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Physics · Moving Charges and Magnetic Field

JEE Main 2026 — 4 April, Morning Shift — Question 15

An insulated wire is wound to form a flat coil with N = 200 turns. Innermost radius r₁ = 3 cm, outermost r₂ = 6 cm. If current 20 mA flows, the magnetic moment is α×10−2\alpha\times10^{-2} A·m². The value of α\alpha is

  1. Option A:

    4.4

  2. Option B:

    2.64

    Correct
  3. Option C:

    3.25

  4. Option D:

    1.2

Answer: B

Step-by-step solution

M = ∫ I dN π r², with dN = (N/(r₂-r₁)) dr, gives M = (IπN/3)(r₂²+r₂r₁+r₁²) = (0.02×π×200/3)×(36+18+9)×10⁻⁴ = 2.64×10⁻²

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
4 April, Morning Shift
Subject
Physics
Chapter
Moving Charges and Magnetic Field
Topic
Force and Torque on Wires and Loops, Magnetic Dipole Moment
An insulated wire is wound to form a flat coil with N = 200 turns.… | JEE Main 2026 PYQ with Solution · DhiX AI