A proton traveling at 3.70 km/s suddenly enters a uniform magnetic field of 0.71
ID: 1594407 • Letter: A
Question
A proton traveling at 3.70 km/s suddenly enters a uniform magnetic field of 0.710 T , traveling at an angle of 55.0 ? with the field lines (see (Figure 1) ).
PART A: Find the direction of the force this magnetic field exerts on the proton. (choices)
- The force is directed into the page.
- The force is directed out of the page.
PART B: Find the magnitude of the force this magnetic field exerts on the proton. Express your answer in newtons to three significant figures.
PART C: If you can vary the direction of the proton's velocity, find the magnitude of the maximum force you could achieve. Express your answer in newtons to three significant figures.
PART D: If you can vary the direction of the proton's velocity, find the magnitude of the minimum force you could achieve. Express your answer in newtons to three significant figures.
PART E: How should the velocity be oriented to achieve the forces in Parts C and D. (choices)
- 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?.
PART F: What would the answer to Part A be if the proton were replaced by an electron traveling in the same way as the proton? (choices)
- Force is directed into the page.
- Force is directed out of the page.
PART G: What would the answer to Part B be if the proton were replaced by an electron traveling in the same way as the proton? Express your answer in newtons to three significant figures.
- F is maximum when v? is perpendicular to B? . F is minimum when v? is perpendicular to F? .Explanation / Answer
A) into the page
B) F = q*v*B*sin(55) = 1.6*10^-19*3.7*1000*0.71*sin(55) = 3.44*10^-16 N
C) Fmax = q*v*B = 1.6*10^-19*3.7*1000*0.71 = 4.2*10^-16 N
D) Fmin = 0 N
E) F is maximum when v is perpendicular to B . F is minimum when V is either parallel or antiparallel to B
F) out of the page
G) F = q*v*B*sin(55) = 1.6*10^-19*3.7*1000*0.71*sin(55) = 3.44*10^-16 N
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