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

Given,

Volume of iron rod is, $V = {10}^{- 3}{m}^{3}$

Relative permeability, ${\mu}_{r} = 1000$

No of turns per unit length of solenoid is, $n = \dfrac{N}{l} = 10turns{cm}^{- 1} = 1000turns{m}^{- 1}$

Current through the solenoid is, $i = 0 . 5A$

Magnetic moment of an iron core solenoid is given by,

$M = ( {\mu}_{r} - 1 ) NiA$

$⇒ M = \left({\mu}_{r} - 1\right)Ni\dfrac{V}{l} = \left({\mu}_{r} - 1\right)\dfrac{N}{l}iV$

$⇒ M = \left(1000 - 1\right) \times 1000 \times 0 . 5 \times {10}^{- 3}$

$⇒ M = 499 . 5 \approx 500A{m}^{2}$