Question

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

• A. $1\times {10}^{-4}N{A}^{-1}{m}^{-1}$
• B. $2\times {10}^{-6}N{A}^{-1}{m}^{-1}$
• C. $4\times {10}^{-2}N{A}^{-1}{m}^{-1}$
• D. $6.64\times {10}^{-8}N{A}^{-1}{m}^{-1}$

Answer 1

The correct option is B $2\times {10}^{-6}N{A}^{-1}{m}^{-1}$

The magnetic field at a point on the axis of a bar magnet is
$B=\frac{{\mu }_0}{4\pi }\frac{2M}{r^3}$
$={10}^{-7}\times \frac{2\times 1.25}{(0.5{)}^3}=2\times {10}^{-6}N{A}^{-1}{m}^{-1}$.