The field dB due to a small element dl of the circle, centered at A has the magnitude

       o       Idl                              Idl     

dB = -----  -------           = ----  ---------

          4 p    |AP|2                    4 p   (R2 + a2)

    This field can be resolved into two components one along the axis OP, and the other (PS) perpendicular to it.  The latter component is exactly cancelled by the perpendicular component (PS) of the field due to a current and centred at A. Field along OP has a magnitude

                       I dl

   dB (along OP) =   ---- ---- { sin }

                4 p      r2

 

                       I dl   a

                                  =   --- --------------

                                    4 p R2 + a2 (R2 + a2)1/2

 

                o          I a

                     =   --- ------- dl

                         4 p (R2 + a2)3/2 

    The magnetic field due to the circular current loop of radius a at a point which is a distance R away, and is on its axis (i.e. on a line perpendicular to the plane of the circle and passing through its center) is

 

                       o         I a2

             B =   ---  -------   i (i is the unit vector along OP, the x-axis)   tesla (Wb/m2)

                2     (R2 + a2)3/2

See also