Q.Derive formula for the magnetic field intensity at a point lies on the broad side-on position due to a magnetic dipole.

Adding the field contributions of the N and S poles at an equatorial point, only the components antiparallel to the dipole moment survive, giving $B_{eq} = \mu_0 m /[4\pi (r^2+l^2)^{3/2}]$.

Consider a short bar magnet (magnetic dipole) with pole strength $q_m$ at each pole, magnetic length $2l$, centred at O, so its dipole moment is $m = q_m (2l)$.

Let P be a point on the equatorial (broad side-on) line, i.e. on the perpendicular bisector of the magnet, at a distance $r$ from the centre O.

Distance of P from each pole:

$d = \sqrt{r^2 + l^2}$

Magnitude of field due to each pole (Coulomb's law for magnetic poles):

$B_N = B_S = \frac{\mu_0}{4\pi}\,\frac{q_m}{r^2+l^2}$

$B_N$ points away from the N pole along PN, and $B_S$ points toward the S pole along PS. By the symmetry of the geometry, the components of $B_N$ and $B_S$ perpendicular to the magnet's axis are equal and opposite, and cancel. The components parallel to the axis but opposite in direction to $m$ add up.

Each field makes an angle $\theta$ with the axis where $\cos\theta = l/\sqrt{r^2+l^2}$. The resultant is: …