Question

A closed loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments PS, SR and RQ are $F_1, F_2, $ and $ F_3$ respectively and are in the plane of the paper and along the directions shown, the force on the segment QP is

• A. $\sqrt{(F_3-F_1)^2-F_2^2}$
• B. $F_3-F_1+F_2$
• C. $F_3-F_1-F_2$
• D. $\sqrt{(F_3-F_1)^2+F_2^2}$

Answer 1

The Correct Option is D

To solve this problem, we need to understand how the magnetic force acts on each segment of a closed loop carrying current in a uniform magnetic field. Let's break it down step-by-step:

1. The magnetic force on a current-carrying conductor in a magnetic field is given by the expression F = I (L \times B), where I is the current, L is the length vector of the conductor, and B is the magnetic field.
2. Each segment (PS, SR, RQ, and QP) of the loop experiences a magnetic force due to the magnetic field, which we denote as F_1, F_2, F_3, and the force on QP.
3. According to the principle of superposition of forces and equilibrium condition (since the loop is closed), the vector sum of forces on all segments must equal zero. Mathematically, this is expressed as: F_{PS} + F_{SR} + F_{RQ} + F_{QP} = 0.
4. Given that F_1, F_2, and F_3 act along specific directions in the plane of the paper, the net force in the plane can be evaluated as:
   • The segment PQ force will compensate the vector sum of the forces F_1, F_2, and F_3.
   • Let’s take the plane forces into account: The force F_{QP} must be opposite to the resultant vector of the forces (F_3 - F_1) \text{ in the horizontal direction and } F_2 \text{ in the vertical direction}, which gives us the net resultant of those vectors.
5. Using vector addition and invoking the Pythagorean theorem, the magnitude of the resulting force is: F_{QP} = \sqrt{(F_3 - F_1)^2 + F_2^2}.

Thus, the force on the segment QP is \sqrt{(F_3-F_1)^2 + F_2^2}, which matches the given correct option.