- #1

kmm

Gold Member

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**E**and

**B**. In the next instant, dt, the charges move around. He then asks the question, "How much work, dW, is done by the electromagnetic forces acting on these charges in the interval dt?" From the Lorentz force law, the work done on a charge q is: [tex] dW= \mathbf{F} \cdot d \mathbf{l} = q( \mathbf{E} + \mathbf{v} \times \mathbf{B}) \cdot \mathbf{v} dt = q \mathbf{E} \cdot \mathbf{v} dt [/tex] But ## q= \rho d \tau ## and ## \rho \mathbf{v} = \mathbf{J}##. So the rate that work is done on all the charges in a volume V is: [tex] \frac{dW}{dt} = \int_{V} ( \mathbf{E} \cdot \mathbf{J}) d \tau [/tex] Great, but the problem with this to me is that this gives a nonzero result for work. But how can the total work done by a system on itself be nonzero? I'm sure it's something simple I'm missing, but I'm not sure.