The magnetic field due to a short bar magnet at a point on its axis, at a large distance \(r\) from its centre, is \(B\). The magnetic field due to the magnet at a point on its equatorial line, at a distance \(r/2\) from the centre, is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(4B\)
Far from a short magnet, the axial field has the size \( \mu_0 2m/4\pi r^{3} \). The equatorial field at the same distance is half of this, \( \mu_0m/4\pi r^{3} \). Both fall as \( 1/r^{3} \).
\( \displaystyle B=\frac{\mu_0}{4\pi}\cdot\frac{2m}{r^{3}} \)
\( \begin{aligned} B_E &= \frac{\mu_0}{4\pi}\cdot\frac{m}{(r/2)^{3}} \\ &= 8\times\frac{\mu_0}{4\pi}\cdot\frac{m}{r^{3}} \\ &= 4\times\frac{\mu_0}{4\pi}\cdot\frac{2m}{r^{3}} \\ &= 4B \end{aligned} \)
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:”
\( \displaystyle \mathbf{B}_E=-\frac{\mu_0\mathbf{m}}{4\pi r^{3}} \)\( \qquad (5.4) \)
“Likewise, the axial field (BA) of a bar magnet for r >> l is:”
\( \displaystyle \mathbf{B}_A=\frac{\mu_0}{4\pi}\frac{2\mathbf{m}}{r^{3}} \)\( \qquad (5.5) \)