If a point charge is at rest in a frame of reference and an observer moves on an inertial frame at a velocity $\vec{v}$ with respect to the frame of reference of the point charge, according to Biot Savart's law the observer should see a magnetic field because of the moving point charge, but from the frame of reference of the charge, there is no magnetic field because the charge is at rest. How do I make sense of this contradiction. It cannot be that there is a space shared by the two frames where one sees a magnetic field and the other doesn't
120561
Suppose we have a magnetic field directed into the computer screen. Assume that there is no gravitational field. Now suppose we have a stationary positively charged particle in the magnetic field. The magnetic force is given by $$\mathbf{F}=q\mathbf{v}\times\mathbf{B}$$  Since $\mathbf{v}=0, \mathbf{F}=0$.  Now pick a reference frame moving to the left with constant velocity. In this frame, the particle is moving to the right with constant velocity. So, the magnetic force is directed upwards.  From what I've learnt in physics, forces are independent of choice of reference frames as long as the frames are inertial. But in the 2 inertial reference frames I have chosen, I get 2 different magnetic forces. How do we reconcile this?
22128
Does a charged rod or a shell have a magnetic potential energy when it is kept in a uniform magnetic field?  Also, can anyone quantitatively explain why a wire loop carrying current would tend to be circular?
5395
I am studying for my quals and came across an old question that reads like the following:      There are two regions in space separated by an infinite conducting plane. Region 1 has a magnetic dipole of moment $\vec{\mu}$ located a distance $d$ from the plane. What is the magnetic field $\vec{B}$ everywhere in region 1?   I know that for a perfect conductor (superconductor), the normal component of $\vec{B}$ is zero–although I don't exactly know why–but I don't know if that is the case when the problem involves a "conductor" (not explicitly a perfect conductor).  TL;DR: Can anybody give me some insight about the magnetic boundary conditions of a stationary magnetic field in the presence of a conductor?  Thanks!
36857
Absence of magnetic charges is reflected in one of Maxwell's fundamental equations: $$\operatorname{div} \vec B = 0. \tag1$$ This equation allows us to introducte concept of vector potential: $$\vec B = \operatorname{rot} \vec A.$$ Using this concept, it is possible to express Maxwell's equations in a graceful symmetric form: $$\nabla^2 \vec A - \frac{1}{c^2}\frac{\partial^2 \vec A}{\partial t^2} = - \frac{\vec j}{\epsilon_0 c^2}, \tag2$$ $$ \nabla^2 \phi -\frac{1}{c^2}\frac{\partial^2 \phi}{\partial t^2} = - \frac{\rho}{\epsilon_0}. \tag3 $$  Noticing, that vector $\vec A$ and scalar $\phi$ potentials, as well as electric current density $\vec j$ and charge density $\rho$, form a 4-vector in Minkovsky space-time. Therefore, Maxwell's equations can be expressed in a covariant form, using d'Alembertian: $$\nabla_{\mu}\nabla^{\mu} A_{\nu} = \frac{j_{\nu}}{\epsilon_0}. \tag4$$  If magnetic monopols exist, Maxwell's equation $(1)$ will look as: $$\operatorname{div} \vec B = \mu_0 c \rho_{\mathrm{magnet}}.$$  As the divergence of $\vec{B}$ isn't equal to zero, it impossible to introduce concept of vector potential. Thus, the equation in the form of $(4)$ will not be possible to express.
22792
This problem is from Griffith's Introduction to Electrodynamics (4th Ed.), Problem 4.31.  A point charge $Q$ is &quot;nailed down&quot; on a table. Around it, at radius $R$, is a frictionless circular track on which a dipole $\mathbf{p}$ rides, constrained always to point tangent to the circle. Use $\mathbf{F} = (\mathbf{p}\cdot\nabla)\mathbf{E}$ and show that the electric force on the dipole is $$\mathbf{F} = \frac{Q}{4\pi\epsilon_0} \frac{\mathbf{p}}{R^3}$$ Notice that this force is always in the &quot;forward&quot; direction (you can easily confirm this by drawing a diagram showing the forces on the two ends of the dipole). Why isn't this a perpetual motion machine?$^{21}$ $^{21}$ This charming paradox was suggested by K. Brownstein.  I have done the first part and shown that the force on the dipole indeed is equal to what is given in the question. But can anyone help me with the perpetual motion bit? And for the circular motion, where is the centripetal force coming from? Is it due to the constraint?
44775
I have studied that  and motional emf are two different lines of thinking but are essentially same. But then, how can Faraday's paradox be explained by Faraday's law of induction? Particularly in paradoxes in which Faraday's law of induction predicts zero EMF but there is a non-zero . The wikipedia article is more confusing rather than enlightening.  I need an explanation regarding  EDIT : I wanted to know how is flux changing if faraday's law is valid .
