12
$\begingroup$

I could see how it would go either way.

On the one hand, there IS moving charge; on the other hand, it is not discernibly moving.

$\endgroup$
11
$\begingroup$

What you call "discernibly" moving, is called a non-stationary current-density. Consider a wire with constant charge-flow at every cross section. Then this current is in the same sense not "discernibly" moving as your charge on the sphere. Still there is a magnetic field around the wire.

You can think of the charged sphere as infinitely many circular wires around the axis of rotation. All these wires create magnetic fields that add up. So yes, there is a magnetic field due to a rotating sphere.

For a calculation approach I suggest this: https://physics.stackexchange.com/a/173022/75518

$\endgroup$
4
  • 2
    $\begingroup$ So a 'spinning' electron would NOT produce a magnetic field, because (as a singular point source) it does not represent a flowing current. Is this correct? $\endgroup$
    – Jiminion
    Apr 3 '15 at 18:47
  • 8
    $\begingroup$ @Jiminion: There is no notion like "spinning point". Still, electrons have a property called Spin, which is kinda similar to angular momentum (this is what u mean by spinning), but still not the same! electrons have a magnetic moment due to this Spin, but it is usually not explained on a classical basis. If you get into the dirac-equation, you'll see that this magnetic moment is actually not caused by any "spinning" at all. Still, sometimes this visualisation is usefull. $\endgroup$
    – image357
    Apr 3 '15 at 19:04
  • $\begingroup$ Did the OP mean theoretically alone in space or somewhere here on Earth? I was under the assumption that the earth generates its own magnetic field; by its outer core moving relative to its solid, inner core. Doesn't this sphere have to spin relative to something? (and if it is, then the answer is, yes?) Or if it's charged it would have a MF, even without spinning? $\endgroup$
    – Mazura
    Apr 4 '15 at 6:19
  • $\begingroup$ @Mazura: You are right, motion is always relative to something. Here, the sphere is thought to be spinning relative to the (inertial) system of an observer, where the sphere's center of mass is not in motion. Furthermore, notions like electric- or magnetic-fields are relative to an observer, too. Think of a charge, that is at rest relative to one observer. This means there is only an electro-static-field. Now consider a second observer that is in uniform movement relative to the first one. He will see the charge in uniform movement and thus observe a magnetic- and electric-field! $\endgroup$
    – image357
    Apr 4 '15 at 11:16
1
$\begingroup$

The things that induce a magnetic field are a current or a changing electric field (with respect to time). In this situation you have charges moving,so you have a J.dS, which means you have a current.

EDIT:

If you prefer a different approach, then consider the sphere to be an infinite amount of circular circuits (each with different radius because of the spherical surface).

So you know that when you have moving charges in a circuit, it means that you have current, and thus you have induced magnetic field.

So now you have infinite circular circuits one next to another. (Think of a solenoid with infinite turns. The same approach with different geometry can be made here-only for intuition though.)

$\endgroup$
0
-6
$\begingroup$

You have a moving charge and a circular moving that is an acceleration for what you need a force. Remember the Lorentz force $ \vec F = q \vec v \times \vec B $. This vector cross product can be rewritten to $$ \vec B = \dfrac{\vec F \times \vec{qv}}{\|\vec{qv}\|^2}$$.

Edit: The equation was edited following https://math.stackexchange.com/questions/1219103/how-to-rewrite-a-vector-cross-product/1219127#1219127

$\endgroup$
15
  • 2
    $\begingroup$ I'm sure this identity is wrong! take vectors $a=(1,1,1), b=(2,0,0), c=(0,2,-2)$ then $a\times b = c$ but $c\times a \ne b$ and $c \times b \ne a$. Furthermore: $c\times a$ and $c\times b$ are not even (anit-)parallel to $b$ and $a$ $\endgroup$
    – image357
    Apr 3 '15 at 18:54
  • 4
    $\begingroup$ (Since this has been flagged) Although MarcelKöpke is completely correct that this identity is wrong, this answer is an attempt to answer the question. Please do not flag wrong answers as "not an answer" or "low quality", but downvote them. $\endgroup$
    – ACuriousMind
    Apr 3 '15 at 19:42
  • $\begingroup$ @MarcelKöpke: Please do your calculations with orthogonal vectors again. $\endgroup$ Apr 3 '15 at 20:21
  • 1
    $\begingroup$ The point is, that Lorentz-foce is the force applied to an particle that "interacts" with the above device by means of its magnetic field. One would usually not apply it to the device particles itself (since they are kept one a certain rotation anyway). $\endgroup$
    – image357
    Apr 3 '15 at 20:40
  • 1
    $\begingroup$ @HolgerFiedler it DOES mean rigid body, but all the FORCES involved are supplied by the structural integrity of the field. Your equation has nothing to do with the field actually generated by the charges. The Lorentz force law does not give you the field generated by the charges. $\endgroup$
    – user12029
    Apr 6 '15 at 5:51

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.