JEE MAIN - Physics (2023 - 31st January Morning Shift - No. 7)
A rod with circular cross-section area $$2 \mathrm{~cm}^{2}$$ and length $$40 \mathrm{~cm}$$ is wound uniformly with 400 turns of an insulated wire. If a current of $$0.4 \mathrm{~A}$$ flows in the wire windings, the total magnetic flux produced inside windings is $$4 \pi \times 10^{-6} \mathrm{~Wb}$$. The relative permeability of the rod is
(Given : Permeability of vacuum $$\mu_{0}=4 \pi \times 10^{-7} \mathrm{NA}^{-2}$$)
$$\frac{5}{16}$$
125
$$\frac{32}{5}$$
12.5
Explanation
Magnetic field in the Solenoid,
$$ \begin{aligned} & \mathrm{B}=\mu_0 \mu_r \mathrm{nI} \\\\ & \text { Magnetic flux, } \phi=\mathrm{N}(\mathrm{BA}) \\\\ & \phi=N\left(\mu_0 \mu_r n I A\right) \\\\ & \Rightarrow 4 \pi \times 10^{-6}=400\left(4 \pi \times 10^{-7} \mu_r \times \frac{400}{0.4} \times 0.4 \times 2 \times 10^{-4}\right) \\\\ & \Rightarrow \frac{1}{40}=\mu_r \times 8 \times 10^{-2} \\\\ & \Rightarrow \mu_r=\frac{100}{320}=\frac{5}{16} \end{aligned} $$
$$ \begin{aligned} & \mathrm{B}=\mu_0 \mu_r \mathrm{nI} \\\\ & \text { Magnetic flux, } \phi=\mathrm{N}(\mathrm{BA}) \\\\ & \phi=N\left(\mu_0 \mu_r n I A\right) \\\\ & \Rightarrow 4 \pi \times 10^{-6}=400\left(4 \pi \times 10^{-7} \mu_r \times \frac{400}{0.4} \times 0.4 \times 2 \times 10^{-4}\right) \\\\ & \Rightarrow \frac{1}{40}=\mu_r \times 8 \times 10^{-2} \\\\ & \Rightarrow \mu_r=\frac{100}{320}=\frac{5}{16} \end{aligned} $$
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