# Motional EMF and forces on negative charges

my doubt is pretty basic but it's complex enough for me not to move on with my studies:

In this example (which is probably known from people who read Serway's book) there is an induced emf in a conductor due to the change in magnetic field and law of Faraday in the clockwise direction. Now the trouble comes when I have to deal with the direction of the magnetic force from Lorentz equation. Following the cross product rule, and aknowledging that the current is in the clockwise direction and the magnetic field is upwards, I get a force pointing to the left. BUT the Lorentz formula tells us that the sign of the charge carriers results in a change in direction, according to F=q(vxB) so the force would be pointing to the left with protons and to the right with electrons. But I am dealing with a conductor, so how is that possible??? Thank you

• The protons, I guess, you mean the atom nuclei of the conductor won't move. The exerted force is smaller than the force which keep the atom nuclei in place. The electrons, however, are in the conduction band of the conductor and can move much easier. – Frederic Thomas Jan 11 '17 at 12:46
• What I meant to say is negative or positive charges, it doesn't really matter what the charge carriers physically represent. The only concern here is idealization of the Lorentz force, which is causing me trouble – George Sailor Jan 11 '17 at 17:53
• I don't understand your point. To what kind of idealization do you refer to ? – Frederic Thomas Jan 12 '17 at 15:11

'Free' electrons (each with charge $$-e)$$ can move in metal conductors. They are the 'mobile charge carriers'.

If the only velocity of the free electrons in the rod were the velocity $$\vec v$$ of the rod itself, the force on each mobile electron would be $$\vec F=(-e)(\vec v \times \vec B)$$ This drives the electrons anticlockwise around the loop.

So the electrons now have another velocity, their drift velocity through the rod, $$\vec v_{dr},$$ added (vectorially of course) to the velocity, $$\vec v,$$ of the rod. This doesn't make things as complicated as you might fear. We can simply work out the additional Lorentz force that this additional velocity gives rise to: $$\vec F_{Lap}=(-e)(\vec v_{dr} \times \vec B)$$ This force is to the left, opposing the motion to the right of the rod. So we have to do mechanical work to move the rod, and this is how we pay for the energy dissipated in the resistor!

[A note – but not for the faint-hearted – on the weird designation, $$\vec F_{Lap}.$$ Because the electrons are confined to the rod by essentially electrostatic forces, the Lorentz force component, $$\vec F_{Lap},$$ is balanced by such a force, whose Newton 3 partner, a force equal to $$\vec F_{Lap},$$ is the electron's contribution to the sideways force on the rod itself. The sideways force on a current-carrying conductor in a magnetic field is called the Laplace force.]

• You made a very important distinction and connection, dividing the problem into two forces, 1) the Lorentz force driving current flow through the circuit, and 2) the Laplace force resisting the movement of the rod (i.e., the force producing displacement at a rate which equals the rate of heat production (i.e., the power dissipated) by the resistor. – Thomas Lee Abshier ND Feb 8 at 12:43
• Good, but I think that the Lorentz force is the resultant force, due to the resultant velocity. Thus $$F_{Lor}=(-e)[(\vec v + \vec v_{dr}) \times \vec B].$$ The component of this force that drives the electrons through the rod should properly be called the 'electromotive force', but this name has been misappropriated for the work per unit charge done by this force! – Philip Wood Feb 8 at 13:02

Positive charge carriers are positive and move clockwise. Negative charge carriers are negative and move counter-clockwise.

Current caused by the motion of the positive charge carriers is a clockwise current.

Current caused by the motion of the negative charge carriers is a clockwise current.

A clockwise current causes a leftwards pointing Lorentz force.