# Magnetic field due to a circular ring

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,

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?

• 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. – Mohammed Arshaan May 13 at 13:01
• 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 $B_{net}=\int dB sin\theta$ – Tojrah May 13 at 13:18