Physics · Moving Charges And Magnetism · NEET
Drop a perpendicular from the point P to the wire (or its line). The foot of this perpendicular is your reference. t1 and t2 are measured from this perpendicular to the lines joining P to the two ends of the wire. If an end lies on the SAME side as you sweep out, its angle is positive (add); if the whole wire is on one side of the foot, one angle can be negative (subtract). For an infinite wire both ends are at 90 degrees, so B = (mu0 I / 4 pi d)(sin90 + sin90) = mu0 I / 2 pi d.
Put the point P right at the level of the wire's end (the end is at the foot of the perpendicular) and let the wire run off to infinity on one side. Then one angle is 0 (the near end sits at the foot) and the other is 90 degrees (far end). So B = (mu0 I / 4 pi d)(sin0 + sin90) = (mu0 I / 4 pi d)(0 + 1) = mu0 I / 4 pi d. The infinite wire is mu0 I / 2 pi d, so semi-infinite is exactly half.
No. If P lies on the extension of a straight segment (the wire points straight at or away from P), then dl and r are parallel, so dl x r = 0 for every element. That segment gives ZERO field at P. This is why in bent-wire problems you often ignore the radial straight pieces that aim at the centre and only add the arc plus the perpendicular-offset segments.
An arc of angle phi (in radians) at its centre gives B = (mu0 I / 4 pi R) x phi. A semicircle is phi = pi, so B = mu0 I / 4R. A full circle (phi = 2 pi) gives mu0 I / 2R. Just take the fraction of a full loop equal to the fraction of 2 pi that the arc covers.
Add the magnitudes only if all contributions point the same way (check with the right-hand rule; here they usually all go out of or into the page at the centre). For the classic NEET figure: semicircle gives mu0 I / 4R, and two semi-infinite straight tails each give mu0 I / 4 pi R, so total = (mu0 I / 4R)(1 + 2/pi).
A very long conducting wire is bent in a semi-circular shape from A to B (radius R), with two long straight tails on either side and point P at the centre of the semicircle. The magnetic field at point P for the steady current configuration is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
B = (mu0 I / 4 pi d)(sin t1 + sin t2), where d is the perpendicular distance from the point to the wire and t1, t2 are the angles subtended by the two ends at that point, measured from the foot of the perpendicular.
B = mu0 I / 4 pi d. It is exactly half the infinite-wire value mu0 I / 2 pi d, because one angle is 90 degrees and the other is 0 degrees.
mu0 I / 4R for just the semicircle. If long straight tails are attached, add 2 x mu0 I / 4 pi R for the two semi-infinite tails, giving (mu0 I / 4R)(1 + 2/pi).
When the point lies on the line of the wire (the wire points directly toward or away from the point). Then dl and the position vector are parallel, so their cross product is zero.
From the perpendicular. Sweep from the foot of the perpendicular to each end. That is why the far end of an infinite wire gives sin 90 = 1.
