56
$\begingroup$

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.

$\endgroup$
7
  • 17
    $\begingroup$ As usual, there's people giving over-your-head non-intuitive explanations designed to show off their knowledge, rather than answering the question. $\endgroup$ Commented May 29, 2012 at 12:06
  • $\begingroup$ There is this analogy arxiv.org/pdf/hep-th/9602081.pdf $\endgroup$
    – The Tiler
    Commented Jul 1, 2022 at 6:01
  • 1
    $\begingroup$ I’m not sure what kind of explanation you could be looking for. This is a fundamental postulate, so which more foundational premises would you be willing to start from in order to justify the equation? $\endgroup$
    – tparker
    Commented Jul 1, 2022 at 15:23
  • $\begingroup$ @Art Brown's answer cites an example from Schwartz's Principles of Electrodynamics explaining the necessity of a force perpendicular to the plane of the velocity and magnetic field in order to offset the electrostatic Coulomb force. But this not-so-simple explanation is far from being intuitive since "...(it) is absolutely deducible from electrostatics and the Lorentz transformation." In my opinion, the search for an intuitive explanation is futile. $\endgroup$
    – Frobenius
    Commented Jul 1, 2022 at 22:19
  • 1
    $\begingroup$ By the way, for an example similar to the one in Schwartz's Principles of Electrodynamics see my answer here Why don't stationary charge feel force from a current carrying wire?. $\endgroup$
    – Frobenius
    Commented Jul 1, 2022 at 22:23

13 Answers 13

29
$\begingroup$

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.)

$\endgroup$
0
19
$\begingroup$

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.

$\endgroup$
9
  • 17
    $\begingroup$ how is this intuitive? there's maths left, right and center $\endgroup$ Commented May 29, 2012 at 11:33
  • 7
    $\begingroup$ @LarryHarson: This is intuitive math. Intuitive doesn't mean not precise. $\endgroup$
    – Ron Maimon
    Commented May 29, 2012 at 16:18
  • 4
    $\begingroup$ It's not intuitive math if it's more complicated than the mathematical expression it's trying to justify!! $\endgroup$ Commented May 29, 2012 at 18:18
  • 3
    $\begingroup$ @LarryHarson: What do you mean? The force is perpendicular to "v" and to "B", and one must argue this without relativity (since it predates relativity, and inspires it). It only becomes non-arbitrary when you have relativity. Perhaps the right thing is to say "relativity" right at the start, but it won't justify why 19th century folks were sure they understood it before Einstein. If you think this is a bad answer, perhaps I'll agree--- the reflection issue is no longer considered fundamental, and the Hamilton formulation might be arbitrary too (but I don't think so). $\endgroup$
    – Ron Maimon
    Commented May 29, 2012 at 20:51
  • 4
    $\begingroup$ Thanks for posting this Ron. No one should expect technical writing to be easy, and no one should demand to understand everything at first reading - one wouldn't learn anything if one weren't challenged. Not all answers work for everybody - so what. Indeed your answers sometimes lose me - so what - I'm damn glad people like you put the effort you do into your posts. $\endgroup$ Commented Oct 3, 2013 at 1:22
12
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ It would help the op enormously if you could show how the electric force of a charge in a static electric field becomes both electric and magnetic in another frame which should only take a few lines. $\endgroup$ Commented May 29, 2012 at 12:20
  • 1
    $\begingroup$ How about if I was to say... pretty please - with cherries on the top? $\endgroup$ Commented May 29, 2012 at 19:57
  • 1
    $\begingroup$ Dear Larry, I am not sure how your proposed addition would answer the original question. For the Lorentz transformation of the electromagnetic field, see en.wikipedia.org/wiki/… - The cross products in the transformation are indeed related to the question but I personally believe that the explanation of the origin and consequences of these transformation laws is more complicated than what the OP was originally asking, so as an attempt to demystify an issue, it's like beating a flu virus with the HIV virus. $\endgroup$ Commented May 30, 2012 at 10:28
  • $\begingroup$ I do appreciate the effort you've clearly put into your answer. Unfortunately this is a bit beyond my knowledge, but I will be trying my best to understand it. Thank you. $\endgroup$ Commented May 31, 2012 at 4:06
  • $\begingroup$ The OP asked for an explanation of a certain fact. The first three paragraphs just restate that fact without explaining why it's true. The fourth, fifth, and sixth paragraphs are again just a statement of the fact, but now in fancier mathematical dress. $\endgroup$
    – user4552
    Commented Oct 2, 2013 at 23:43
12
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ This the example I use in my lectures. Though I start with a charged particle at rest net to a current carrying wire - no force. Then look at the fields in the moving frame - there is both magnetic and electric fields. The particle must not move towards the wire in the moving frame, so there must be a force qvB and it must be perpendicular to v and B. As you say - just an example, but a powerful one. $\endgroup$
    – ProfRob
    Commented Oct 31, 2014 at 0:18
  • $\begingroup$ @RobJeffries, Nice! Thanks for the info. "Nulling" set-ups are very attractive; er, well, you know what I mean... $\endgroup$
    – Art Brown
    Commented Oct 31, 2014 at 0:38
7
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ This is at completely the wrong level compared to the level at which the OP asked the question, and it's only very tangentially related to the question. $\endgroup$
    – user4552
    Commented Oct 2, 2013 at 23:53
  • 1
    $\begingroup$ Okay, maybe this isn't the most focused answer (hey, I would NEVER go off on a tangent, just ask my friends :) ) but it sure is rivetingly interesting, and not altogether unrelated. At an abstract level it IS exactly the kind of thing the OP is asking for: deeper, intuitive motivation. Thanks David for posting this - I learnt something. And it isn't too different in level from the ideas in Ben's book (which I also think are damned interesting). $\endgroup$ Commented Oct 3, 2013 at 1:12
4
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Very insightful. Thank you! Couldn't edit the last comment so had to remove it and entered a new one. $\endgroup$
    – verdelite
    Commented Apr 1, 2019 at 20:30
3
$\begingroup$

