A proton traveling at 4.50 km/s suddenly enters a uniform magnetic field of 0.73

Question

A proton traveling at 4.50 km/s suddenly enters a uniform magnetic field of 0.730 T , traveling at an angle of 55.0 ? with the field lines (see (Figure 1) ).

Problem 20.03 Part A A proton traveling at 4.50 km/s suddenly enters a uniform magnetic field of 0.730 T, traveling at an angle of 55.0 with the field lines (see (Fiqure 1) Find the direction of the force this magnetic field exerts on the proton Please Choose Submit My Answers Give Up Part B Find the magnitude of the force this magnetic field exerts on the proton. Express your answer in newtons to three significant figures Submit My Answers Give Up 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 max Submit My Answers Give Up 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 min Submit My Answers Give Up Part E Figure 1 of 1 How should the velocity be oriented to achieve the forces in Parts C and D. F is maximum when v is perpendicular to B. F is minimum when v is perpendicular to F O F is maximum when is perpendicular to B. F is minimum when v is either parallel or antiparallel to B O F is maximum when v is parallel to B. F is minimum when v is perpendicular to B Submit 55.0°B My Answers Give Up Proton Part F

Explanation / Answer

v =4.5 km/s B =0.73 T theta =55 degrees

proton charge q = +1.6*10^-19 C

Part A:

from fleming right hand rule into the page
Part B : F = qvBsn(theta)

F = 1.6*10^-19*4.5*1000*0.73*sin(55)

F = 4.31*10^-16 N

Part C: Fmax = qvB sin(90)

Fmax = 1.6*10^-19*4.5*1000*0.73

Fmax = 5.26*10^-16 N

Part D: Fmin = qvBsin(0)

Fmin =0

Part E: F is maximum v is perpendicular to B . F is minimum when v is either parallel or antiparallel to B

Part F: F = -q(vxB)

out of page

Part G

F' = 1.6*10^-19*4.5*1000*0.73*sin(55)

F' = 4.31*10^-16 N

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