An iron rod of volume `10^(-4)m^(3)` and relative permeability 1000 is placed inside a long solenoid wound with `5 "turns"//"cm"`. If a current of `0.5 A` is passed through the solenoid, then the magnetic moment of the rod is

Options

A. `10 A m^(2)`
B. `15 A m^(2)`
C. `20 A m^(2)`
D. `25 A m^(2)`

Answer

D

Explanation

B in magnetic material
`B = mu_(0) H + mu_(0)I`
`rArr I= (B- mu_(0)H)/(mu_(0))= (mu H- mu_(0)H)/(mu_(0))= [(mu)/(mu_(0))-1]H`
`rArr I= (mu_(r)-1)H`
Now, H= ni (Solenoid)
`rArr I= (mu_(r)-1)ni = (1000-1) xx 500 xx 0.5`
`rArr I= 2.5 xx 10^(5) A//m`
Magnetic moment
`M= IV= 2.5 xx 10^(5) xx 10^(-4)= 25 A m^(2)`