A proton traveling at 3.60 km/s suddenly enters a uniform magnetic field of 0.72

Question

A proton traveling at 3.60 km/s suddenly enters a uniform magnetic field of 0.720 T , traveling at an angle of 55.0 with the field lines (see (Figure 1) ).

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

Part B: If you can vary the direction of the proton's velocity, find the magnitude of the maximum force you could achieve.

Part C: If you can vary the direction of the proton's velocity, find the magnitude of the minimum force you could achieve.

Part D: What would the answer to Part B be if the proton were replaced by an electron traveling in the same way as the proton?

Explanation / Answer

A:

magnitude of the magnetic force:

F = Bvqsin@ = 0.72*3.6x10^3*1.6x10^-19*sin55 = 3.4e-16 N

B.

magnitude of the maximum force:

  F = Bvqsin@ = 0.72*3.6x10^3*1.6x10^-19*sin90= 4.15e-16 N

C:

magnitude of the minimum force:

F = Bvqsin@ = Bvqsin0 = 0

D:

both electron and proton have a same charge, so the result in part B is not chnaged.

that is,

F = Bvqsin@ = 0.72*3.6x10^3*1.6x10^-19*sin90= 4.15e-16 N

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