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By Lorentz force law, $F_m$ is always perpendicular to velocity of charge $v$. But $F_m$ is not necessarily perpendicular to the displacement of charge caused by $F_m$ .

Then how can we say that magnetic force does no work?

Edit:

One thing to clear: Is work done = displacement just before the force was applied dotted with force OR displacement just after the force was applied dotted with force? If the former were true, you would be correct. But if the latter were true, I don't see any reason why $F_m$ is perpendicular to the displacement just after the force was applied as there should be a component of displacement in the direction of force.

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The force from a magnetic field is given by the magnetic part of the Lorentz force equation: $$\mathbf{F} = q \mathbf{v} \times \mathbf{B}.$$

The work done by any force is given by the path integral: $$W = \int_{\mathrm{start}}^{\mathrm{end}} \mathbf{F} \cdot \mathrm{d}\mathbf{x}.$$ If we parameterize the path in terms of time we can rewrite the work integral as: $$\begin{align} W &= \int_{t_0}^{t_f} \mathbf{F}\cdot \frac{\mathrm{d} \mathbf{x}}{\mathrm{d} t} \,\mathrm{d}t \\ & = \int_{t_0}^{t_f} \mathbf{F}\cdot \mathbf{v}\, \mathrm{d}t. \end{align}$$ The integrand in the second line is one way of writing the power (rate of exchange of energy).

The point being, since the magnetic force is always perpendicular to the velocity, the integrand is always zero, so the work is zero, too. It doesn't matter if the magnetic force is the only one acting or not.

The magnetic field can store energy, though, but that energy is added to it and removed from it indirectly through the electric field according to Faraday's law.

Take the Faraday-Maxwell equation, and dot both sides with $\mathbf{B}$: $$\mathbf{B}\cdot \left(\nabla \times \mathbf{E}\right) = - \mathbf{B}\cdot \frac{\partial \mathbf{B}}{\partial t}. $$ The right hand side is equal to $- \frac{1}{2} \frac{\partial}{\partial t} \left(\mathbf{B}\cdot \mathbf{B}\right) = -\mu_0 \frac{\partial}{\partial t} u_B $, with $u_B$ the energy density in the magnetic field. Thus: $$\Delta u_B = -\frac{1}{\mu_0} \int_{t_0}^{t_f} \mathbf{B}\cdot \left(\nabla \times \mathbf{E}\right) \, \mathrm{d} t. $$

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  • $\begingroup$ One thing to clear: Is work done = displacement just before the force was applied dotted with force OR displacement just after the force was applied dotted with force? If the former were true, you would be correct. But if the latter were true, I don't see any reason why $F_m$ is perpendicular to the displacement just after the force was applied as there should be a component of displacement in the direction of force. $\endgroup$ – stack exchange Apr 11 '17 at 17:16
  • $\begingroup$ @stackexchange This is calculus. The force is proportional and perpendicular to the velocity, and the rate of change of energy (rate work is done) is proportional to the part of the force parallel to the velocity. $\endgroup$ – Sean E. Lake Apr 11 '17 at 17:20

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