A proton traveling at 4.50 km/s suddenly enters a uniform magnetic field of 0.70
ID: 3163928 • Letter: A
Question
A proton traveling at 4.50 km/s suddenly enters a uniform magnetic field of 0.700 T , traveling at an angle of 55.0 o with the field lines (see the figure (Figure 1) ).
A. Find the direction of the force this magnetic field exerts on the proton.
ANSWER: The force is directed into the page.
B.Find the magnitude of the force this magnetic field exerts on the proton.
C. 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.
D.How the velocity should be oriented to achieve the forces in part (C).
F is maximum when v is parallel to B . F is minimum when v is perpendicular to B .
E. Find the direction of the force this magnetic field exerts on the proton.
ANSWER: The force is directed into the page.
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
F is maximum when v is perpendicular to B . F is minimum when v is perpendicular to F . F is maximum when v is perpendicular to B . F is minimum when v is either parallel or antiparallel to B .F is maximum when v is parallel to B . F is minimum when v is perpendicular to B .
E. Find the direction of the force this magnetic field exerts on the proton.
ANSWER: The force is directed into the page.
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
55.0 ProtonExplanation / Answer
a) into the page
b) F = q v B sin theta
= 1.6 * 10-19 * 4.5 * 103 * 0.700 * sin 55 = 4.13 * 10-16 N
c) Maximum force on proton is observed when theta = 90 deg
F = q v B sin theta = 1.6 * 10-19 * 4.5 * 103 * 0.700 * sin 90 = 5.04 * 10-16 N
Minimum force on proton is observed when theta = 0 deg
F = q v B sin 0 = 0 N
f) magnitude of force remains same
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