A short bar magnet has a magnetic moment of \(0.5\ \mathrm{A\,m^{2}}\). The magnetic field due to it at a point on its equatorial line is \(5\times10^{-5}\ \mathrm{T}\). The distance of this point from the centre of the magnet is (take \(\mu_0/4\pi=10^{-7}\ \mathrm{T\,m\,A^{-1}}\))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 10 cm
At a large distance \( r \) on the equatorial line of a short magnet, the field has the size \( B_E=\mu_0m/4\pi r^{3} \). We solve this for \( r \).
\( \begin{aligned} r^{3} &= \frac{\mu_0}{4\pi}\cdot\frac{m}{B_E} \\ &= \frac{10^{-7}\times0.5}{5\times10^{-5}} \\ &= 10^{-3}\ \mathrm{m^{3}} \end{aligned} \)
\( \begin{aligned} r &= 0.1\ \mathrm{m} \\ &= 10\ \mathrm{cm} \end{aligned} \)
NCERT (2026-27):
The electrostatic analog – “In particular, we can write down the equatorial field (BE) of a bar magnet at a distance r, for r >> l, where l is the size of the magnet:”
\( \displaystyle \mathbf{B}_E=-\frac{\mu_0\mathbf{m}}{4\pi r^{3}} \)\( \qquad (5.4) \)