Q. A domain in a ferromagnetic substance is in the form of a cube of side length $1\ \mu m$. If it contains $8 \times 10^{10}$ atoms and each atomic dipole has a dipole moment of $9 \times 10^{- 24}\ A\ m^{2}$, then magnetization of the domain is
A) $7.2 \times 10^{5}\ A\ m^{- 1}$
B) $7.2 \times 10^{3}\ A\ m^{- 1}$
C) $7.2 \times 10^{9}\ A\ m^{- 1}$
D) $7.2 \times 10^{12}\ A\ m^{- 1}$
Explanation
(a) The volume of the cubic domain is
$V\left( 10^{- 6}m \right)^{3} = 10^{- 18}\ m^{3}$
Net dipole moment $m_{net} = 8 \times 10^{10} \times 9 \times 10^{- 24}A\ m^{2}$
$= 72 \times 10^{- 14}A\ m^{2}$
Magnetization, $M = \frac{m_{net}}{Domain\ volume}$
$= \frac{72 \times 10^{- 14}A\ m^{2}}{10^{- 18}\ m^{3}} = 72 \times 10^{4}\ A\ m^{- 1} = 7.2 \times 10^{5}\ A\ m^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.