Is there an intuitive explanation for why Lorentz force is perpendicular to a particle's velocity and the magnetic field?

Question

The Lorentz force on a charged particle is perpendicular to the particle's velocity and the magnetic field it's moving through. This is obvious from the equation:

$$ \mathbf{F} = q\mathbf{v} \times \mathbf{B} $$

Is there an intuitive explanation for this behavior? Every explanation I've seen simply points at the equation and leaves it at that.

I can accept mathematically why $\mathbf{F}$ will be perpendicular to $\mathbf{v}$ and $\mathbf{B}$ (assuming the equation is correct, which it is of course). But that doesn't help me picture what's fundamentally going on.

Trying to create an analogy with common experiences seems useless; if I were running north through a west-flowing "field" of some sort, I wouldn't expect to suddenly go flying into the sky.

I'm hoping there's a way to visualize the reason for this behavior without a deep understanding of advanced theory. Unfortunately, my searching for an explanation makes it seem like something one just has to accept as bizarre until several more years of study.

Answer 1

Trying to create an analogy with common experiences seems useless; if I were running north through a west-flowing "field" of some sort, I wouldn't expect to suddenly go flying into the sky.

This is a reasonable expectation, since the electric and gravitational fields do make forces that are in the direction of the field. So let's try to see what goes wrong if we write down a force law for magnetism that behaves in the same way. The first thing we could try would be

$$ \textbf{F}=q\textbf{B} \qquad (1) $$

Well, this doesn't work, because such a force would behave in exactly the same way as the electric force, and it would therefore be the electric force, not a separate phenomenon. Magnetic forces are supposed to be interactions of moving charges with moving charges, so clearly we need to include $\textbf{v}$ on the right-hand-side. One way to do this would be the standard Lorentz force law, but we're looking for some alternative that is in the direction of the field. So we could write down this:

$$ \textbf{F}=q\textbf{B}|\textbf{v}| \qquad (2) $$

As an example of what's wrong with this one, suppose we have identical charges $q$ bound together with a spring. If they're sitting at rest in equilibrium, equation (2) says there's no magnetic force on them. But suppose we start them vibrating just a little bit. Now they're going to start shooting off in the direction of the magnetic field. This violates conservation of energy and momentum.

Fundamentally, this comes down to an algebraic issue. The vector cross product has the distributive property $(\textbf{v}_1+\textbf{v}_2)\times\textbf{B}=\textbf{v}_1\times\textbf{B}+\textbf{v}_2\times\textbf{B}$, and in the example of the charges on a spring, with the actual Lorentz force, this guarantees that the magnetic forces on the two charges cancel out. We really need this distributive property, and in fact it can be proved that the vector cross product is the only possible form of vector multiplication (up to a multiplicative constant) that produces a vector result, is rotationally invariant, is distributive, and commutes with scalar multiplication. (See my book http://www.lightandmatter.com/area1sn.html , appendix 2.)

Answer 2

Pseudovector argument

There is an intuitive argument, but the first thing to do is to take the Poincare dual of B. In 3 dimensions, there is an epsilon tensor $\epsilon_{ijk}$ which is invariant--- it doesn't change under rotations. It has $\epsilon_{123}=1$ and all interchanges give a minus sign, so that the value of $\epsilon$ is zero of two of the indices are equal, and the sign of the permutation to get to 123 if they are all different. The epsilon tensor contracted with three vectors $v_1,v_2,v_3$ gives the signed area spanned by the parallelepiped they form. Because the signed area is the determinant of the matrix of v's put together as 3 columns, it changes sign under reflection of all three coordinate axes.

The fundamental quantity in electromagnetism is $B_{\mu\nu}=\epsilon_{\mu\nu\sigma}B^{\sigma}$, the epsilon tensor contracted with B. This thing is a rank 2 antisymmetric tensor. Because the $\epsilon$ tensor is invariant, an antisymmetric tensor is equivalent to a vector under rotations, but it isn't equivalent under reflections. The reason is that although a vector changes sign under reflections, a tensor doesn't. This is also true of B--- it's a pseudovector, if you reflect a space with a current carrying wire, the direction of B does not reverse.

