# 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

## 1 Answer

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.

• May you provide me with references for me to study? I wonder why not used current I instead of the density J? Why cant we used the power x time to figure out the total energy used? – Pupil Apr 19 '14 at 9:18
• The above reasoning is in Landau Lifshitz, Electrodynamics of continuous media, Pergamon Press, but that is an advanced book. You can start with Purcell's book on electricity and magnetism (Berkely physics course) and with Griffiths' standard textbook. Total energy transferred from the battery to the wire can be written as power x time only if this power is constant in time. That will happen when the current and the velocity of the wire reach stable constant values. That may happen, for example, in electromotor transmitting constant torque and turning with constant angular velocity. – Ján Lalinský Apr 19 '14 at 19:53
• I assume in my example both current and velocity are not stable, if so why aren't they? – Pupil Apr 19 '14 at 20:11
• Yes. But in practice power sources do not fix power by themselves. Battery, for example, usually fixes voltage. In case of electrical circuit made of rod on rails in magnetic field, connecting the voltage source to the circuit will initiate change in the system. Electric current will begin to flow and the rod will start to move. This is a non-stationary process where power will increase from zero to some maximum value and then will probably oscillate as in RLC circuit. – Ján Lalinský Apr 20 '14 at 10:28
• Only after some time, if the rails have negligible resistance, we can expect that due to constant resistance of the rod, stationary motion will establish itself and then the power supplied by the battery will be constant. – Ján Lalinský Apr 20 '14 at 10:29