Magnetic flux is given by ϕ=NBA, here, N is number of turns, B is magnetic field and A is area.
Given here, ϕ=4π×10−6Wb,A=2cm2,N=400,I=0.4A.
Putting the values,
ϕ=400×(μrμ0nI)×A
⇒4π×10−6=400[μr×4π×10−7×0.4400×0.4]×2×10−4
⇒μr=125
JEE Main 2023 — Physics Electromagnetism
A rod with circular cross-section area 2cm2 and length 40cm is wound uniformly with 400 turns of an insulated wire. If a current of 0.4A flows in the wire windings, the total magnetic flux produced inside windings is 4π×10−6Wb. The relative permeability of the rod is
(Given : Permeability of vacuum μ0=4π×10−7NA−2)
Held on 31 Jan 2023 · Verified 6 Jul 2026.
12.5
532
125
165
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