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Physics · Moving Charges and Magnetic Field

JEE Main 2026 — 6 April, Morning Shift — Question 13

A small cube of side 1mm1\mathrm{mm} is placed at the centre of a circular loop of radius 10cm10\mathrm{cm} carrying a current of 2A2\mathrm{A}. The magnetic energy stored inside the cube is α×10−14J\alpha \times 10^{-14}\mathrm{J}. The value of α\alpha is (μ0=4π×10−7Tm/A\mu_0 = 4\pi\times10^{-7}\mathrm{Tm/A}, π=3.14\pi = 3.14)

  1. Option A:

    6.28

    Correct
  2. Option B:

    6.28×10−66.28\times10^{-6}

  3. Option C:

    628

  4. Option D:

    6.28×10−46.28\times10^{-4}

Answer: A

Step-by-step solution

Magnetic field at centre: B=μ0I2R=4π×10−7×22×0.1=4π×10−6B = \frac{\mu_0 I}{2R} = \frac{4\pi\times10^{-7}\times2}{2\times0.1} = 4\pi\times10^{-6} T. Energy density u=B2/(2μ0)u = B^2/(2\mu_0). Volume of cube V=(10−3)3=10−9V = (10^{-3})^3 = 10^{-9} m³. Energy U=uV=(4π×10−6)22×4π×10−7×10−9=16π2×10−128π×10−7×10−9=2π×10−14U = uV = \frac{(4\pi\times10^{-6})^2}{2\times4\pi\times10^{-7}} \times 10^{-9} = \frac{16\pi^2\times10^{-12}}{8\pi\times10^{-7}} \times 10^{-9} = 2\pi\times10^{-14} J = 6.28×10−146.28\times10^{-14} J. So α=6.28\alpha = 6.28.

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
6 April, Morning Shift
Subject
Physics
Chapter
Moving Charges and Magnetic Field
Topic
Force and Torque on Wires and Loops, Magnetic Dipole Moment
A small cube of side 1 mm is placed at the centre of a circular loop… | JEE Main 2026 PYQ with Solution · DhiX AI