If be the position vector of point P relative to the elementdl, then from Biot-Savart s law, the field at point P due to the current element is I dl 4 dB 2 r = , therefore Since dl s, i.e., Idl dB 4 s2 The field lies in the plane of paper and is perpendicular to , as shown by PQ. dB s be the angle between OP and CP. Then dB can be resolved into two rectangular components. 1. the axis, 2. perpendicular to the axis. dB For any two diametrically opposite elements of the loop, the components perpendicular to the axis of the loop will be equal and opposite and will cancel out. Their axial components will be in the same direction, i.e., along CP and get added up. Therefore, total magnetic field at the point P in the direction CP is = sin a Idl dB and 4 s s2 B .Idl.a Since and are constant, and and same for all points on the circular loop, we have I s aare I 0Ia [ = 2 3 3 3 s (r s 2 As, I a2 0 B 2 2
If s be the position vector of point P relative to the elementdl, then from Biot-Savart law, the field at point P due to the current element is dB I dl r 2 = , therefore Since dl s, i.e., dB Idl s2 The field dB lies in the plane of paper and is perpendicular to s , as shown by PQ. be the angle between OP and CP. Then dB can be resolved into two rectangular components. 1. dB the axis, 2. perpendicular to the axis. For any two diametrically opposite elements of the loop, the components perpendicular to the axis of the loop will be equal and opposite and will cancel out. Their axial components will be in the same direction, i.e., along CP and get added up. Therefore, total magnetic field at the point P in the direction CP is = sin a and dB Idl s s2 B .Idl.a Since and I are constant, and s and aare same for all points on the circular loop, we have = 0Ia I [ 3 3 2 s 3 As, s 2 B 0 I a2 2 2 (r 4 4 4
physics-xii-15-16.docx