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Is there a reason why two magnetic fields perpendicular to each other do not interact? If they are parallel or at a non-90 degree angle they interact. Is it because magnetic field lines can be viwed as forces (perpendicular forces have no effect on each other).

Edit: Sorry I didn't mention that the book was talking about a current-carrying wire: A current-carrying wire that is parallel to an external magnetic field will not experience a force from the magnet field because the field lines of the wire are perpendicular to those of the magnetic field.

Thanks!

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    $\begingroup$ How they do not interact? Take two permanent magnets and hold them under 90° to each other. Did you feel an action when moving them closer? $\endgroup$ – HolgerFiedler May 5 '15 at 13:37
  • $\begingroup$ In my book it says: A current-carrying wire that is parallel to an external magnetic field will not experience a force from the magnet field because the field lines of the wire are perpendicular to those of the magnetic field. $\endgroup$ – user45220 May 5 '15 at 14:53
  • $\begingroup$ That is right according to Lorentz force. $\endgroup$ – HolgerFiedler May 5 '15 at 14:55
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It is not that there is no "interaction" - at any point in space, the two magnetic fields will add up with the resultant pointing in another direction. In other words, the magnetic field caused by the wire will appear to distort the externally imposed field.

However, this distortion is radially symmetrical: you can think of the magnetic field lines as turning from closed circles into a spiral. There is no preferred direction because of symmetry.

Another way of looking at it is this:

The energy stored in a magnetic field is proportional to (the volume integral of) $B^2$. If I have two magnetic field vectors $\vec B_1$ and $\vec B_2$, with a sum $\vec B_{12}=\vec B_1 + \vec B_2$, then the energy can be thought of as

$$E_1 \propto B_1^2\\ E_2 \propto B_2^2\\ E_{12} \propto B_1^2 + B_2^2 + 2 B_1 B_2 \cos \theta$$

When the fields are perpendicular, the total energy stored is the same as the sum of the individual energies. That is a nice way of saying "one field doesn't know the other one is there". And that only happens when $\cos\theta$ is zero everywhere...

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  • $\begingroup$ "one field doesn't know the other one is there" This helped me understand it! $\endgroup$ – user45220 May 5 '15 at 15:21
  • $\begingroup$ Why is there a minus sign for the interaction term? Is it due to the way that you have defined $\theta$$? E_{12} \propto B_1^2 + B_2^2 - 2 B_1 B_2 \cos \theta$ $\endgroup$ – Farcher Aug 21 '17 at 11:17
  • $\begingroup$ @Farcher it is because I wrote down the "cosine rule" without thinking about the direction of the sum vector. Yes with the "right" definition of $\theta$ one would have a minus sign. But it should be plus. Some the entire argument was about "the cross term disappearing" the sign doesn't really matter... still thanks for spotting it, will edit. $\endgroup$ – Floris Aug 21 '17 at 11:21

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