While studying magnetism, I learned that force on a charged particle inside a magnetic field is $$\mathbf{F}=q(\mathbf{v}\times\mathbf{B})$$ where all symbols have their usual meaning, which implies that a charged particle must be in motion to experience a force in this magnetic field. Now in case of electric field there was no such condition, so why here? Why is it necessary for a charge to be in motion, that also not parallel or antiparallel to the magnetic field as $F=0$ if angle between the velocity vector and the magnetic field vector becomes $0°$ or $180°$?

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    $\begingroup$ Veritasium put out a video How Special Relativity Makes Magnets Work. But see this one instead - Veritasium's 'How Special Relativity Makes Magnets Work' - EXPLAINED (better) $\endgroup$
    – mmesser314
    Commented May 22 at 3:08
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    $\begingroup$ Physics has no answer to fundamental 'why?' questions. $\endgroup$
    – my2cts
    Commented May 22 at 9:17
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    $\begingroup$ Moving charge produces own magnetic field which interacts with external magnetic field applied. So to say only electric field can interact with electric field and magnetic field - with magnetic field directly. That's why. Different field type interactions can happen only indirectly, as in electromagnetic wave propagation for example. $\endgroup$ Commented May 22 at 11:20
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    $\begingroup$ I think its because the magnetic force is actually an electro static force of charge distributions in the past, given the finite speed of light. Then effects of time dilation and such add a layer of complexity, effectively weighting past distributions to yield the real time field. $\endgroup$
    – R. Romero
    Commented May 22 at 18:50
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    $\begingroup$ Thank you guys ::). Concept cleared! $\endgroup$
    – Advait K
    Commented May 23 at 4:54

7 Answers 7


The question "why?" in Physics may have different answers, depending on the depth one wants and is able to dig.

Lorentz force and the development of special relativity

So, assuming you know nothing/little about special relativity:

  • one answer could be "it is what it is". This was the answer for the best scientists investigating electromagnetism until the early years of the $20^{th}$ century;
  • another answer could tell you that Einstein's special relativity provides a more detailed explanation:
  1. your question is one of those that could have made Einstein start thinking about relativity, especially the expression of Lorentz force on electric charges in an electromagnetic field $$\mathbf{F} = e \left( \mathbf{e} + \mathbf{v} \times \mathbf{b}\right) $$

  2. despite its name, Einstein's relativity may be thought as a theory about invariance/absolute nature of physics: physics is independent from the observer

  3. whenever we speak about velocity, we should tell "velocity with respect to ..."; since it's possible to choose an arbitrary reference for evaluating velocities, it's very unlikely that a force can be a function of velocity

  4. since:

    • 2 inertial observers (in uniform relative motion) see the very same physics (and no "fictitious force" due to relative acceleration), the force measured by these 2 observers must be the same
    • the velocity of the particle w.r.t. these two observers differs by the relative motion of the observers


    • these 2 observers measure different components of the electromagnetic field as well: what an observer sees as components of a magnetic field, the other observer may see them as components of the electric field

Low-speed example

Let's do an example with negligible special relativity effects on space and time, i.e. in the "low-speed" limit, $v \ll c$. Let's take two observers consider an electric charge in a uniform linear motion with velocity $\mathbf{v}_{/0} = v_{/0} \mathbf{\hat{x}}$ with respect to an observer "0"; and an observer "1" at rest with the electric charge (w.r.t. the electric charge has zero velocity). The observer "0" measures a magnetic field $\mathbf{b}_{/0} = B_{/0} \mathbf{\hat{y}}$ and no electric field $\mathbf{e}_{/0} = \mathbf{0}$. Both observers measure the same force, whose general expression is the Lorentz force,

$$\mathbf{F} = e \left( \mathbf{e}_{/k} + \mathbf{v}_{/k} \times \mathbf{b}_{/k}\right) \ ,$$

being the quantities $\{\}_{/k}$ the quantities measured by observer $k$ and the charge $e$ independent from the observer.

Observer "0" measures a force determined by the magnetic field $\mathbf{b}_{/0}$,

$$\mathbf{F} = e \left( \underbrace{\mathbf{e}_{/0}}_{=\mathbf{0}} + \mathbf{v}_{/0} \times \mathbf{b}_{/0} \right) = e \mathbf{v}_{/0} \times \mathbf{b}_{/0} = e v_{/0} B_{/0} \mathbf{\hat{z}} \ .$$

Now, observer "1" measures the same force on the electric charge that has no relative velocity,

$$\mathbf{F} = e \left( \mathbf{e}_{/1} + \underbrace{\mathbf{v}_{/1}}_{=\mathbf{0}} \times \mathbf{b}_{/1} \right) = e \mathbf{e}_{/1} = e v_{/0} B_{/0} \mathbf{\hat{z}} \ .$$

Thus, he can only interpret the force as the result of an electric field acting on the charge, and that electric field must be $\mathbf{e}_{/1} = E_{/1} \mathbf{\hat{z}} = v_{/0} B_{/0} \mathbf{\hat{z}}$.