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.

The Lorentz force visualized as an interaction between 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

$\endgroup$
8
  • $\begingroup$ I don't think this works as written. Field lines aren't physical objects that attract and repel each other, and the charge doesn't experience forces from fields that occur at some distance from it. $\endgroup$
    – user4552
    Commented Oct 2, 2013 at 23:40
  • $\begingroup$ I know that magnetic lines doesn't exist, because the real explanation is relativist. However, it's an useful abstraction (tpub.com/neets/book1/chapter1/1i.htm) to improve Physics intuition. If you follow carefully the above explanation, you notice that it matches with hand rule convention. $\endgroup$ Commented Oct 3, 2013 at 2:21
  • $\begingroup$ Look at faculty.polytechnic.org/physics/3%20A.P.%20PHYSICS%202009-2010/…. Use the magnetic field lines analogy it's useful to draw the right force direction. See the magnets and parallel wires examples. $\endgroup$ Commented Oct 3, 2013 at 3:00
  • $\begingroup$ I know that magnetic lines doesn't exist, because the real explanation is relativist. This doesn't really make sense. They do exist, but they aren't physical objects. The powerpoint page you linked to actually doesn't support the analysis in your answer. It describes a rule for telling whether physical objects attract or repel. Your answer talks about field lines attracting or repelling. $\endgroup$
    – user4552
    Commented Oct 3, 2013 at 3:45
  • $\begingroup$ I'm talking about special relativity explaining magnetism. Supposes two parallel wires A and B with positive current to the right. Get a electron in A, distance of protons in B decrease, so this electron will be attracted. Now get a proton in A, distance of electrons in B decrease, so the proton will be attracted. That force explain the Magnetism. $\endgroup$ Commented Oct 3, 2013 at 5:05
2
$\begingroup$

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 $v$.

