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I can do the math using Lorentz force equation and it turns out that the angle is $0$. I understand that the magnetic field lines of force generated by the charge will be circular and perpendicular to the the those of the uniform magnetic field at all times. But I don't get how will those field lines interact and why will the charge experience no force at all.

Could you please give me an intuitive explanation involving magnetic field lines and how will they interact in this case and why will the charge experience no force at all?

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  • $\begingroup$ Note that a charge at rest WRT a magnetic field also experiences no force. So you shouldb't be expecting simply that "charge + magnetic field = force"; the situation is obviously more complicated than that. But I suspect that to push below the level of "this is simply an experimental fact" will require you to deal with the relativistic treatment in terms of the electromagnetic field tensor. $\endgroup$ – dmckee --- ex-moderator kitten Mar 2 '19 at 18:42
  • $\begingroup$ @dmckee I said the charge is moving...I mean with uniform velocity (I forgot to say that). Is there no intuition ? $\endgroup$ – user8718165 Mar 2 '19 at 18:43
  • $\begingroup$ I know. I was asking you to consider a second situation in which you have a charge and a field and a lack of force. The point is to kick you out of the mindset (if indeed you have such a mindset) that you should automatically expect an electric charge to experience a force from a magnetic field. Such a force is conditional and the conditions are non-trivial. $\endgroup$ – dmckee --- ex-moderator kitten Mar 2 '19 at 18:45
  • $\begingroup$ 0 velocity is also uniform velocity. $\endgroup$ – Rudi_Birnbaum Mar 2 '19 at 19:41
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An intuitive understanding is somewhat subjective, so...

I think the 1st thing to do is forget about magnetic field lines, especially "field lines of force". That's not really a thing. I think the intuitive upside of field lines is that it is a graphical way to show an irrotational magnetic field (in free space). Likewise, for the electric field, one can show that the divergence is proportional to the charge density. Regarding forces on charges, they don't add much intuition.

For the Lorentz force law:

$$ \vec F = q(\vec E + \vec v \times \vec B )$$

all that matters is $\vec E$ and $\vec B$ at the (point) location of $q$, in the reference frame where its velocity is $\vec v$.

So all that matters is what $q$ sees locally. Of course, $q$ always sees itself at rest, so only its definition of $\vec E$ matters regarding the $\vec F$ it experiences.

Now that is strange, since it completely discounts $\vec B$. It's not a paradox, but it is an "oddity", and the resolution of the oddity requires understanding Special Relativity.

If you have a $1/r^2$ central force, such as:

$$ F = \frac 1 {4\pi\epsilon_0}\frac{q_1q_2}{r^2} $$

or

$$ F =G\frac{m_1m_2}{r^2} $$

you will find that they don't hold up under Galilean transformations. Analysis of either interaction in a moving frame leads to inconsistencies.

For gravity, one must introduce curved spacetime and General Relativity to get a consistent picture.

For electromagnetism, the inconsistencies are resolved by working in Minkowski space and seeing that the electric field is part of an antisymmetric 4-tensor (with $c=1$):

$$ F^{\mu\nu} = \left [\begin{array}{cccc}0&-E_x&-E_y&-E_z\\E_x&0&B_z&-B_y\\E_y&-B_z&0&B_x\\E_z&B_y&-B_x&0\end{array}\right ]$$

The covariant form of the Lorentz force law can then be expressed as:

$$\frac{dp^{\mu}}{d\tau} = qF^{\mu\nu}u_{\nu} $$

At this point, it appears we are moving away from intuition; however, when we pick a reference frame, we split spacetime into space and time and $F^{\mu\nu}$ in $\vec E$ and $\vec B$, recovering:

$$ \vec F = q(\vec E + \vec v \times \vec B )$$

Since in the frame where $\vec v = 0$, only $\vec E$ matter, I somewhat subjectively see the existence of $\vec B$ as being required to maintain covariance (or consistency, if you will) in other reference frames.

The fact the magnetic field of a moving charge is:

$$ \vec B = \frac 1{4\pi\epsilon_0 c^2}\frac{q\vec v \times \hat r}{r^2}$$

ensures that this interpretation works.

That:

$$ \vec F_B \equiv q\vec v \times \vec B $$

can be zero for non-zero velocity and non-zero magnetic field only if they are parallel is a lot like the Coriolis Force:

$$ \vec F_C = 2(\vec v \times \omega)$$

If the velocity is aligned, there is no effect.

Note that in both these cases, neither $\vec B$ nor $\vec \omega$ are true vectors. They are axial-vectors (aka pseudovectors). That is, they rotate like vectors but do not change sign under reflection.

Under reflection (parity inversion):

$$ \vec F \rightarrow -\vec F$$ $$ \vec E \rightarrow -\vec E$$ $$ \vec v \rightarrow -\vec v$$

while

$$ \vec B \rightarrow +\vec B$$ $$ \vec \omega \rightarrow +\vec \omega$$

so that any force that isn't zero when the velocity is parallel to the magnetic field (or rotation axis) would be a parity violating force, and no-one's intuition can accept that.

Note that the cross product:

$$ \vec v \times \vec B $$

is a true vector, so that a non-parity violating force can be proportional to it.

So in summary, the entire thing has a profound geometric symmetry, and that is the source of useful intuition.

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