The fact that B is fundamentally a tensor, not a vector, means that when it is interacting with a particle with velocity v, it can only form a force when one of the indices are contracted with something. The only thing one can contract with is the velocity, so you get $B_{\mu\nu}v^\mu$ as the force, and this is $v\times B$

In relativity, this is seen to be the only natural thing, since the E and B fields together make an antisymmetric 2-tensor, and the four-Lorentz force is this tensor contracted with the 4-velocity. This form is so natural and intuitive, that it does not require a detailed justification.

More physical restatement of the argument

The above is sort-of formal sounding, but it is just saying this: the magnetic field doesn't change sign under reversing the coordinates of space. To see this physically, consider a solenoid of current stretching along the z axis from -a to a, with the current mostly in the x-y plane along each winding, and reflect this solenoid in the x-y-z axes. Reflecting x reverses the current, reflecting y gets it back to where it started, and reflecting z doesn't change the solenoid.

Since the current is the same, B is the same! So the B from a solenoid doesn't change under reflection. So the force on a particle can't be along the direction of B, because force reverses direction under reflection and B doesn't. The force can only be in the direction of a quantity which does reverse direction, and the simplest such quantity is $v\times B$. Under a reflection, v reverses direction and B doesn't, so the Lorentz force properly reverses.

This argument assumes reflection symmetry, which is a symmetry of electromagnetism, but is actually not a fundamental symmetry in our universe. The same reflection argument shows that magnetic charge is not properly symmetric with electric charge, since magnetic charge changes sign under reflection (reflect all the coordinates with the charge at the origin--- the field moves to a new location, but points in the same direction, so the sense of the magnetic charge is reversed). This property means that magnetic monopoles were an early sign that nature is not parity invariant, and may explain why Dirac was not surprised when the weak interactions were shown to violate parity.

The other assumption is that the force is the simplest reflection invariant combination of B and v. If you abandon the idea that the force is linearly proportional to B, there are more complicated combinations that also work to give a reflection invariant force-law. These combinations generally fail to obey conservation of energy.

In order to have automatic energy conservation (and an automatic phase space with the symplectic properties), you should derive your equations of motion from the action.

Hamiltonian argument and gauge invariance

The best argument is from the concept of momentum potential (or vector potential). Like the energy has a potential energy added to it, which is $e\phi$, the momentum has a potential added which is $eA$, where A is the vector potential.

The Lagrangian for the interaction is

$$ mA\cdot \dot{x} $$

Which makes the conjugate momentum $mv + eA$, so that the kinetic energy is $(p-eA)^2\over 2m$ and the potential energy is $e\phi$. The Hamilton equations for this energy give the Lorentz force law. The Hamilton equations are:

$$ \partial_t p = \partial_x {(p-eA)^2\over 2m + \phi}$$ $$ \partial_t x = {p - eA \over m}$$

And combining the equations to a second order equation for the acceleration of x gives the Lorentz force law. The same replacement in the Hamiltonian, $p$ to $p-eA$, works in relativity to give the correct four dimensional Lorentz force law.

The identification of B with $\nabla \times A$ can be justified from the invariance of the equations under adding a gradiant to A. The classically physical part of A is its curl, and this is sensible to identify with the B in Maxwell's equations.

This argument is fundamentally sound, because it doesn't depend on reflection invariance (any argument relying on reflection symmetry is really bogus, since we know this is not a symmetry of nature in any fundamental sense), and is correct quantum mechanically when you interpret the gauge invariance as a freedom in the local phase redefinition of a charged particle wavefunction. It's only drawback is that it requires some familiarity with Hamilton's principle.

Answer 3

The Lorentz force is orthogonal to the velocity which is equivalent to the proposition that the force does no work on the charged particle; it only changes the direction of the velocity, not its magnitude.

The force is also orthogonal to the magnetic field. It follows from the formula and this fact – and the whole formula – may be derived in various methods, e.g. from the special theory of relativity.

This feature – the force's being perpendicular to the field – makes the magnetic field different from the electrostatic and gravitational fields. It is different in this respect: unlike the electrostatic and gravitational field, the field strength isn't a gradient of any "scalar potential". But there is no paradox. Different effects in Nature may follow different mathematical formulae and they often do.