100113
Remark: This is not a homework question...It is pure out of theoretical interest. I asked this the mathematics-community a couple days ago and got no answer, so I figured I'd try here.  Most standard physics textbooks compute the force two infinite wires exert on each other, but they remain silent about the case where the wires are finite. Let's say we have two parallel wires carrying a current of equal magnitude in the same direction, both of which have a length $d$ and also seperated by a distance $d$. I now want to find out the force one wire exerts on another, using the Biot-Savart Law.  Let the left wire be positioned at the origin of the $xy$-plane, going along the $y$-axis, and let the other wire be a distance $d$ to the right. We assume the currents are flowing in the positive $y$-direction. Then we first choose a source element (on the left wire) of infinitesimal length $dy$ described by the position vector $\mathbf{r_0} = y_0 \hat{j}$. This constitutes a current source of $I d\vec{l} = (Idy) \hat{j}$.  We then pick an arbitrary field point $P$ on the other wire with position vector $\mathbf{r_p} = x\hat{i} + y \hat{j}$. Then the position vector $\mathbf{r}$ pointing from the source point to the field point is given as  \begin{align*}  \mathbf{r} = \mathbf{r_p} - \mathbf{r_0} = x\hat{i} + (y - y_0) \hat{j}, \end{align*} with $\sqrt{x^2 + (y-y_0)^2}$ being the length of this vector. If we now calculate the crossproduct $d\vec{l} \times \mathbf{r}$, we can write it as \begin{align*} dy \hat{j} \times (x\hat{i} + (y - y_0)\hat{j}) = -dy x \hat{k} \end{align*} Now comes the tricky part. I think I need to setup a double integral, because we are working with infinitesimal force elements $d\mathbf{F}$, each which is given as  $d\mathbf{F} = I d\vec{l} \times \mathbf{B}$. But we also have that \begin{align*} d\mathbf{B} = \frac{\mu_0 I}{4 \pi} \frac{d \vec{l} \times \hat{r}}{r^2}  = \frac{\mu_0 I}{4 \pi} \frac{d \vec{l} \times \mathbf{r}}{r^3} = -\frac{\mu_0 I}{4 \pi} \frac{dyx}{\sqrt{x^2 + (y-y_0)^2}} \hat{k} \end{align*} Hence I need to somehow integrate over $d\mathbf{F}$ and $d\mathbf{B}$. Does anyone have an idea how to do this?
68485
In the Wikipedia article , it says     After Newtonian gravitation was found to be inconsistent with special relativity, . . .   I don't see how Newtonian gravitation itself is inconsistent with special relativity.  After all, Newton's Universal Law of Gravitation is completley analogous to Coulomb's Law, so it would seem that, if there were an analogous "gravitational magnetic field", one could formulate a theory of gravitation in exact analogy with Maxwell's Theory of Electromagnetism, and of course, this would automatically be consistent with Special Relativity.  What about this approach to gravitation does not work?  The only problem I could see with this is the lack of evidence for a "gravitational magnetic field".  That being said, gravity is "weak" as it is, and my guess is that it would be extremely difficult to set up an experiment in which a massive body were moving fast enough for this "magnetic field effect" to be obsevable.  EDIT:  As has been pointed out to me in the answers, Newton itself is inconsistent with SR.  Similarly, however, so is Coulomb's Law, yet, electromagnetism is still consistent with SR.  Thus, how do we know that it is not the case that Newton's Law is the special "static" case of a more general gravitomagnetic theory exactly analogous to Maxwell's Theory:  Let $\mathbf{G}$ be the gravitation field, let $\rho$ be the mass density, and $\mathbf{J}$ be the mass current density, let $\gamma _0$ be defined so that $\frac{1}{4\pi \gamma _0}=G$, the Universal Gravitation Constant, let $\nu _0$ be defined so that $\frac{1}{\sqrt{\gamma _0\nu _0}}=c$, and suppose there exists a field $\mathbf{M}$, the gravitomagnetic field, so that the following equations hold: $$ \vec{\nabla}\cdot \mathbf{G}=-\frac{\rho}{\gamma _0} $$ $$ \vec{\nabla}\cdot \mathbf{M}=0 $$ $$ \vec{\nabla}\times \mathbf{G}=-\frac{\partial \mathbf{M}}{\partial t} $$ $$ \vec{\nabla}\times \mathbf{M}=\nu _0\mathbf{J}+\gamma _0\nu _0\frac{\partial \mathbf{G}}{\partial t} $$ where theses fields would produce on a mass $m$ moving with velocity $\mathbf{v}$ in our inertial frame the Lorentz force $\mathbf{F}=m\left( \mathbf{G}+\mathbf{v}\times \mathbf{M}\right)$.  This theory would automatically be consistent with SR and would reduce to Newton's Law of Gravitation for the case of gravitostatics:  $\frac{\partial \mathbf{G}}{\partial t}=\frac{\partial \mathbf{M}}{\partial t}=\mathbf{J}=0$.  (To be clear, you can't just set the time derivatives equal to $\mathbf{0}$ as it seems I have done.  Upon doing so, you obtain the corresponding static theory, which is technically incorrect (as you can easily see because it won't be relativistically invariant), but is nevertheless often a useful approximation.)  My question can thus be phrased as:  without appealing directly to GR, what is wrong with this theory?
