An iron rod of volume 10-3 m3 and relative permeability 1000 is placed as core in a solenoid with 10 turns/cm. If a current of 0.5 A is passed through the solenoid, then the magnetic moment of the rod will be:

Bnet=Binduced+Bexternal⇒μ0μrH = μ0I +μ0H⇒1000μ0ni=μ0ni+I       here , I=intensity of magnetization and H =ni =Magnetizing field⇒I=999ni=999×1000×0.5 ⇒ magnetic moment of the rod =I×vol=999×500×10−3=9992=499.5≈5×102Am2