# 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 '19 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 '19 at 13:18
Yes, the radial component due to a current element on the ring always points outwards. But, you have to consider this "outward" direction with reference to coordinates in the $$x$$-$$y$$-plane. Consider the current element sitting on the positive side of the x-axis on the ring at point $$(a,0,0)$$. This will produce a radial component of the $$B$$-field at point $$(0,0,z)$$ in $$-x$$ direction. Whereas the current element sitting on the negative side of the $$x$$-axis at point $$(-a,0,0)$$ on the ring will produce a radial $$B$$-field component pointing in $$+x$$ direction. Both components have equal strengths and such cancel out. For every current element on the ring you will find a current element on the opposite side of the ring, and in any case the generated radial $$B$$-field components will cancel out for them.