Derive the equation of torque on a magnetic needle in a uniform magnetic field.

(N/A) Consider a magnetic needle (magnetic dipole) of magnetic moment $m$ and length $2l$ placed in a uniform magnetic field $\vec{B}$ at an angle $\theta$ with the direction of the field.
Let $q_m$ be the pole strength of each pole of the needle.
The magnetic dipole moment is given by $m = q_m \times 2l$.
The force acting on the North pole $(N)$ is $\vec{F}_N = q_m \vec{B}$ (along the direction of the field).
The force acting on the South pole $(S)$ is $\vec{F}_S = -q_m \vec{B}$ (opposite to the direction of the field).
These two equal and opposite forces form a couple that exerts a torque $(\tau)$ on the needle, tending to rotate it to align with the magnetic field.
Torque $\tau = (\text{Magnitude of either force}) \times (\text{Perpendicular distance between the forces})$.
From the geometry of the triangle $NDS$, the perpendicular distance $ND = 2l \sin \theta$.
Therefore, $\tau = (q_m B) \times (2l \sin \theta)$.
Since $m = q_m(2l)$, we have $\tau = m B \sin \theta$.
In vector form, this is expressed as $\vec{\tau} = \vec{m} \times \vec{B}$.