Bar Magnet As A Magnetic Dipole

NCERT Class 12, Chapter 5 (Magnetism and Matter) treats a bar magnet as a magnetic dipole of moment $m$ that, in a uniform field $B$, experiences a torque $\tau =mB\sin \theta$, where $\theta$ is the angle between the dipole moment and the field. This torque tends to align the magnet with the field, analogous to an electric dipole in an electric field. Substituting $m=0.8$ A·m${}^2$, $B=0.25$ T, $\theta =30^{\circ }$ so $\sin 30^{\circ }=0.5$: $\tau =(0.8)(0.25)(0.5)=0.1$ N·m. The value $0.2$ N·m wrongly uses $\sin \theta =1$. The value $0.17$ N·m uses $\sin 60^{\circ }$ by confusing the angle. The value $0.05$ N·m halves the moment. Plausibility check: torque is maximum at $90^{\circ }$ and zero at $0^{\circ }$, so a value at $30^{\circ }$ should be well below the maximum of $0.2$ N·m, and $0.1$ N·m being exactly half fits $\sin 30^{\circ }=0.5$ perfectly.