16523
When an electron moves in uniform external magnetic field, with velocity not perpendicular, I understand that the magnitude of force is only due to the perpendicular component of velocity and is perpendicular to both components of velocity. But this force, though only due to the perpendicular component, acts on the particle. So why is it that the parallel component of velocity is not changed whereas the perpendicular component of velocity keeps changing its direction.   Is there a way to prove that its path is helical, using only vectors, and to show that it does not affect the parallel component? I've tried looking everywhere but haven't found any proof using vectors.
89443
I don't know how this "paradox" can be solved. I'm given the following system: A permanent magnet with a magnetic field given by ($\hat{a}$ are unit vectors in the x and y directions)  $$\vec{H}=H_0\hat{a}_y$$  and a parallel plane capacitor with an electric field  $$\vec{E}=E_0\hat{a}_x$$  Poynting's vector is given by:  $$\vec{S}=\vec{E}\times\vec{H}=H_0E_0\hat{a}_z \neq 0$$  The funny part comes when the professor told that in a system like that "clearly" there is not propagation (wich I know will imply some short of energy flux) in the $z$ direction, hence the "paradox". Is there or is there not propagation of energy?  Any hint will be appreciated, thank you for your time.
22381
From Landau &amp; Lifshitz The Classical Theory Of Fields it is said:     To determine the action integral for a free material particle (a particle not under the influence of any external force), we note that this integral must not depend on our choice of reference system, that is, it must be invariant under Lorentz transformations.   This seems understandable. But in comments on  Ján Lalinský says that "there is no good physical reason why action should be invariant". Further he suggests another Lagrangian than that given in L&amp;L, namely, if we denote L&amp;L Lagrangian as $L_0$, then the example could be $L_0+Cv_x$, which is clearly anisotropic. Clearly, the equations of motion must not change with this Lagrangian, because $Cv_x$ is a total time derivative (of $Cx$).  On the other hand, in  Luboš Motl says that "the invariance of the action follows from special relativity – and special relativity is right (not only) because it is experimentally verified. [snip] If $S$ depended on the inertial system, so would the terms in the equations $\delta S=0$, and these laws of motion couldn't be Lorentz-covariant".  How could I connect L&amp;L and Luboš's arguments with Ján's example? Can both sides be simultaneously right? They seem to contradict each other.
41277
Here are some depictions of electromagnetic wave, similar to the depictions in other places:      Isn't there an error? It is logical to presume that the electric field should have maximum when magnetic field is at zero and vise versa, so that there is no moment when the both vectors are zero at the same time. Otherwise one comes to a conclusion that the total energy of the system becomes zero, then grows to maximum, then becomes zero again which contradicts the conservation law.
114005
In an electromagnetic theory class, my professor introduced the concept of "action at a distance in physics". He said that:     If two charges are at some very large distance, and if any one of the charge moves, then the force associated with the charges changes instantaneously.   But according to Einstein, no information can travel faster than the speed of light.   So photons (the information carriers in electromagnetic force) cannot instantaneously deliver information.   So that we associate a field with the two charges and if any charge moves, there is a deformation in that field and this deformation travels with the speed of light and conveys the information.   If the field deformation information cannot travel more than the speed of light, how does the force instantaneously change at very large distances?
465
What's the difference between magnetic field and electric field lines?  Does $d\vec B$ point in the direction you would experience a force if you were a moving charged particle at that point? I know for a fact the electric field and electrostatic force are parallel, but since $F = q\ \vec v \times \vec B$ for magnetic forces. Does that actually mean though that if, say, in the following graphic:    Where the dot is a wire going in and out of the page, that the $d\vec B$ vectors are not showing how a moving charged particle will move in the field?   Basically, I feel justified in thinking electric field lines communicate how a test charge will move at any given point in space.  How can I juxtapose this with magnetic field lines>
59885
So my physics teacher assigned us an article about how special relativity causes magnetism in a wire with a current, even with the low drift velocities of electrons in a current.  It seemed that the basis of the article was that magnetism is just relativistic electricity, so I was wondering how a permanent magnet worked? It makes sense to me that a moving charged particle attracts unmoving particles of the opposite charge, but how do the orbits of electrons in a magnet cause it to have a magnetic field?
121906
Say you have a current $I(t)$ (notice the time dependence) flowing through a Toroid with $N$ total loops and all the usual approximations: $(b-a) \ll r,\; B=0$ outside.   (source: )    You are asked to calculate the magnetic field $B$. Could you apply Ampere's Law to obtain the familiar form of $B$ found for example ?:  $$B~=~\frac{\mu N I}{2\Pi r}.$$
39543
The principle of equivalence - that, locally, you can't distinguish between a uniform gravitational field and a non-inertial frame accelerating in the sense opposite to the gravitational field - is dependent on the equality of gravitational and inertial mass. Is there any deeper reason for why this equality of "charge corresponding to gravitation" (that is, the gravitational mass) and the inertial mass (that, in Newtonian mechanics, enters the equation $F=ma$) should hold? While it has been observed to be true to a very high precision, is there any theoretical backing or justification for this? You could, for example (I wonder what physics would look like then, though), have the "charge corresponding to electromagnetic theory" equal to the the inertial mass, but that isn't seen to be the case.