11
$\begingroup$

According to relativity, If magnetic field is just an electric field viewed from a different frame of reference, why is the magnetic field around the wire is circular?

enter image description here

$\endgroup$
  • $\begingroup$ I don't know enough to provide you with why the magnetic field is circular, but I don't think your terminology is quite correct. The magnetic field is certainly not the electric field viewed from another Gaililean frame. The magnetic field arises when the electric field is transformed under relativity. The exact transformation in your example above will have to come from someone else. $\endgroup$ – Reid Erdwien Mar 21 '15 at 5:16
  • $\begingroup$ why do you think it is circular? $\endgroup$ – Skaperen Mar 23 '15 at 9:23
  • $\begingroup$ what if the wires are charged but no current is flowing? $\endgroup$ – Skaperen Mar 23 '15 at 9:25
  • $\begingroup$ If I assume the essence of the question is why the magnetic field force lines do not originate from anywhere ? That’s a question that just cannot be explained from “credible and/or official sources”. In other words: there is no answer yet. $\endgroup$ – Leon Sprenger Mar 24 '15 at 19:04
  • $\begingroup$ @Skaperen: If the wires are charged with nu current flowing there is no magnetic field. $\endgroup$ – Leon Sprenger Mar 25 '15 at 16:07
6
$\begingroup$

Your statement is not really true, since if you only have a magnetic field in one frame of reference, then it can never be viewed as just an electric field in another frame of reference. And vice-versa.

As described here, the magnetic field can be defined as (e.g. in Jackson's Classical Electrodynamics) the field that is responsible for the Lorentz force $q\vec{v} \times \vec{B}$. Since in the example you show, the force would always be directed radially for charges moving parallel to a current carrying wire, then the field must circulate around the wire.

The reason that the force associated with the magnetic field is radial in such circumstances is one of the set-piece arguments in most textbooks that deal with these things, but arises from the requirement that a charge that is radially stationary with respect to the wire in one frame of reference is also stationary in any other frame of reference moving parallel to the wire. It goes something like this:

Consider the electric/magnetic fields due to a current carrying wire in the stationary frame and a frame moving uniformly, but parallel to the wire.

In the stationary frame, the wire is overall neutral, so there can only be a magnetic field. In the moving frame there is some transformed magnetic field and an electric field radial to the wire, caused by a difference in length contraction for the positive and negative charges in the wire, which due to the current flow, must be moving in opposite directions in the stationary frame.

This electric field in the moving frame clearly exerts a radial force on any test charge originally at rest with respect to the wire in the stationary frame. But, given that there is no radial force or acceleration in the stationary frame, there also cannot be a net radial force on the charge when it is in the moving frame either. The force that counteracts the radial electric field in the moving frame is the Lorentz force due to a mystery (B-)field. As the Lorentz force due to the mysterious (B-)field is observed to be both proportional to and perpendicular to the velocity, then it is natural to define it in terms of a vector product. And in that case, in order to act radially for a charge at any point around the wire, the B-field must circulate around the wire.

$\endgroup$
  • $\begingroup$ Thanks. Why does B,according to Biot-Savart Law, have a cross product? Magnetic has a different direction from the magnetic force. Electric field and force have same direction, but this not true for magnetic field and force. I am trying to understand the logic behind the cross product in magnetism $\endgroup$ – user50322 Mar 24 '15 at 16:41
  • $\begingroup$ @user50322 The Lorentz force is defined in that way. I suppose what you need to think about is if you wanted to define an intrinsic field that produced a radial force on a particle moving parallel to the wire, what else could it be but circulating around the wire. It must be a vector and it must produce a force perpendicular to the velocity for all positions around the wire. $\endgroup$ – Rob Jeffries Mar 24 '15 at 19:59
  • $\begingroup$ Same direction with magnetic force. Like electric field and electric force. $\endgroup$ – user50322 Mar 24 '15 at 22:16
  • $\begingroup$ @user50322 I understand (I think) your comment; but then how do you incorporate the fact that the magnitude of the force is proportional to the velocity? The force due to the electric field doesn't depend on the particle velocity. That is how they differ. In addition there are no sources or sinks of B-field, which precludes a radial field. $\endgroup$ – Rob Jeffries Mar 24 '15 at 22:19
  • $\begingroup$ So can we say, magnetic field doesn't have physical meaning and we don't have any other way to define it besides defined in that way. So we define it that because we can not make any definition any other way. Right? Can we say that? $\endgroup$ – user50322 Mar 24 '15 at 22:47
3
$\begingroup$

According to relativity, If magnetic field is just an electric field viewed from a different frame of reference

It is true that a pure electrostatic field in an inertial reference frame (IRF) will be observed as a mix of electric and magnetic fields in some relatively moving IRFs.

