# Magnetostatics – the Lorentz force law

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 5.1.

Up to now, we’ve looked only at problems in electrostatics, that is, situations where we have a stationary distribution of charge.

The other half of electromagnetism is, of course, magnetism, so we’ll start by looking at magnetostatics, which is the study of static magnetic fields (possibly mixed in with static electric fields).

The starting point in the study of magnetism is the Lorentz force law, which states that a charge ${q}$ moving at velocity v through a magnetic field B feels a force of

$\displaystyle \mathbf{F}=q\mathbf{v}\times\mathbf{B} \ \ \ \ \ (1)$

This law is not derived; it is merely an expression of what is observed in experiments.

Magnetism is an unusual force in several ways. First, it acts on a charge only if the charge is moving relative to the field. Second, it produces a force perpendicular both to the field and the direction of motion. Third, as a consequence of the second point, a magnetic force cannot do any work. This is because work is defined as the integral of ${\mathbf{F}\cdot d\mathbf{l}}$, in which only the component of force in the direction of motion appears. Since the magnetic force is always perpendicular to the direction of motion, ${\mathbf{F}\cdot d\mathbf{l}=0}$ always, so no work is done. The magnetic force can therefore change only the direction of motion, and not its speed.

As a simple example of this, suppose we send a charged particle into a region with a constant magnetic field. The particle’s velocity is perpendicular to the field. The particle will feel a constant force of magnitude ${qvB}$ perpendicular to its direction of motion, so that this force acts as a centripetal force, and the particle moves in a circle. Equating these two forces, we get

$\displaystyle qvB=\frac{mv^{2}}{r} \ \ \ \ \ (2)$

This is known as the cyclotron formula, since it describes the principle used in a cyclotron, where charged particles are made to travel in circles by shooting them between the poles of a large electromagnet.

Using this formula, the radius of the circle is

$\displaystyle r=\frac{mv}{qB} \ \ \ \ \ (3)$

The momentum of the particle can be expressed in terms of the radius, charge and field:

 $\displaystyle p$ $\displaystyle =$ $\displaystyle mv\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle qrB \ \ \ \ \ (5)$

Suppose the field points along the ${+y}$ direction and the particle starts off moving in the ${+x}$ direction. After moving in the field a horizontal distance ${a}$, we find the particle is deflected in the ${-z}$ direction by a vertical distance ${d}$. By applying the right hand rule for cross products, we find that the particle must be positively charged. We can also work out its momentum (and thus its mass, assuming we measure its speed) if we can find the radius of the circular path.

If we draw a triangle with one vertex at the centre of the circle, another where the particle has reached its displacement of ${d}$ below the ${xy}$ plane, and then draw a horizontal line from the second vertex to intersect the ${z}$ axis, we get a right angled triangle with sides of ${r}$, ${r-d}$ and ${a}$. From Pythagoras, we have

 $\displaystyle r^{2}$ $\displaystyle =$ $\displaystyle \left(r-d\right)^{2}+a^{2}\ \ \ \ \ (6)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle -2rd+d^{2}+a^{2}\ \ \ \ \ (7)$ $\displaystyle r$ $\displaystyle =$ $\displaystyle \frac{a^{2}+d^{2}}{2d} \ \ \ \ \ (8)$

Therefore, the momentum is

 $\displaystyle p$ $\displaystyle =$ $\displaystyle qrB\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{qB}{2d}\left(a^{2}+d^{2}\right) \ \ \ \ \ (10)$