If you have a problem with that, just appreciate that $(B_x,B_y,B_z)$ which looks like a vector is just a shorthand for $(F_{yz},F_{zx},F_{xy})$, three components of an antisymmetric tensor with three indices (the components are subsets of the relativistic tensor $F_{\mu\nu}$ which also contains the electric field).

For example, if $\vec B=(0,0,B_z)$, then the nonzero third component may be written as $B_{z}=F_{xy}$ and instead of an arrow in the $z$-direction, you may visualize the field by an oriented loop (with an arrow) in the $xy$-plane (to which the $z$-directed vector is normal). It's the same information.

This loop in the $xy$ plane really tells you what the magnetic field does to the charged particles: it rotates their velocities clockwise (or counter-clockwise, depending on the sign of $B_z$ and the charge $Q$) in the $xy$-plane.

Answer 4

Here's an example from Schwartz's Principles of Electrodynamics, based on relativity. (Have I just disqualified this answer as "not intuitive"? Carrying on regardless...)

Imagine an infinite, straight wire with a constant current that is composed of equal numbers of positive and negative charges flowing in opposite directions (so the wire's net charge density is 0).

Now add a particle with charge q moving parallel to the wire with constant velocity v (lab frame K). What's the force on the particle?

To answer, Schwartz transforms the problem to the rest frame K' of the particle. Applying the corresponding Lorentz boost to the wire's charge-current four-vector, one finds that in K' it has a non-zero charge density. The charge is therefore attracted towards the wire by an electrostatic field.

(There are two assumption/empirical facts here: 1) in the particle's rest frame, the force is given by the electric field as the particle sees it, and 2) because the charge and current distributions are time independent, one can calculate the electric field in K' by the usual integration-over-charge-density approach.)

Boosting back to the lab frame, one finds the answer to the question, which is that the moving charge feels a force perpendicular to its velocity.

If you now separately calculate the B-field for the current by the usual formula, you also find that the force calculated above satisfies F=q vxB.

Of course, the above is just one example (and would not be applicable if the particle were moving towards the wire instead of parallel to it, because assumption 2 would be violated.) There's a more elaborate induction process to get to the full relativistic apparatus. However, the above example does establish the existence of a force perpendicular to a charged particle's velocity and the magnetic field.

Answer 5

I'll give you here a very short argument based on quantum mechanics. This argument actually has a profound physical and historical origin which I'll try to give in the sequel. In quantum mechanics, the velocity operator of a particle in an external magnetic field is (under minimal coupling):

$ \mathbf{v} = \frac{p - q \mathbf{A}}{m}$

Where $\mathbf{A}$ is the vector potential. This implies that the magnetic field is given using the canonical commutation relations:

$ \mathbf{B} = - i m^2 \mathbf{v} \times \mathbf{v}$

Now, it is not difficult to check that

$ \mathbf{v}.\mathbf{F} \propto \epsilon_{ijk} [v_i, [v_j, v_k]]$

which vanishes by the Jacobi identity ($ \mathbf {F} $ is the Lorentz force), which is the required orthogonality relation.

This "derivation" is actually a part of a very interesting argument by Feynman, behind which there is a very interesting story. Actually, the argument doesn't require quantum mechanics but only the notion of Poisson brackets. Feynman didn't take the argument seriously and didn't publish it. It was not until 1990 after his death when this argument was published (in his name) by Dyson: F.Dyson,Am.J.Phys.58,209(1990).

Feynman's argument is very profound because it allows the derivation of the whole Maxwell theory (together with the Lorentz force equation) starting from very simple and basic assumptions:

1. The canonical Posisson brackets of the position and velocity.

2. Minimal coupling: The electromagnetic force on a charged point particle depends on the position and velocity only.

Please, see the introductory section in the following article by Carinena, Ibort, Marmo and Stern.

One of the generalization of this procedure is the derivation of the Yang-Mills equations according to the same principles

Answer 6

I suppose one could think of this in terms of source and response. Electric and magnetic fields are produced by sources, namely electric charges and currents. One can think of currents as a relative drift between oppositely charged particles.

