An iron rod of volume 10^-3 m^3 and relative permeability 1000

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

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

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

  4. $5 \times 10^{2} \mathrm{Am}^{2}$


Correct Option: , 4

Solution:

$\mathrm{M}=\mu_{\mathrm{r}} \mathrm{NiA}$

Here

$\mu_{\mathrm{r}}=$ Relative permeability

$\mathrm{N}=$ No. of turns

$\mathrm{i}=$ Current

$\mathrm{A}=$ Aea of cross section

$\mathrm{M}=\mu_{\mathrm{r}} \mathrm{NiA}=\mu_{\mathrm{r}} \mathrm{n} \ell \mathrm{i} \mathrm{A}$

$\mathrm{M}=\mu_{\mathrm{r}} \mathrm{niV}=1000(1000) 0.5\left(10^{-3}\right)$

$=500=5 \times 10^{2} \mathrm{Am}^{2}$

Leave a comment