This last expression provides you the low-speed limit rule of transformation of the electromagnetic field components between the two observers of this example, preserving the invariant nature of physics.

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    $\begingroup$ Thank you so much for this explanation, beyond amazing! I understood what you said :) $\endgroup$
    – Advait K
    Commented May 23 at 4:52
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    $\begingroup$ Glad it helped. P.S. Remember to accept the answer if you're fine with it $\endgroup$
    – basics
    Commented May 23 at 7:00
  • $\begingroup$ How is the electric field zero? E is non-zero in any frame $\endgroup$ Commented May 26 at 15:26
  • $\begingroup$ @ThomasTiger this is not the way Nature works. The electric field is not an invariant quantity, but it's a contribution of the invariant electromagnetic field (mathematically a tensor field) along with the magnetic field. If special relativity holds, the components of this tensor transforms with Lorentz transformations $\endgroup$
    – basics
    Commented May 28 at 13:16
  • $\begingroup$ @basics I think that what is missing in your explanation is the assumption that the E and B are the result of unseen charge carriers (current in a wire). In the reference frame of the wire the charges cancel so that the net electric field is zero. With this clarification, then your explanation makes perfect sense $\endgroup$ Commented May 29 at 19:43

If there is a field force, then it can do work on a particle that is being affected by the field. Therefore, the energy corresponding to the work done is stored or removed from the field between the two particles. Basically, for field force between two particles, both particles must have there own field which can interact with each other.

For example:1) Gravitational force acts because both masses have their own fields.

2)If a charge is kept in an electric field, the charge's electric field interacts with external field.

In case of magnetic force, the magnetic field of a charge is produced when the charge is in motion. Therefore if the charge was at rest, then it does not have its own magnetic field which can interact with the external field.

You may have heard of gravitational potential energy. If you take the Earth and the block as a system, then you lift the block, you lose some energy. That energy goes into the system. Gravitational potential energy is not the energy of block, neither the earth. This energy is the energy of the combined system, which is stored in the interacting field lines, since interacting field lines are the reason for force (work-energy theorem).

Similarly, if a charge is stationary, it has no magnetic field of its own, therefore there is no interaction between the charge and the field. Thus no force should act on the charge.

To answer your question about why there is no force when v and B are parallel. enter image description here

If a charge is moving straight towards a magnetic field, the field of charge and the external field is perpendicular. Therefore, there is no interaction that causes force. You can say that the green field of the charge are like small bar magnets, aligned in the tangential direction of green rings. If you place a magnetic perpendicular to a magnetic field, there is no external force. (although there will be a torque on one bar magnetic, but since the field is circular, all the torque gets cancelled out. (also it was an analogy, there are no actual bar magnets, but it helps to tell the direction of force.)

In summary, for force you need two interacting field which can store energy and also the fields should not be perpendicular. This will generaly work for any field like gravitational, electric, etc.

  • $\begingroup$ Thank you!!! I understood! $\endgroup$
    – Advait K
    Commented May 23 at 4:53
  • $\begingroup$ Happy to help :) $\endgroup$ Commented May 23 at 15:18

a charged particle must be in motion to experience a force in this magnetic field.

This is not entirely true; a macroscopic charged particle may experience force in purely magnetic field, even when the particle is at rest, if it carries non-zero magnetic moment $\mathbf m$ and magnetic field varies with position. The force then can be expressed as $$ \mathbf F = \mathbf m \cdot \nabla \mathbf B $$ where $\nabla \mathbf B$ is called gradient of magnetic field and denotes matrix of spatial derivatives of magnetic field components. Such force pulls the particle to regions of space where magnetic field is stronger.

We believe electrons do carry some magnetic moment, but when in a beam in macroscopic EM field, predicted value of this gradient force is too small to detect and verify, and the motion is well explained by the Lorentz force

$$ \mathbf F = q\,\mathbf v\! \times\! \mathbf B. $$

Now in case of electric field there was no such condition, so why here?

