JEE Main 2019PhysicsMagnetic Effects of CurrentEasyMCQ

JEE Main 2019 — Magnetic Effects of Current Question with Solution

JEE Main 2019 (09 Apr Shift 1)

Question

A rigid square loop of side ‘a’ and carrying current I2 is lying on a horizontal surface near a long current I1 carrying wire in the same plane as shown in figure. The net force on the loop due to the wire will be:

Choose an option

Show full solutionCorrect option: B
Correct answer
BRepulsive and equal to μ0I1I24π

Step-by-step explanation


Force on segment 1-2 and 3-4 will get cancel out.
Force on segment 4-1,
F1=μ0I1I22πa×a toward Right
Force on segment 2-3:
F2=μ0I1I22π(2a)×a  toward left
Net force on loop:
Fnet=F1-F2 toward Right
=μ0I1I22π-μ0I1I24π
=μ0I1I24π toward Right or repulsive

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About this question

This is a previous-year question from JEE Main 2019, covering the Magnetic Effects of Current chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.