$\endgroup$
10
  • 2
    $\begingroup$ No, this doesn't rule out that F could have a component anti-parallel to v. $\endgroup$ Commented Mar 2, 2013 at 5:45
  • 1
    $\begingroup$ @RetardedPotential then the velocity would slow down, without the energy being dissipated into heat. Still violates the first law. $\endgroup$
    – N. Virgo
    Commented Mar 2, 2013 at 9:11
  • 2
    $\begingroup$ Also if F had an antiparallel component I could just change the sign of the charge could I not? $\endgroup$
    – Ryan
    Commented Mar 2, 2013 at 17:47
  • $\begingroup$ How does this argument explain why $B$ has to be perpendicular to $F$? (You meant $F$ at the end? $B$ does not have to be perpendicular to $v$.) $\endgroup$ Commented Mar 2, 2013 at 22:25
  • 1
    $\begingroup$ I would move this to chat but I can't. After thinking about it some more I have an idea . We have established that F must be perpendicular to v. Thus, F can only cause the velocity vector to change direction. If the F is not perpendicular to B then the force will cause the velocity vector to align with the B field more and more. However, the magnitude of the force is proportional to the velocity vector component perpendicular to v. Thus, if F is not perpendicular to B then the force will cause v to align with B thereby decreasing the magnitude of the force. $\endgroup$
    – Ryan
    Commented Mar 3, 2013 at 4:54
2
$\begingroup$

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!

$\endgroup$
1
+100
$\begingroup$

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.

Energetic figure in 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.

enter image description here

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

$\endgroup$
1
$\begingroup$

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

That means $𝐁$ 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 $𝐢$, $𝐣$ and $𝐤$ the respective unit vectors. Then $𝐅 = F_x𝐢$, $𝐯 = v^y𝐣$ and $𝐁 = B^z𝐤$, 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: $𝐁 = ∇×𝐀$, 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, $𝐋$, which is defined in terms of position $𝐫$, velocity $𝐯$ and mass $m$ by $𝐋 = m𝐫×𝐯$ is also a bi-vector; with $L_z = L^{xy} = m\left(xv^y - yv^x\right)$, and $L^{xy} = -L^{yx}$.

$\endgroup$
1
$\begingroup$

There is an intuitive way to answer your question without the need to look at math and just by using some logical reasoning and using the relativistic nature of magnetic field specifically contraction effects.

If we see the visual representation of the contraction from Lorentz transformation of the electric field of a moving particle, we will see that the electric lines will become denser in the region perpendicular to particle motion:

1

and since every disturbed electric field vector has a magnetic vector that accompanies it, the particle's own generated magnetic field will be densest also in the perpendicular direction to the motion of the charge.

(An analogy can be just like a Danser where she contracts her arms so that she turns faster around herself due to the redistribution of her energy the electric field lines contract to redistribute the energy of the charged particle and as a result create a denser magnetic field perpendicular to its motion.)

And since the reason of Lorentz force is actually due to interaction between the external magnetic field and the magnetic field of the particle, and since the particle has its own field densest perpendicular to its motion as stated earlier, for the external magnetic field to interact most efficiently with it, it will need to be able to affect the densest region thus the external magnetic field also needs to be exerted perpendicular to the particle line of motion. and since this rotation depends on how perpendicular the magnetic and velocity vectors are and also since it results in a circular path so its a centripetal force than it must be a product vector between both velocity and magnetic vectors thus perpendicular to both of them.

To summarize, the reason this perpendicular orientation maximizes the Lorentz force is because the external magnetic field can most effectively "grab" onto the moving charge by interacting with its own generated magnetic field. Since from the way the electric fields of charge are contracted and are densest in perpendicular direction to the axis of motion than its own magnetic field is densest there so the external magnetic field can only maximize his interaction in the perpendicular direction to the particle.

In this way, it will maximize the torque it will exert on the particle own generated magnetic field so the particle will precess to conserve the angular momentum of it s own field at a maximum frequency. This precession, combined with the particle's initial movement in straight line, will result in it going in a circular path due the maximized torque that is generated by the external magnetic field.

A good analogy will be a spinning toy. The toy, if it doesn't spin, it falls down instantly and don't move. Just like the particle, if it's not initially moving, then the external magnetic field will not apply Lorentz force on it. But now when you spin the toy it will not fall. Instead, over time, as it loses its energy, it will start to wobble around its axis and with the way you throw the toy initially (assuming you threw it in a straight line just like the initial electron straight path of motion), the initial tendency to go in straight line plus the wobbling of the toy (which is called precession if we want to get more technical) will result in the toy going in a circular path just like the electron did. The gravity is analogous to the external magnetic field and the spin of the toy is analogous to the angular momentum of the particle own generated magnetic field. The precession to conserve angular momentum of particle is ultimately why currents tends to oppose the external change in magnetic field.

$\endgroup$
0
$\begingroup$

enter image description here

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.