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QJEEM23Ph302a: JEE Main 2023 Physics - Class 12 JEEM23Ph302a Solution: A rod with circular cross-section area and length is wound…

A rod with circular cross-section area 2 cm22 \mathrm{~cm}^{2} and length 40 cm40 \mathrm{~cm} is wound uniformly with 400 turns of an insulated wire. If a current of 0.4 A0.4 \mathrm{~A} flows in the wire windings, the total magnetic flux produced inside windings is 4π×10−6 Wb4 \pi \times 10^{-6} \mathrm{~Wb} . The relative permeability of the rod is

(Given : Permeability of vacuum μ0=4π×10−7NA−2\mu_{0}=4 \pi \times 10^{-7} \mathrm{NA}^{-2} )

Multiple Choice Options

  • A 516\frac{5}{16}
  • B 125
  • C 325\frac{32}{5}
  • D 12.5

Step-by-Step Verified Solution

Magnetic field in the Solenoid,

amp;B=μ0μrnIamp; Magnetic flux, ϕ=N(BA)amp;ϕ=N(μ0μrnIA)amp;⇒4π×10−6=400(4π×10−7μr×4000.4×0.4×2×10−4)amp;⇒140=μr×8×10−2amp;⇒μr=100320=516\begin{aligned}& \mathrm{B}=\mu_0 \mu_r \mathrm{nI} \\\\& \text { Magnetic flux, } \phi=\mathrm{N}(\mathrm{BA}) \\\\& \phi=N\left(\mu_0 \mu_r n I A\right) \\\\& \Rightarrow 4 \pi \times 10^{-6}=400\left(4 \pi \times 10^{-7} \mu_r \times \frac{400}{0.4} \times 0.4 \times 2 \times 10^{-4}\right) \\\\& \Rightarrow \frac{1}{40}=\mu_r \times 8 \times 10^{-2} \\\\& \Rightarrow \mu_r=\frac{100}{320}=\frac{5}{16}\end{aligned}

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