Let us consider an electron with velocity in the xy-plane (V$_{\phi}$) under the influence of a magnetic field along +$\hat{z}$, or B$_{z}$. To produce B$_{z}$, we could use a circular wire of current, I$_{\phi}$, in the xy-plane that had a vector direction in the positive azimuthal sense (or counterclockwise). Currents are defined by the motion of positive charges, so electrons move in the opposite direction.

Thus, the acceleration of an electron due to the influence of B$_{z}$ would cause the particle's trajectory to look like a counterclockwise circle, similar to I$_{\phi}$ mentioned earlier. However, the current due to this electron, I$_{e}$, is in the opposite direction to I$_{\phi}$, or I$_{\phi}$ ~ -I$_{e}$ (ignoring magnitudes, just worry about sign here).

The purpose of this somewhat elaborate answer is to illustrate that the response of an electron is to effectively counteract the field acting on it. It is similar to the concept of induction in Faraday's law, whereby a current is induced to try and prevent magnetic flux from changing. The idea is similar with the electron, where its orbit is effectively an induced current and this current acts counter to the B$_{z}$.

This should make sense since energy/momentum need to be conserved. One can think of this in two ways: B$_{z}$ loses energy/momentum to accelerate the electron or B$_{z}$ loses energy/momentum because the current induced (by the electron's orbit) acts against it. Hmm, this last part is more confusing than I first thought. I wonder if the "or" should be an "and"? Regardless, the particle responds to the field producing an effective current in the opposite sense to the one that produced the field.

Answer 7

I don't like intuitive explanations that are not intuitive! Intuitive explanations cannot contains formulas and math.

It should make an analogy, which, although not totally accurate, helps the reader to feel something behind a dry formula or theorem.

I search for some intuitive explanation for a Lorentz Force for a long time and now I've found one that I like very much.

Let's start with a figure (below), that shows the Lorentz force visualized as an interaction between imaginary magnetic tubes.

_{(source: conspiracyoflight.com)}

Imagine one that is looking at a vertical magnet, south part to the left side and north part to the right side. The magnetic lines go from right to left (N->S)

Now imagine a positive charge moving vertically through the magnetic lines. It generates a magnetic field around itself by the right hand rule. The lines of this field are horizontal and counterclockwise.

Remember that parallel magnetic lines of force traveling in the same direction are normally consequence of a repulsion force. Parallel magnetic lines of force traveling in opposite directions are usually consequence of an attraction force.

If you are looking at the magnets and the moving charge in the vertical, in the back (far side) the magnetic lines (external and charge generated) are at the same direction, that is typically produced by a forward force. In the front (near side) the magnetic lines are at opposite direction, that is normally generated by a additional forward force.

As a consequence, the particle experiences a force from the far side to the near side, with the dark arrow shown in the figure.

Finally, if the force had a component at the same direction to the speed, the force will generates a continuous speed increase. It will create kinetic energy increase forever, because the magnets don't need to be loaded. If the force had a component in the opposite direction to the speed, the charges will stop and there's no possible electric current inside a magnetic field.

With the advent of Einstein’s treatment of electromagnetism, the magnetic lines of force has been relegated to an imaginary entity. However, it's a useful approach to explain concepts!

Source: https://www.conspiracyoflight.com/Lorentz/Lorentzforce.html

Answer 8

How about this? Assume that the Lorentz force is not perpendicular to $v$ (that is $v$ has a nonzero component parallel to $F$). The force acts on the charge causing it to accelerate...which in turn increases the force ($qvB$ where $v$ is the component parallel to $F$) this in turn increases the acceleration which increases the force and so on ad infinitum. This would obviously violate conservation of energy and therefore $v$ must be perpendicular to $F$. The same argument explains why $B$ must be perpendicular to $v$.