Because the Lorentz force (dependent on particle velocity) is observed to be the major component of total magnetic force when the particle moves. Being proportional to velocity and having direction perpendicular to it is the differentiating property of the observed force due to magnetic field we've discovered in experiments, such as when studying electron beams in a cathode ray tube in controlled macroscopic EM field. These experiments show that observed force due to $\mathbf B$ depends strongly on voltage creating the electron beam entering the region with magnetic field, and is perpendicular to the beam, while force due to electric field does not depend on that voltage, and its direction does not depend on direction of the beam.

Mathematically, these characteristics of force due to $\mathbf B$ are not necessary - in another world, force due to $\mathbf B$ could have been independent of velocity. For example, force as described by the expression

$$ \mathbf F = q_m\mathbf B + \mathbf m \cdot \nabla \mathbf B $$ allows for the particle to carry "magnetic charge", and is independent of particle's velocity. Or the force could have been dependent on velocity, but in a different, more complicated way that would make magnetic force do work on the particle (e.g. the Landau & Lifshitz force they proposed to describe the effect of "radiation reaction" on motion of the charged particle).

But these alternative laws of force, if the forces were strong enough, would not describe the world as we observe it. For example, it is not clear how the first alternative could be consistent with theory of relativity; to make the statements of law Lorentz-invariant, Maxwell's equations would have to change, to account for magnetic charge, and to recover consistency with special relativity. The second alternative could be made consistent with theory of relativity easily, but it breaks conservation of energy of matter+EM field. So these alternatives describe different worlds, or particles and fields we have not detected yet.

  • $\begingroup$ ‘We believe electrons do carry some magnetic moment’ We know. ‘when in a beam in macroscopic EM field, predicted value of this gradient force is too small to detect’ Except in the Stern-Gerlach experiment. $\endgroup$
    – my2cts
    Commented May 28 at 11:44
  • $\begingroup$ @my2cts Stern-Gerlach experiment was done with silver atoms, so we know atoms have magnetic moment. As far as I know it is hard to do it with electron beams, the required magnetic field gradient is too high. $\endgroup$ Commented May 28 at 12:38

I could answer your question with a lot of demonstrations or formulas, but I think it is much more intuitive to say that the formula describes the world we live in. If the particle has to be in motion it is because based on experimentation we have seen that if a particle does not move when it is under the action of a magnetic field no force acts on it. Moreover, if it is perpendicular to the magnetic field lines, it does not experience any force either. To describe these experimentally measured data, formulas are developed, in this case $\vec{F}=q \vec{v} \wedge \vec{B}$.


Instead of thinking of it as an electric field and a magnetic field, think of it as a single electromagnetic field has that affects the particle whether it is moving or not. The Lorentz force is $\mathbf{F}=q(\mathbf{E} + \mathbf{v}\times\mathbf{B})$. If there is no Electric field and the test particle has velocity v relative to the observer at rest in a magnetic field, the force is $\mathbf{F}=q(\mathbf{v}\times\mathbf{B})$, but in the rest frame of the particle, the force is simply $\mathbf{F}=q(\mathbf{E})$ and the magnetic field has vanished and is replaced by an electric field. There is no way to definitively say that the force is due to motion relative to the magnetic field, because the fields are frame dependent.


A stationary electric charge creates an electric field around it. An electric charge in motion creates a magnetic field around it. Two magnetic fields against one another create a force. Two charges in proximity to one another, with their own electric fields, creates a force between them. Use the mathematics as your guide. From the formula for force above, $\textbf F$, $\textbf v$, and $\textbf B$ are all vectors. Vectors have magnitude and direction associated with them. The charge, $\textit q$ is a scalar, which means just magnitude. The reason angles are considered is because of the cross product:

$$\textbf v \times \textbf B = vBsin (\theta ).$$

In the right hand side of this equation, $\textit v$, and $\textit B$ are now scalars. Using this, and the right had rule, is all you need from this point.


A magnetic charge would move in a magnetic field even if stationary at the beginning but an electric charge would not. You can view them as a pair like this: (magnetic charge, magnetic field), (electric charge, electric field). A magnetic charge would only be affected by an electric field if in motion too. There is a certain symmetry between the pairs. However there is no magnetic charge or at least we could not find any so far.

  • $\begingroup$ There is no such thing as magnetic charge. Thus is not sn answer. $\endgroup$
    – my2cts
    Commented May 28 at 11:39
  • $\begingroup$ @my2cts You just repeated what I have already said. Did you not read my last sentence? And you cannot prove that magnetic charge will never be found. $\endgroup$ Commented Jun 5 at 14:24

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