Work Done by Magnetic Field?

[boaz]

From EK Physics, 30min exam, lecture 7, answer to question 140:

"... You should know that the force on a moving particle due to a magnetic field is perpendicular to the velocity of the charge. This means that a magnetic field can do no work on a moving charge, which in turn, means that it cannot change its kinetic energy or its speed."

Excuse me? The force due to the field actually causes a change of the direction of the velocity. Since the force acts over a certain distance x, it must have done some work.

Anyone?

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[261766]

Work is defined as Fdcos(theta); only the distance along which the force is applied counts as work.

The direction which magnetic fields work are 90degrees. In essence, F is qvbsin(theta). Hence force applied is perpendicular to distance moved, leading to zero work.

 

[mcat45]

I don't think that causing a change in the direction of velocity necessarily corresponds to doing work. consider the slightly analogous situation of a satellite orbiting the earth. during such a situation, the satellite is experiencing a gravitational force that is like a centripetal force that points toward earth's center. as the satellite is orbiting it is simultaneously experiencing a change in the direction of the velocity vector. however, if one were to calculate the work done by the force of gravity they would calculate a value of 0 because the gravitational force would always be perpendicular to the velocity vector and also the displacement vector. work is strictly dependent on having a force and a displacement that occurs parallel to the force. if the force and displacement vector are perpendicular NO work will be done.

 

[261766]

I don't think that causing a change in the direction of velocity necessarily corresponds to doing work. consider the slightly analogous situation of a satellite orbiting the earth. during such a situation, the satellite is experiencing a gravitational force that is like a centripetal force that points toward earth's center. as the satellite is orbiting it is simultaneously experiencing a change in the direction of the velocity vector. however, if one were to calculate the work done by the force of gravity they would calculate a value of 0 because the gravitational force would always be perpendicular to the velocity vector and also the displacement vector. work is strictly dependent on having a force and a displacement that occurs parallel to the force. if the force and displacement vector are perpendicular NO work will be done.

haha what i was trying to say but not expressed as well.

 

[boaz]

Yea, so I just looked back at my textbook and it says so very clearly.

However, what was bothering me before is still nagging me. Consider the following example:

1) Object A is traveling with a constant velocity in the +x direction.
2) Object B then comes along and nudges the object with a force F directed in the +y direction, perpendicular to the original velocity vector of A.
3) As a result, the velocity vector of object A is now displaced slightly towards the +y direction, the displacement vector being parallel to the to the force vector.

Would you say that by the virtue of the fact that object A already had a velocity there's no displacement? If it were at rest it surely would fit the definition of work. How does the parallel velocity change that?

But the odd thing about the force in a magnetic field is that there would be no force if the particle weren't moving ...

No wonder why I forgot this ...

 

[mcat45]

c7 places your explanation was fine, no worries. i just wanted to include a visual. ishchayill, in part 3 of your question you stated that the velocity vector is now displaced. remember, the definition of velocity is derived from a change in displacement. so, to say that the velocity vector is displaced is not a proper statement as any change that occurs in displacement is already covered by a change in either the magnitude or (in your example) the direction of the velocity vector itself. also, in your example object A does have a displacement in the y direction and since the dislacement in this direction is parallel to the original force from B, then B does work on A. The fact that A was originally moving in the x direction does not matter because the push from B causes A to move in the y-direction and subsequently to have a velocity in the y direction (parallel to the original force). The push of B on A essentially causes the x-component of the velocity to be reduced to 0, since no further displacement occurs in the x-direction following the push.