contestada

A proton moving to the left at 10m⋅s−1 enters a magnetic field of strength 5.0×10−2T directed into the page. The proton is deflected into a circular path. During half a revolution, how much work is done on the proton by the magnetic force?

Respuesta :

Poltergeist

Answer:

zero.

Explanation:

The work done by the magnetic force on the proton is zero. This is because direction of the magnetic force, which is towards the center of the circular path, is perpendicular to to the displacement of the proton, which is tangent to the path.

In other words,

[tex]W = F\cdot d = Fdcos(\theta)[/tex],

and since the angle [tex]\theta[/tex] between the magnetic force [tex]F[/tex] and the displacement [tex]d[/tex] is 90°

[tex]W = Fdcos(90) =0[/tex].

This is a general result, true for all magnetic forces: The work done by a magnetic force is zero.

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