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What happens if I rotate a current-carrying wire?

  1. May 31, 2008 #1
    Let's assume that we have a straight current-carrying wire, along the z axis of the cartesian (or cylindrical) coordinates' system. Biot - Savart law tells as that the magnetic field at a distance r from the wire is [itex]dB=\frac{\mu_0}{4\pi}I\frac{d\boldmath{l}\times r}{r^3}[/itex], where dl is the wire element.

    After we integrate over the whole wire, (assume it's infinite), we get the well known expression [itex]B=\frac{\mu_0 I}{2\pi r}[/itex].

    My question is: What happens to the magnetic field, in case we "rotate" the current carrying wire, so it's no longer perpendicular to the xy (or [itex]r\varphi[/itex]) plane, but it has an angle theta ?

    My first idea was that the only change is at the [itex]dl[/itex] element of the law, since it will no longer be along the z axis, but "analyzed" in two perpendicular vectors that will be added... vector-like. The new dlz will be the only element that will be giving a dBz field, whereas the new "dlx" (or "dlr") will not give a field, since the cross product dlr x r will be zero (the two vectors are along the same axis). So, the field B' in comparison with the initial field B will be B' = Bsin(theta). (Only dlz contributes to the magnetic field)

    My second idea (which was chronologically first, though :P) was pretty simplistic: I rotated the whole well known image of the current wire and the "rings" of magnetic field lines (the contours), well given by the "right hand rule", so every tiny "part" of the field, which was only Br first, will be now analyzed in a Br' and a Bz' components. So, this means that Br' will be Bcos(theta) and Bz' = Bsin(theta). If theta = 0 (ie our wire is straight and perp again, as it was initially) we have Br' = B and Bz' = 0 (pretty convenient...)

    Is any of these two ideas correct? If not, does anybody have a really good solution for this? :)
  2. jcsd
  3. May 31, 2008 #2


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    Your second idea is basically correct. If you angle the wire with respect to your coordinate system, the field from the wire will also be "angled." In other words, the field will have the same orientation with respect to the wire, no matter what direction this wire is pointing in.
  4. May 31, 2008 #3


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    If you write the vector direction as [tex]{\bf I\times{\hat r}}[/tex], you don't need a coordinate system.
  5. Jun 5, 2008 #4
    Electromagnetic induction in transformer

    Hello!I would like to ask about Electromagnetic induction in transformer.As we learn that an alternating voltage is connected to the primary coil,the alternating current in the primary coil produces a magnetic flux linkage with the secondary coil which constantly changes.[tex]\phi[/tex]=magnetic flux linkage in each coil of primary and secondary coil
    and the way that this formula(turn ratio)work out.Considering the transformer is ideal,why "the magnetic flux linkage in primary coil is same to secondary coil-using to work out the turn ratio formula" this assumption is true?We study of the solenoid that produce the magnetic flux density is affected by the number of turns per unit length of solenoid and yet for transformer,the solenoid is just like primary coil.With constantly changes of magnetic flux,the magnetic flux induced in the secondary coil is [tex]\phi[/tex]=NBA,B=magnetic flux density from primary coil,there is a 'N' inside the formula,so why should be the magnetic flux linkage in primary coil same as secondary coil?
  6. Jun 5, 2008 #5
    Sorry!I am the new member for this forum,I post it wrong place.
  7. Jun 12, 2008 #6
    an interesting problem relationated with that, is consider a wire, like yours, but with cross section area and besides it's infinit and have current, it is rotating about it axis with an constant angular velocity and calcule the B everywhere...:P
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