## 29 thoughts on “Magnetostatics – the Lorentz force law”

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2. Kevin Moore

Thanks again for posting all these. The extra insight is helpful. One thing that is bothering me is the idea of a magnetic force not doing work. Well sort of. It’s completely clear to me that the magnetic field does no work since the magnetic force is always perpendicular to the displacement. What’s made me confused is I’ve come across some claimed perpetual motion machines. I understand that true perpetual motion machines cannot exist because something that does work without the input of energy violates conservation of energy. But I’m hung up on what could be called a “perpetual motion machine” such as a water wheel, but using magnets. A water wheel can do work, but obviously energy is being put into it by the flowing water. When it comes to magnets, a magnet can indirectly do work such as lifting a paper clip or another magnet. But in that case I assume that the work is actually done by the electric field of displaced of moving charges(electrons) towards the magnet. The magnetic force changes the directions of the electrons towards the magnet, then the electric field from the electrons pulls the lattice of the wire with it, so it’s the electric field doing work. This brings up a question I’m struggling to answer. Could some arrangement of magnets then be put together to do work(even if it isn’t done directly by the magnetic field)? If so, when would it stop producing work? I know the energy density associated with a magnetic field is attributed to the energy required to create the field, so I would think that the magnetic field would continue to diminish as work was being done. But could thermal excitation of the electrons from outside the system continue to provide energy to the electrons and therefore the magnetic field so that this thermal excitation would act as the water in the water wheel?

Thanks,
Kevin

3. Kevin

I thought I should ask something a little more specific that may get to the heart of my first post. Suppose I’m holding a magnet and then lift a paperclip(or another magnet) with the magnet. Now the energy density of a magnetic field or the energy that went into making the magnetic field is ${W=\frac{1}{2\mu_{o}}\int B^{2}d\tau}$.

but since work was done in lifting the magnets, what happenes to the energy of the electrons in the magnet that produced the B field? Would their energy decrease by ${\Delta W}$ ? Here, I’m assuming the energy density of the magnetic field has decreased.

I’ve assumed that the work done when a magnet lifts a paper clip is due to the E field generated by the electrons pulling the lattice with them as the electrons are deflected from the lattice. This seems to contradict my first question where I assumed the energy is lost in the electrons producing the B field. My first question also seems to imply that it is actually the current loops in the magnet that’s doing the work. So which is it? Or is it both?

In example 5.3 from Griffiths he gives is a rectangle loop of wire, with some current ‘I’, in a magnetic field with some mass suspended to it. If the current is increased the mass is lifted and the work is done by the battery ${W_{battery}=\lambda aB\int uwdt}$.

where a is the width of the loop and u and w are the upward and downward velocity components respectively. This doesn’t include anything about the work done by the electrons on the lattice of the wire. So does this mean ALL the work is done by whatever generates the magnetic field?

1. growescience

I’ve edited your comment to make the equations show up properly. Your first attempt was closest to the correct syntax, but I think you had an extra } at the end before the final \$.

2. growescience

The
impression I get from Griffiths is that any work associated with a magnetic field must be done by the electric field that ultimately generates the magnetic field.. In the case of an electromagnet, then, yes, all the work is done by the source of the electric field (battery, generator, whatever).
If you use a bar magnet to pick up a paperclip, though, then it’s your arm that is doing the work, not the magnetic field. If you want to get down to the molecular or atomic level, I’d guess you would really need to use quantum mechanics, and I haven’t read much of anything yet on quantum electrodynamics, so I’m not sure how it’s dealt with there.

1. Kevin

Yes, I guess it depends on how far down the rabbit hole we want to go. I don’t think Griffiths mentioned anything about potential energy for two magnets in the field of each other. Although equation 6.3 in Griffiths which is: ${\mathbf{F} = \nabla (\mathbf{m} \cdot \mathbf{B})}$ Where ${\mathbf{m} \cdot \mathbf{B})}$ Is apparently referred to as the potential energy according to this paper I came across: academic.csuohio.edu/deissler/PhysRevE_77_036609.pdf
It sounds like it is common to say the magnetic field itself does work on a dipole, at least to the electron spin contribution to the magnetic moment.
The paper gives an alternate explanation of how the magnetic field doesn’t actually do work in this circumstance. Even if the a magnetic field does in fact do work on the dipole, it seems as though we have a situation like gravitational potential energy, where the energy you would get back still depends on the work done to put the dipole at that potential energy. I guess I still need to process all of this more!

4. Kevin

I thought I’d give my code another shot to see if I got it:

${W = \frac{1} {2 \mu_o} \int B^2 d \tau}$

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