Magnetic Dipole Moment And Torque On A Loop

As explained in NCERT Class 12, Chapter 4 (Moving Charges and Magnetism), a current loop of $N$ turns, area $A$, carrying current $I$ in a field $B$ has a magnetic dipole moment $m=NIA$ and experiences a torque $\tau =mB\sin \theta =NIAB\sin \theta$, where $\theta$ is the angle between the field and the coil's normal. When the coil's plane is parallel to the field, its normal is perpendicular to the field, so $\theta =90^{\circ }$ and $\sin \theta =1$, giving maximum torque. Substituting $N=50$, $I=1.5$ A, $A=2\times 10^{-2}$ m${}^2$, $B=0.4$ T: $\tau =(50)(1.5)(2\times 10^{-2})(0.4)=0.6$ N·m. The value $0.3$ N·m arises from halving the turns. The value $1.2$ N·m doubles the area. The 'zero' choice wrongly assumes the plane parallel to the field gives no torque, confusing plane orientation with normal orientation. Plausibility check: units give $\text{A}\cdot {\text{m}}^2\cdot \text{T}=\text{Ncdotpm}$, confirming a torque, and a sub-newton-metre value is typical for a small multi-turn coil. Note also that if the coil's plane were instead perpendicular to the field, the normal would be aligned with the field, $\theta$ would be zero, and the torque would vanish, which is the exact opposite of the maximum-torque situation described here, reinforcing that this configuration gives the largest possible twist.