The magnetic field at a point on the axis of a short bar magnet, 10 cm from its centre, is \(2.4\times10^{-4}\ \mathrm{T}\). The magnetic moment of the magnet is \((\mu_0/4\pi=10^{-7}\ \mathrm{T\,m\,A^{-1}})\)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(1.2\ \mathrm{A\,m^{2}}\)
For a short bar magnet, the field at a distance \( r \) on its axis is \( B_A=\frac{\mu_0}{4\pi}\frac{2m}{r^{3}} \). We rearrange it to find \( m \).
Given:
\( B_A=2.4\times10^{-4}\ \mathrm{T} \)
\( r=10\ \mathrm{cm}=0.1\ \mathrm{m} \)
\( \begin{aligned} m &= \frac{B_Ar^{3}}{2\times(\mu_0/4\pi)} \\ &= \frac{2.4\times10^{-4}\times(0.1)^{3}}{2\times10^{-7}} \\ &= \frac{2.4\times10^{-7}}{2\times10^{-7}} \\ &= 1.2\ \mathrm{A\,m^{2}} \end{aligned} \)
NCERT (2026-27):
The electrostatic analog – “… Likewise, the axial field (BA) of a bar magnet for r >> l is:”
\( \displaystyle B_A=\frac{\mu_0}{4\pi}\frac{2m}{r^{3}} \)\( \qquad (5.5) \)