Physics · Moving Charges And Magnetism · NEET
The current splits into two parallel paths, so it divides inversely with resistance (i.e. inversely with arc length). Field of an arc is B = (mu0 * i_arc * theta)/(4*pi*R). The shorter arc carries more current but subtends a smaller angle; the longer arc carries less current but a bigger angle. Since i1*theta1 = i2*theta2, both fields have EQUAL magnitude. But the two arcs run in opposite rotational senses about the centre (one clockwise, one anticlockwise), so one field points into the page and the other out. Equal and opposite means net field = 0.
The SHORTER arc carries MORE current. The two arcs are like two resistors in parallel. Resistance is proportional to length. The short arc has less length, so less resistance, so more current flows through it. Rule: current divides inversely with arc length, so i1/i2 = theta2/theta1 (the shorter angle gets the larger current).
No. The cancellation is general for ANY split of a single circle into two arcs, as long as the current enters at one point and leaves at another point on the same circle. It does not matter if the split is 90/270, 120/240, or any pair. As long as i1*theta1 = i2*theta2 holds (which it always does for parallel arcs), the fields are equal and opposite, so the net field is zero.
For a single arc of radius R carrying current i and subtending angle theta (in radians) at the centre, B = (mu0 * i * theta)/(4*pi*R). If you plug theta = 2*pi (full circle) you recover the full-loop result B = (mu0 * i)/(2*R). This arc formula is the tool you use for split-loop problems.
A straight conductor carrying current i splits into two parts as shown in the figure. The radius of the circular loop is R. The total magnetic field at the centre P of the loop is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The clean zero result is for two arcs of the same circle sharing the two junction points. The same idea (current inversely with resistance, then add fields as vectors) extends to more paths, but for NEET the two-arc case giving zero is the standard result.
Then they are not the same circle and the fields will not generally cancel. This concept is specifically about ONE circular loop of radius R split into two arcs. If radii differ, compute each arc field B = (mu0 * i_arc * theta)/(4*pi*R) separately and add with signs.
In the standard figure the straight parts point along the diameter (radially, or toward the centre line), so a straight segment whose line passes through the centre contributes zero field there (because the element and the position vector are parallel, dl x r = 0). So only the arcs matter, and they cancel.
A full loop has one current i going all the way around, giving B = mu0*i/(2R). A split loop has the current divided between two arcs going opposite ways around, so their contributions cancel and the centre field is zero instead.
