Magnetic Effects & Magnetism

JEE Main level, single correct, medium

6 of 276

Question 6 of 276

A rod with circular cross-section area 2cm22{cm}^{2} and length 40cm40cm is wound uniformly with 400400 turns of an insulated wire. If a current of 0.4A0.4A flows in the wire windings, the total magnetic flux produced inside windings is 4π×10−6Wb4\pi \times {10}^{-6}Wb. The relative permeability of the rod is
(Given : Permeability of vacuum μ0=4π×10−7NA−2{\mu }_{0}=4\pi \times {10}^{-7}N{A}^{-2})

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Answer: (C) 125125

Magnetic flux is given by ϕ=NBA\phi =NBA, here, NN is number of turns, BB is magnetic field and AA is area.
Given here, ϕ=4π×10−6Wb,A=2cm2,N=400,I=0.4A\phi =4\pi \times {10}^{-6}Wb,A=2{cm}^{2},N=400,I=0.4A.
Putting the values,
ϕ=400×(μrμ0nI)×A\phi =400\times \left({\mu }_{r}{\mu }_{0}nI\right)\times A
⇒4π×10−6=400[μr×4π×10−7×4000.4×0.4]×2×10−4\Rightarrow 4\pi \times {10}^{-6}=400\left[{\mu }_{r}\times 4\pi \times {10}^{-7}\times \dfrac{400}{0.4}\times 0.4\right]\times 2\times {10}^{-4}
⇒μr=125\Rightarrow {\mu }_{r}=125

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