A small iron cube of side 2 cm is uniformly magnetised. Its magnetisation is \(1.0\times10^{6}\ \mathrm{A\,m^{-1}}\). The net magnetic moment of the cube is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(8\ \mathrm{A\,m^{2}}\)
Magnetisation is the net magnetic moment per unit volume, \( M=m_{net}/V \). So a uniformly magnetised sample has a net moment \( m_{net}=MV \).
Side of the cube, \( a=2\ \mathrm{cm}=0.02\ \mathrm{m} \)
Magnetisation, \( M=1.0\times10^{6}\ \mathrm{A\,m^{-1}} \)
\( \begin{aligned} V &= a^{3} \\ &= (0.02)^{3} \\ &= 8\times10^{-6}\ \mathrm{m^{3}} \end{aligned} \)
\( \begin{aligned} m_{net} &= MV \\ &= 1.0\times10^{6}\times8\times10^{-6} \\ &= 8\ \mathrm{A\,m^{2}} \end{aligned} \)
NCERT (2026-27):
Magnetisation and Magnetic Intensity – “We have seen that a circulating electron in an atom has a magnetic moment. In a bulk material, these moments add up vectorially and they can give a net magnetic moment which is non-zero. We define magnetisation M of a sample to be equal to its net magnetic moment per unit volume:”
\( \displaystyle \mathbf{M}=\frac{\mathbf{m}_{net}}{V} \)\( \qquad (5.7) \)