90. If the dipole moment of a short bar magnet is $1.25 \mathrm{~A}-\mathrm{m}^2$, find the magnetic field on its axis at a distance of $0.5 \mathrm{~m}$ from the centre of the magnet.

Question

Aicharya · Accepted Answer

Dipole moment,
$\begin{aligned}
M & =1.25 \mathrm{~A}-\mathrm{m}^2 \\
r & =0.5 \mathrm{~m}
\end{aligned}$
Magnetic field on the axial position of bar magnet is given as
$\begin{aligned}
B & =\frac{\mu_0}{4 \pi} \times \frac{2 M}{r^3}=10^{-7} \times \frac{2 \times 1.25}{(0.5)^3} \\
& =10^{-7} \times 2 \times \frac{1.25}{0.125}=2 \times 10^{-6} \mathrm{NA}^{-1} \mathrm{~m}^{-1}
\end{aligned}$

90. If the dipole moment of a short bar magnet is $1.25 \mathrm{~A}-\mathrm{m}^2$, find the magnetic field…

Solution