Q.State the expression for the magnetic field intensity on the equatorial line of a short bar magnet of dipole moment $m$ at distance $d$ from its centre, and explain why this field is exactly half the axial-line field at the same distance and points opposite to $\vec{m}$.

At an equatorial point $P$, distance $\sqrt{d^2+l^2}$ from EACH pole, the two pole fields have equal magnitude $B_N=B_S=(\mu_0/4\pi)q_m/(d^2+l^2)$. Resolving each into components parallel and perpendicular to the axis: the perpendicular (equatorial-direction) components cancel by symmetry, while the parallel (axial-direction) components -- both pointing opposite to $\vec{m}$ -- add, each of size $B_N\cos\theta$ with $\cos\theta=l/\sqrt{d^2+l^2}$:

$B_{\text{eq}} = 2B_N\cos\theta = \frac{\mu_0}{4\pi}\cdot\frac{q_m(2l)}{(d^2+l^2)^{3/2}} = \frac{\mu_0}{4\pi}\cdot\frac{m}{(d^2+l^2)^{3/2}} \approx \frac{\mu_0}{4\pi}\cdot\frac{m}{d^3} \quad (d\gg l)$ …

Resolve each pole's field into axial/equatorial components, cancel the equatorial pair, add the ax …

• Forgetting to resolve into components and instead adding the two pole-field magnitudes directly (they are NOT collinear here, unlike the axial case). …