0o with the field lines (see the figure(Figure 1) ). Part A Find the direction of the force this magnetic field exerts on the proton. Part B Find the magnitude of the force this magnetic field exerts on the proton. If you can vary the direction of the proton's velocity, find the magnitude of the maximum and minimum forces you could achieve. Answer the question in the order indicated. Separate your answers using comma. 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? 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

Force due to a magnetic field is
F = q(v x B), where (v x B) is the cross product of the velocity vector and magnetic field lines.

cross product = |v|*|B|*sin(theta), |x| means magnitude.

so,
F = q|v|*|B|*sin(theta)

q = 1.6x10^-19C
v = 3.8km/s = 3800m/s
B = 0.780T
theta = 55degrees

F = 3.91x10^-16N

Varying the direction varies theta. sin(theta) is at a maximum at theta = 90degress and at a minimum at theta = 0 or 180degrees

sin(90) = 1
sin(0 or 180) = 0

therefore maximum will occur when the proton is perpendicular the field lines.

F = q|v|*|B|*sin(90)
= 4.7424x10^-16

theta = 90, F = 4.7424x10^-16N

minimum magnitude is when the velocity is parallel or anti parallel the field lines
theta = 180 or 0, F = 0 because sin(theta) then = 0.

c.
the magnitude's would still be the same but in opposite directions because of the opposite charge (negative charge for electrons).

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