An iron rod of volume
Question:

An iron rod of volume $10^{-3} \mathrm{~m}^{3}$ and relative permeability 1000 is placed as core in a solenoid with 10 turns $/ \mathrm{cm}$. If a current of $0.5 \mathrm{~A}$ is passed through the solenoid, then the magnetic moment of the rod will be :

  1. (1) $50 \times 10^{2} \mathrm{Am}^{2}$

  2. (2) $5 \times 10^{2} \mathrm{Am}^{2}$

  3. (3) $500 \times 10^{2} \mathrm{Am}^{2}$

  4. (4) $0.5 \times 10^{2} \mathrm{Am}^{2}$


Correct Option: , 2

Solution:

(2) Given,

Volume of iron rod, $V=10^{-3} \mathrm{~m}^{3}$

Relative permeability, $\mu_{r}=1000$

Number of turns per unit length, $n=10$

Magnetic moment of an iron core solenoid,

$M=\left(\mu_{r}-1\right) \times N i A$

$\Rightarrow M=\left(\mu_{r}-1\right) \times N i \frac{V}{l} \Rightarrow M=\left(\mu_{r}-1\right) \times \frac{N}{l} i V$

$\Rightarrow M=999 \times \frac{10}{10^{-2}} \times 0.5 \times 10^{-3}=499.5 \approx 500$

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