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Say I consider an amperian loop surrounding the sphere, wouldn't the net current through it be zero as the current entering the loop due to one half of the sphere and the current exiting in the opposite direction are equal in magnitude? However this contradicts the fact that magnetic field outside such a system is not zero. What am I doing wrong?

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  • $\begingroup$ The loop is in the plane perpendicular to the rotation axis of the sphere? $\endgroup$
    – Christophe
    Commented Mar 21 at 6:48
  • $\begingroup$ No, the loop is in the same plane as the rotation axis of the sphere, you could also say the area vector of the loop is perpendicular to the axis of rotation. $\endgroup$ Commented Mar 21 at 6:50

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As you have figured in your question, the net current through the area of the loop would be $0$. Therefore also the magnetic field integrated over the entire loop is indeed $0$. If the loop is oriented symmetrical around the rotation axis, you can assume, that the magnetic field in direction of the loop is the same for each point of the loop due to rotational symmetry and therefore it would be $0$ at each point. Looking at a more general loop geometry for example the one mentioned in your comment, such symmetry argumentation doesn’t hold anymore. So the magnetic field integrated over the entire loop would still be $0$, but it does not have to be at every point, in general there will be positive contributions in some parts of the loop and negative in others.

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  • $\begingroup$ I did not get why you would say such symmetric arguments wouldn't hold anymore? Could you explain a bit more? $\endgroup$ Commented Mar 24 at 4:51
  • $\begingroup$ Well your rotating ball, that generates the field, is completely symmetrical around its rotation axis, therefore you expect the fields also to be rotational symmetrical around the rotation axis. That means, that, if the loop is oriented symmetrically around the axis of rotation, you expect the field to be the same with regard to the orientation of the loop at each point of the loop. If it is not oriented symmetrically around the rotation axis, you just have no reason to assume such thing anymore. $\endgroup$
    – Zaph
    Commented Mar 25 at 12:05

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