The ratio of the magnetic fields of a short bar magnet at the same (large) distance from its centre on the axial and equatorial lines is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 2 : 1
Take a short bar magnet of magnetic moment \( m \). At a large distance \( r \) from it, the field on the axial line is twice the field on the equatorial line in magnitude. The same is true for an electric dipole.
\( \displaystyle B_A=\frac{\mu_0}{4\pi}\,\frac{2m}{r^{3}} \)
\( \displaystyle B_E=\frac{\mu_0}{4\pi}\,\frac{m}{r^{3}} \qquad \text{(magnitude)} \)
For the same \( r \), the ratio of the magnitudes is:
\( \begin{aligned} \frac{B_A}{B_E} &= \frac{2m/r^{3}}{m/r^{3}} \\ &= 2 \end{aligned} \)
Therefore, \( B_A:B_E=2:1 \).
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:
\( \displaystyle B_E=-\frac{\mu_0m}{4\pi r^{3}} \)\( \qquad (5.4) \)
Likewise, the axial field (BA) of a bar magnet for r >> l is:
\( \displaystyle B_A=\frac{\mu_0}{4\pi}\,\frac{2m}{r^{3}} \)\( \qquad (5.5) \)”