JEE Main 2022 · 28 July · evening · Single correct · +4 / −1

A triangular shaped wire carrying 10 A current is placed in a uniform magnetic field of 0.5 T, as shown in figure. The magnetic force on segment CD is
(Given BC=CD=BD=5 cm.)

JEE Main PYQ figure
  1. (a) 0.126 N
  2. (b) 0.312 N
  3. (c) 0.216 N
  4. (d) 0.245 N
Answer and solution

Answer (c) 0.216 N

Solution

Physics sir · 3 steps

Only the part of CD perpendicular to the field feels a force, so find CD's angle with B first.

  1. Angle with the field. BCD is equilateral, so every interior angle is 60°. BD lies along the horizontal field, so CD makes θ = 60° with B⃗.

    F = I L B sin θ

  2. Put in numbers. With I = 10 A, L = 0.05 m and B = 0.5 T: F = 10 × 0.05 × 0.5 × sin 60° = 0.25 × 0.866 = 0.2165 N.
  3. So. The force on CD is about 0.216 N, directed perpendicular to the plane of the wire. That is option (c).

The idea to keep. For a straight wire in a uniform field, F = ILB sin θ. Here θ is the angle between the wire and B⃗, so read it carefully from the geometry before you substitute.

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

Study Moving Charges & Magnetism

Asked about 200 times in JEE Main since 2020. This question is from 2022.