Figure shows a coil of conductor carrying in clockwise direction. The point P is marked at a distance x from the center of the coil and it is desired to find strength at the point due to current carried by the coil. Let ds and ds’ be elements of the conductor and the distance of point from these opposite elements be r making an angle θ with the axis. The field strength at a distance r at right angles to the tangent at ds is given by—
dH = I ds/4πr2
This is along the direction of P A. the horizontal component of this H along the axis is PC given by—
dH = (I ds/4πr2)sinθ
and its direction is towards the center of the coil if the direction of the current in the coil is clockwise as shown. The component of field strength at right angles to the axis is PB.
If we take field strength due to another element of conductor just opposite to ds on the circular conductor, it will produce field to add to the component along the axis but component at right angle to the axis will be exactly equal and opposite to that produced by ds. If we consider the circular conductor as composed of a number of small elements opposite to each other and take their resultant field strength, it will be seen that there will only be addition of their components along the axis. The perpendicular components of dH = (I ds/4πr2)cosθ will cancel each other.
Therefore,
