To calculated the field due to a circular coil at point P on it axis (see figure) we take a element AB of the circular coil. The field due to this element is
We resolve this field into perpendicular component (dBcosβ) and parallel component (dBsinβ) . Now if we take a symmetrically situated element A’B’ on the opposite end of the diameter and similarly resolve into parallel and perpendicular components and add the field component’s due to these two elements we find that the perpendicular components cancels out. Whatever is true for one pair of elements is also trace for every other pair. Hence we conclude that the field due to a circular coil at a point other on its axis is along the axis of the coil and is given by
Putting these values we obtain, after integration.
There is a simple rule for finding the polarity of the two faces of a current carrying circular coil. According to the rule when an observer looking at the circular coil, finds the current to be flowing in the anticlockwise sense that face of the coil behaves like the N–pole of the equivalent magnet. On the other hand, if the current is seem to flow in the clockwise sense that face of the coil behaves like the S–pole of the equivalent magnet.