However, in the general (time varying) case, it is not possible to find an IRF in which the magnetic field vanishes.

why is the magnetic field around the wire is circular?

Consider the field of an isolated point charge at rest; a purely radial, static electric field.

From a relatively moving IRF, there is a magnetic field component in addition to the electric field. This magnetic field is perpendicular to the velocity vector and electric field in the rest frame and is given by

$$\mathbf {{B}_{\bot}}'= \gamma \left(-\frac{1}{c^2} \mathbf{ v} \times \mathbf {E} \right)$$

A little reflection on the above should convince you that, looking at the charge along the direction of motion, the magnetic field lines form circles centered on the charge.

The extension to a line of charge is straightforward.

$\endgroup$
  • $\begingroup$ Why does B have to be equal to v x E? Why does B have to be circular? Don't have the same direction with the magnetic(in fact electric) force? Thanks... $\endgroup$ – user50322 Mar 23 '15 at 23:07
  • $\begingroup$ @user50322, what do you mean by "why?". If you are asking why nature is the way it is and not some other way, I don't have and answer (and there may not be one). $\endgroup$ – Alfred Centauri Mar 23 '15 at 23:15
  • $\begingroup$ I mean "how". How does B equal to v x E? $\endgroup$ – user50322 Mar 23 '15 at 23:17
  • 1
    $\begingroup$ @user50322, this is how the fields transform between relatively moving inertial reference frames according to special relativity. This is not something that can be explained in a comment or an answer for that matter. Simply put, Maxwell's equations, the equations that correctly describe classical electromagnetism, are relativistically covariant. $\endgroup$ – Alfred Centauri Mar 23 '15 at 23:33
1
$\begingroup$

Your question consists actually of two parts, I will answer them one-by-one:

Why is the magnetic field circular?

Any vector field $\vec F$ can be decomposed into a rotational part and a divergent part, according to the Helmholtz decomposition theorem
$$\vec F = - \vec \nabla \Phi + \vec \nabla \times \vec A $$
This is a purely mathematical statement and has nothing yet to do with physics. Physics comes into play when considering Maxwell's second equation
$$\vec \nabla \cdot \vec B = 0$$ Which means that $\vec B$ is divergence-free or source-free. Because we can use the decomposition theorem to calculate
$$\vec \nabla \cdot \vec B = \vec \nabla \cdot (-\vec \nabla \Phi + \nabla \times A) = \nabla \cdot \nabla \times A$$ We see that $B$ must be a purely rotational field and this results because the divergence of a gradient ($\vec \nabla \cdot \vec \nabla \Phi$) always vanishes. The latter statement also follows from math and is not a physical modell.

the second thing is:
How are magnetic and electric field related?

This now follows from Maxwells equations 3+4 other equations, that we didn't use until now. It's them that imply relativity and the transformations that my fellow posters noticed.

Concluding

So really the circularity of the B-field has nothing to do with relativity. But its relativity that allows us to transform between both, no matter how their geometries are.

$\endgroup$
  • $\begingroup$ Yes but B is related with v x r (according to biot-savart). I understand why F is perpendicular to v(Lorentz Force). It is because energy conservation. But I don't understand why B and current must be perpendicular(Biot-Savart force) $\endgroup$ – user50322 Mar 25 '15 at 10:10
  • $\begingroup$ The current is provided by the movement of the electrons. $\vec v \sim \vec j$ in metals. $\endgroup$ – AtmosphericPrisonEscape Mar 25 '15 at 12:36
0
$\begingroup$