Answer 9

Don't be fooled by the "strange" definition of the magnetic field $\boldsymbol{B}$. Consider two particles moving with velocities $\boldsymbol{v}_1$ and $\boldsymbol{v}_2$, separated by a distance d (relatively close to ignore delay effects in the field propagation) and such that at the instant the separation vector between them is $\boldsymbol{r} = d\ \boldsymbol{\hat{u}}_{12}$ (being $\boldsymbol{\hat{u}}_{12}$ a unit vector). The force of particle 1 on particle 2 is:

$$\mathbf{F}_{1\to 2}= q_2 \mathbf{v}_2\times \mathbf{B}_1 = \frac{\mu q_2q_1}{4\pi}\ \frac{\mathbf{v}_2\times (\mathbf{v}_1\times\mathbf{\hat{u}}_{12})}{d^2}$$

while the force of particle 2 on particle 1 is:

$$\mathbf{F}_{2\to 1}= q_1 \mathbf{v}_1\times \mathbf{B}_2 = \frac{\mu q_2q_1}{4\pi}\ \frac{\mathbf{v}_1\times (\mathbf{v}_2\times(-\mathbf{\hat{u}}_{12}) )}{d^2}$$

By using the vector identity $\mathbf{a}\times(\mathbf{b}\times\mathbf{c}) = (\mathbf{a}\times\mathbf{c})\times\mathbf{b} = (\mathbf{a}\times\mathbf{b})\times\mathbf{c}$, it can be seen that the first force $\mathbf{F}_{1\to2}$ is in the plane formed by $\mathbf{\hat{u}}_{12}$ and $\mathbf{v}_1$. On the other hand, the second force $\mathbf{F}_{2\to1}$ is in the plane formed by $\mathbf{\hat{u}}_{12}$ and $\mathbf{v}_2$.

Therefore, the force made by each particle is on the plane defined by the separation and its own velocity: it is a bit as if each particle pushes in the direction of its own velocity and the direction towards the other particle!

Answer 10

For intuition perhaps you can think about it from an experimental perspective. If you look at a charged particle moving in a uniform magnetic field the motion is circular. For example looking at the path of a particle in a cloud chamber.

You can then do things like measure the radius and since it moving in a circle, you realize that there needs to be an inward force that is at right angles to the direction of the velocity motion.

Since there is no other charge present at the center of the circle for the particle to orbit around, or mass or gravitation force for the particle to orbit around something must be generating the force (we call it the Lorentz force).

Thinking back to your classical mechanics you know the

You can do other experiments like change the mass of the charge particle and see that radius of the circle changes. When you do that you find the radius is

$\rho=\frac{mv}{qB}$

or thinking about the acceleration for a particle of mass in a circle is $\frac{mv^2}{\rho}$ you come up with the force

$f=qvB = \frac{mv^2}{\rho}$

At that point you f= qvB is the same as qvBsin(90) since the angle is 90 degrees and that the definition of the cross product can be written as

$\vec{v} \times \vec{B} = |v||B|\sin\theta\hat{n}$

And you do some more experiments and find out that the angle between the velocity of the particle and magnetic field matters and writing the vector equations in terms of a cross product is a nice short hand for that.

So $F=q(\vec{v} \times \vec{B})$

Answer 11

In the force law $ = q×$ if $q$ is a positive charge, $$ points north, and $$ points up, then $$ points east. So a positive charge goes clockwise in a circular motion, from the vantage point looking down at its motion.

That means $$ is not actually a vector, but stands for circulation along the horizontal plane. It doesn't point up, it circulates horizontally. That defines a bivector.

Clockwise, though; at least with positive charges. Negative charges go counter-clockwise.

Call the $x$, $y$ and $z$ directions, respectively, east, north and up and $$, $$ and $$ the respective unit vectors. Then $ = F_x$, $ = v^y$ and $ = B^z$, and $B^z$ is actually more accurately written as $B_{xy} = -B_{yx}$, so that $F_x = q B_{xy} v^y$.

A strong hint of that fact is already staring everybody in the face: $ = ∇×$, therefore $$B_{xy} = \frac{∂A_y}{∂x} - \frac{∂A_x}{∂y},$$ or as it is more usually written: $B_{xy} = ∂_x A_y - ∂_y A_x$.

The "orbital" part of angular momentum, $$, which is defined in terms of position $$, velocity $$ and mass $m$ by $ = m×$ is also a bi-vector; with $L_z = L^{xy} = m\left(xv^y - yv^x\right)$, and $L^{xy} = -L^{yx}$.

Answer 12

The charged partial would never stay in place. You are trying to solve without the whole equation, but your answer should look like this for a working model. Instead if the positive monopole disk you would replace it with your positive charged particle.