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If the magnetic force does no work on a particle with electric charge, then: How can you influence the motion of the particle? Is there perhaps another example of the work force but do not have a significant effect on the motion of the particle?

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Related: physics.stackexchange.com/q/16326/2451 –  Qmechanic Mar 18 '12 at 19:31
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thanks again Mr. Qmechanic. –  jormansandoval Mar 18 '12 at 19:32
    
related: physics.stackexchange.com/questions/67826/… –  Ben Crowell Jun 28 '13 at 14:59
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4 Answers

The energy of a freely moving particle is its kinetic energy $E=\dfrac{m v^2}{2}$. If the energy of the particle remains unchanged after someone acts on it, it means that no work has been done. However, the direction of ${\bf v}$ could have changed, provided that $v^2$ and hence $E$ is kept the same.

This means that one can affect the motion of a particle by changing the direction of its velocity and making no work for it.

If you do make work, however, you change $E$ and hence $v^2$. Hence, you must change ${\bf v}$, and affect the motion.

To conclude, the work cannot be done without affecting the motion of a particle.

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I believe your first question is quite adequately answered by Qmechanic’s link. Regarding force that do not work, i.e. are perpendicular to motion, there are indeed other examples:

  • In a circular motion, a force acting towards the center of the circle is always perpendicular to velocity, and therefore does zero work.
  • An object rolling on the floor: the reaction force exerted by the floor on the object, equal to the opposite of the object’s weight, does zero work.

In all these cases, the force does impact the object’s movement: zero work means it does not change its energy, but it does constraint its movement.

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Work performed by forces acting on a particle is equal to the change in particle's energy. If the forces acting on a particle perform zero work on it, particle's energy does not change.

In particular, whenever a force acting on a particle is perpendicular to the particle's displacement as is the case with magnetic component of the Lorentz force, the work performed will be zero and particle's energy will not change.

Note that the direction of the particle's velocity may change without affecting its energy.

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The equation for the magnetic force $\mathbf{F}$ on a particle with electric charge $q$ moving with velocity $\mathbf{v}$ in a magnetic field $\mathbf{B}$ is given by $$\mathbf{F} = q\mathbf{v} \times \mathbf{B}$$ Because of the nature of the cross product, $\mathbf{F}$ is always perpendicular to $\mathbf{v}$. Since the force is always perpendicular to the velocity, the result is that the charge experiences uniform circular motion. In uniform circular motion, the speed of the particle stays constant. In other words, while the direction of $\mathbf{v}$ is changing due to the magnetic force, the speed, or $| \mathbf{v} |$, stays constant. Since the kinetic energy of the particle is independent of direction, and only depends on speed ($E_k = \frac{1}{2} m | \mathbf{v} |^2$). Since the speed is unchanging, the kinetic energy doesn't change. Since the kinetic energy of the particle doesn't change, there is no work done on the charge, as work done on the charge must change the kinetic energy. So the short answer is: The magnetic force causes a charged object to experience circular motion. Circular motion doesn't do any work on an object. Even though the charge's motion is influenced, no work is done on it.

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