In Jackson's Classical Electrodynamics, the rate of work done by fields in a finite volume is defined as $$\int _{v}\vec{J}\cdot\vec{E}\,d^{3}x^{'}$$ How?
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This is an assumption one makes, reasonable when the current goes through homogeneous metal (if different metals meet, or there are temperature gradients, other electromotive forces than macroscopic electric field may come into play). What exactly are you asking? – Ján Lalinský Aug 26 '15 at 20:33
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Jackson states why this is the case beforehand:
For a single charge $q$ the rate of doing work by external electromagnetic fields $\mathbf E$ and $\mathbf B$ is $q\mathbf v\cdot\mathbf E$, where $\mathbf v$ is the velocity of the charge.... If there exists a continuous distribution of charges and current, the total rate of doing work by the fields in a finite volume $V$ is $$\int_V\mathbf J\cdot\mathbf E\,d^3x$$
For continuous distributions of charges. $$ q=\int\rho\,d^3x $$ which is used to change the charge $q$ into the volume charge density $\rho$. The fact that magnetic forces of point charges do no work allows us to ignore the magnetic component. Otherwise, this is an application of power $$ \frac{dW}{dt}=\mathbf F\cdot \mathbf v $$ for the Lorentz force.
Kyle Kanos
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