Question

An iron rod of volume ${10}^{-3}{m}^3$ and relative permeability $1000$ is placed as the core in a solenoid with $10\raisebox{1ex}{$turns$}\!\left/ \!\raisebox{-1ex}{$cm$}\right.$. If a current of $0.5A$ is passed through the solenoid, then the magnetic moment of the rod will be:

• A. $0.5\times {10}^{2}A{m}^2$
• B. $50\times {10}^{2}A{m}^2$
• C. $5\times {10}^{2}A{m}^2$
• D. $500\times {10}^{2}A{m}^2$

Answer 1

The correct option is C

$5\times {10}^{2}A{m}^2$

Step 1: Given data.

The volume of the rod, $V={10}^{-3}{m}^3$

Number of turns per, $n=10\raisebox{1ex}{$turns$}\!\left/ \!\raisebox{-1ex}{$cm$}\right.=1000\raisebox{1ex}{$turns$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$

Current through solenoid, $i=0.5A$

Relative permeability, ${\mu }_r=1000$

Step 2: Find the magnetic moment of the rod.

Formula used:

$M={\mu }_{r}NiA$

Where, $M=$Magnetic moment, ${\mu }_r=$Relative permeability, $N=$ No. of turns, $i=$ Current and $A=$ Area of cross-section

Now,

$M={\mu }_{r}NiA$

$\Rightarrow$$M={\mu }_{r}nliA$ $\left[\because N=nl\right]$

Where $l$ is the length of the conductor

$M={\mu }_{r}niV$ $\left[\because lA=V\right]$

$\Rightarrow$$M=1000\times 1000\times 0.5\times {10}^{-3}$

$\Rightarrow$$M=500A{m}^2$

$\Rightarrow$$M=5\times {10}^{2}A{m}^2$

Hence, option C is correct.