A rigid wire consists of a semicircular portion of radius R and two straight sections. The wire is partially immerged in a perpendicular magnetic field B=B₀ k̂ as shown in figure. The magnetic force on the wire if it has a current i is:

Solution

Physics sir · 3 steps

In a uniform field, any bent wire feels the same force as the straight line joining its ends.

1. Straight sections cancel. Both straight sections have the same length inside the field. In the left one the current runs along +ĵ, and in the right one it runs along -ĵ. Their forces, ± iLB î, are equal and opposite, so they cancel.
2. Replace the curve. In a uniform B⃗, the force on a curved wire is iL⃗×B⃗. Here L⃗ is the straight vector from the start of the wire to its end. The semicircle runs from the top of the left section to the top of the right section, so L⃗ = 2R î.

   F⃗ = i(2R î)×(B₀ k̂) = 2iB₀R(î×k̂) = -2iB₀R ĵ

3. So. Since î×k̂ = -ĵ, the net force is -2iBR ĵ, which is option (b).

The idea to keep. In a uniform field, do not integrate along the curve. Join the start and end points with a straight vector L⃗ and use F⃗ = iL⃗×B⃗. A closed loop feels zero net force.

Answer from the NTA final answer key. Solution by PrepWiser.