A proton traveling at 4.10 km/s suddenly enters a uniform magnetic field of 0.80

Question

A proton traveling at 4.10 km/s suddenly enters a uniform magnetic field of 0.800 T, traveling at an angle of 55.0 degree with the field lines (see the figure (Figure 1)). How the velocity should be oriented to achieve the forces in part (c). 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. What would the answers to part (A) be if the proton were replaced by an electron traveling in the same way as the proton? 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

The force on the electron will be = F = -evBsin550 = -1.6 x 10-19 x 4.10 x 1000 x 0.8 x sin550

F = -4.3 x 10-16 N

The only change will be the direction of the force and hence it will act in upward direction.

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