# The Energy involved in the work done here?

When a wire that has current $I$ flowing within it and its in a magnetic field, the wire experience the Lorentz force, and that force moved the wire over a certain distance $x$(no matter how small), can we state that work is done by the Lorentz force on the wire?

If so...

What kind of energy is transferred here? And, what potential energy was converted for this wire to move? What is the source of energy?

• The force on the wire can be thought of as a sum of elementary forces acting on the particles that compose it. These elementary forces may be assumed to be given by the Lorentz formula $\mathbf F = q\mathbf E + q\frac{\mathbf v}{c} \times \mathbf B$ In order to answer you other questions, you will have to specify better the setup. How is the current in the wire maintained? By pushing the wire through the magnetic field or by some source of voltage? – Ján Lalinský Apr 18 '14 at 20:06
• By a source of voltage, any kind you would like: Generator, battery, etc... – Pupil Apr 18 '14 at 23:20

By assumption, there is electric current (with density $\mathbf j$) flowing in the wire (think of a circuit or a rod on rails). When placed in the magnetic field, there will be magnetic force $\int \frac{\mathbf j}{c} \times\mathbf B \,dV$ acting on the wire (on the nuclei it is composed of) so some of its parts will begin to move (with velocity $\mathbf v$), and their kinetic energy will be increased with rate $$dE_k/dt =\int \mathbf v \cdot\left(\frac{\mathbf j}{c} \times\mathbf B \right)\,dV$$ If the current is due to source of voltage (battery), the kinetic energy of the wire (and its internal energy) comes from this source. There is electric field in the wire $\mathbf E$ directed along it maintaining the current and the rate at which the battery loses energy is $$R = \int \mathbf j\cdot \mathbf E\,dV .$$
In the limit of an ideal conductor, it is known that the magnetic electromotive intensity balances the electric intensity: $$\frac{\mathbf v }{c}\times \mathbf B = -\mathbf E.$$ The power lost by the battery is thus $$R = \int -\mathbf j \cdot \left(\frac{\mathbf v}{c}\times\mathbf B\right) \,dV .$$ Permuting the terms, we obtain $$R = \int \mathbf v \cdot \left(\frac{\mathbf j}{c}\times\mathbf B\right) \,dV .$$ which is the same as $dE_k/dt$, so all battery's energy goes into kinetic energy of the wire. If the conductor has some resistance, the electric intensity won't be balanced entirely by the magnetic electromotive intensity $\mathbf v/c \times \mathbf B$ and $dE_k/dt$ will be lower than $R$ as some energy of the source will dissipate into internal energy of the wire.