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JEE Main 202624 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Evening Shift
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Main 2026 · 24 January · Evening Shift · Physics · Magnetic Effects of Current

Two identical circular loops P and Q each of radius r are lying in...

Two identical circular loops P and Q each of radius r are lying in parallel planes such that they have common axis. The
Two identical circular loops PP and QQ each of radius rr are lying in parallel planes such that they have common axis. The current through PP and QQ are II and 4I4 I respectively in clockwise direction as seen from OO. The net magnetic field at OO is:

Options

  1. A3μoI42r\frac{3 \mu_{\mathrm{o}} I}{4 \sqrt{2} r} towards QQ
  2. BμoI42r\frac{\mu_{\mathrm{o}} I}{4 \sqrt{2} r} towards QQ
  3. C3μ0I42r\frac{3 \mu_{0} I}{4 \sqrt{2} r} towards PP
  4. DμoI42r\frac{\mu_{\mathrm{o}} I}{4 \sqrt{2} r} towards PP

Correct answer

A. 3μoI42r\frac{3 \mu_{\mathrm{o}} I}{4 \sqrt{2} r} towards QQ

Step-by-step solution

Two identical circular loops P and Q, each of radius rr, are separated by distance rr along their common axis. Current in P is II and in Q is 4I4I (both clockwise as seen from O).
Point O is at the midpoint. The magnetic fields from each loop at O:
B1=μ0IR22(R2+R2)3/2B_1 = \frac{\mu_0 I R^2}{2(R^2 + R^2)^{3/2}} (from P, pointing toward Q)
B2=4μ0IR22(R2+R2)3/2B_2 = \frac{4\mu_0 I R^2}{2(R^2 + R^2)^{3/2}} (from Q, pointing toward P)
Bnet=B2−B1=3μ0IR22(2R2)3/2=3μ0I42RB_{net} = B_2 - B_1 = \frac{3\mu_0 I R^2}{2(2R^2)^{3/2}} = \frac{3\mu_0 I}{4\sqrt{2}R}
Direction: towards Q.
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