In the EMFT notes of MIT Course-ware, the derivation of the magnetic field due to a circular ring at its axis, using Biot-Savart's Law and the cylindrical coordinate system is done as follows,

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I am unable to understand how they calculate the $a_r$ vector component. Mainly the line, 'the radial vector changes direction as a function of $\phi$, being oppositely directed at $-\phi$, so that the total magnetic field due to the whole in the radial direction is zero.'

Why does the radial vector change direction isn't it always going to point outwards, thus in the positive direction? Can someone please explain this?


There are actually two components of magnetic field due to an element on the ring. The radial itself means along the radius (radially outwards). It is perpendicular to axis of the ring for an element and it's direction depends on position of element on the ring. As you add the vectors the net will be zero. The other component is along the axis of the ring and they sum up to produce a net field along the axis of the ring.

  • $\begingroup$ I know. But why does the direction of the radial vector become negative as we move along the $\phi$ angle? I know that the field along the axis of the ring will remain. $\endgroup$ – Mohammed Arshaan May 13 at 13:01
  • $\begingroup$ The radial vector is along the radius of the ring. The radius vector changes direction at different points on ring $\endgroup$ – Tojrah May 13 at 13:08
  • $\begingroup$ Can you give me a mathematical equation that can represent the change so that the math works out? $\endgroup$ – Mohammed Arshaan May 13 at 13:09
  • 1
    $\begingroup$ Suppose an element on the ring. The line joining that element to the point P on the axis makes angle $\theta $with the axis of the ring. (We have to determine field at the point P on the axis). The field due to element dB is perpendicular to that line. Then $\vec {dB}$ makes $\theta$ with vertical in the plane containing the line. So the radial field component is $dBcos \theta $and axial component is $dB sin \theta$. Note that$ \int dB cos \theta=0$ and $B_{net}=\int dB sin\theta$ $\endgroup$ – Tojrah May 13 at 13:18

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