Given a certain four-current $J^\mu = (c \varrho, \vec{j})$, that is a charge density $\varrho$ and current density $\vec{j}$. the four-potential $A^\mu = (\Phi / c, \vec{A})$ is given by: $$ A^\mu(\vec{r},t) \propto \int \frac{j^\mu(\vec{r}\ ', t_r)}{|\vec{r}-\vec{r}\ '|} d^3r'$$ with $t_r = t - \frac{|\vec{r}-\vec{r}\ '|}{c}$ if one takes the retarded solution. Therefore the B-Field is: \begin{align} \vec{B} = \nabla \times \vec{A} & \propto \int \frac{\vec{j}(\vec{r}\ ', t_r) \times (\vec{r}-\vec{r}\ ')}{|\vec{r}-\vec{r}\ '|^3} d^3r' + \int \frac{\nabla \times \vec{j}(\vec{r}\ ', t_r)}{|\vec{r}-\vec{r}\ '|} d^3r' \\ & = \int \frac{\vec{j}(\vec{r}\ ', t_r) \times (\vec{r}-\vec{r}\ ')}{|\vec{r}-\vec{r}\ '|^3} d^3r' + \int \frac{\frac{\partial \vec{j}(\vec{r}\ ', t_r)}{\partial t_r} \times (\vec{r}-\vec{r}\ ')}{c|\vec{r}-\vec{r}\ '|^2} d^3r' \end{align}

The first term is which you would call "circular", since $\vec{j}(\vec{r}\ ', t_r) \times (\vec{r}-\vec{r}\ ')$ points always in a direction perpendicular to the current density and the point of interest (relative to the current density). The second term is zero if the current is stationary, that is if it is not time dependant. For example, this is the case if one looks at magnetic fields induced by a wire of constant charge flow.

So in general, the magnetic field is not circular around a wire.

$\endgroup$
  • $\begingroup$ Yes but B is related with v x r (according to biot-savart). I understand why F is perpendicular to v(Lorentz Force). It is because energy conservation. But I don't understand why B and current must be perpendicular(Biot-Savart force) $\endgroup$ – user50322 Mar 25 '15 at 10:11
  • $\begingroup$ Do you accept the maxwell equations? If so, my answer above shows you why. $\endgroup$ – image Mar 25 '15 at 14:47
0
$\begingroup$

since magnetic field in a wire carrying current is due to the movement of electrons, I assume that the magnetic field of a single isolated moving electron, or any other charged particle, can also be deduced by the right hand thumb rule i.e. the field will be much like Saturn's ring around it, where we assume saturn to be a charged particle, and its ring as its field. Now my question is: How is this field deduced. Why is it circular? Is it the result of superposition, if any? Secondly, I know that magnetic force exerted on a moving charge consists of a cross product of B and v. But I think that this force is the result of interaction between the charge's own magnetic field and the applied field, B. My question is: How does this interaction exactly happen? How does it result in this force? Most importantly, what causes this force to be perp to both B and v (I need an explanation other than that it is just the result of cross product)? I raised a similar question elsewhere but didn't get much help. I'd be grateful if somebody explains.

Reference https://www.physicsforums.com/threads/why-is-the-magnetic-field-of-a-wire-circular.180845/

$\endgroup$
0
$\begingroup$

According to relativity, If magnetic field is just an electric field viewed from a different frame of reference...

Relativity doesn't quite say this. Take a look at Minkowski's Space and Time: "In the description of the field caused by the electron itself, then it will appear that the division of the field into electric and magnetic forces is a relative one with respect to the time-axis assumed; the two forces considered together can most vividly be described by a certain analogy to the force-screw in mechanics; the analogy is, however, imperfect". Also see Jackson’s Classical Electrodynamics section 11.10 where he says "one should properly speak of the electromagnetic field Fμv rather than E or B separately". The field of the electron is the electromagnetic field, and it has a "screw" nature, which you can trace back to Maxwell. When you have two charged particles with no initial motion, you see linear electric force only. When you throw one past the other, you also see rotational magnetic force, as per positronium. IMHO the electron's electromagnetic field isn't totally unlike the frame-dragged gravitomagnetic field, and you can depict it by thinking "spinor" and combining radial electric field lines with concentric magnetic field lines, a bit like Maxwell's convergence + curl sketch on page 7 of this paper:

enter image description here

Why is the magnetic field around the wire circular?

Because it's rotationally symmetrical. Think of the wire as a column of electrons interleaved with a column of protons. Their electromagnetic fields cancel. But when you turn on the current and move the electrons, the fields don't quite cancel any more. The residual field has a cylindrical disposition, such that a charged particle thrown past the wire will loop around the magnetic field lines. We call it a magnetic field, but it's just one aspect of the greater whole that is the electromagnetic field.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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