Physics · Moving Charges And Magnetism · NEET
The wire's field B = mu0*I/(2*pi*r) gets weaker as distance r grows. The loop's near side sits in a stronger field than the far side, so it feels a bigger force. The two forces point opposite ways but are unequal, so they do not fully cancel. A net force is left, always directed toward the stronger (near) side.
Only the two sides parallel to the wire. For a side parallel to the wire, the whole side is at one fixed distance r, so its force = (mu0*I*i)/(2*pi*r) * L is clean and easy. The near side and far side are at different r, so their forces differ, and that difference is the net force.
On a side perpendicular to the wire, every small piece is at a different distance, but the top perpendicular side and the bottom perpendicular side are mirror images carrying current in opposite directions. Their force contributions are equal in size and opposite in direction, so they add to zero. You can ignore them for the net force.
Use the parallel-wire rule on the NEAR side only, because it dominates. Parallel currents attract, antiparallel currents repel. If the near side of the loop carries current in the same direction as the wire, the loop is pulled toward the wire; if opposite, it is pushed away.
In that figure the near side of the square (side L) is at distance L from the wire, and since the square has side L, the far side is one more square-width away, at L + L... but the marked geometry places the near side at L and far side at 3L. Then F_near = mu0*I*i/(2*pi) and F_far = mu0*I*i/(6*pi), giving net = 2*mu0*I*i/(3*pi). Always read the figure distances; do not assume they are r and 2r.
A square loop ABCD carrying a current i is placed near and coplanar with a long straight conductor XY carrying a current I. The net force on the loop will be:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
No, the side length L cancels out. Each parallel side gives force proportional to L/r, and because r itself scales with the geometry set by L, the L cancels. That is why the NEET 2016 answer 2*mu0*I*i/(3*pi) has no L, and any option containing L is wrong.
Same method. Only the two long sides parallel to the wire (length b, say) give a net force. Net force = (mu0*I*i*b)/(2*pi) * (1/r_near - 1/r_far), directed toward the near side. The two shorter perpendicular sides still cancel.
The near side, because it feels the stronger force. If the near side current is parallel (same direction) to the wire current, the loop is pulled in. If antiparallel, it is pushed away. This follows the parallel-wire attraction and repulsion rule.
If the loop is coplanar with the wire, the field lies in the plane of the loop and gives a net force but the standard torque tau = m x B analysis is for a uniform field; here the field is non-uniform. For NEET, these questions ask only for the net force. Torque in a uniform field is a separate topic.
On the far side, the current direction is opposite to the near side (it is the return path of the loop). If the near side is parallel to the wire (attraction), the far side is antiparallel (repulsion). So the far side always opposes the near side, but weaker because it is farther, so the near side wins.
