$A$ uniform magnetic field of $3000 \; G$ is established along the positive $z-$direction. $A$ rectangular loop of sides $10 \; cm$ and $5 \; cm$ carries a current of $12 \; A$. What is the torque on the loop in the different cases shown in Figure? What is the force on each case? Which case corresponds to stable equilibrium?

(E) Magnetic field strength,$B = 3000 \; G = 3000 \times 10^{-4} \; T = 0.3 \; T$.
Length of the rectangular loop,$l = 10 \; cm = 0.1 \; m$.
Width of the rectangular loop,$b = 5 \; cm = 0.05 \; m$.
Area of the loop,$A = l \times b = 0.1 \times 0.05 = 50 \times 10^{-4} \; m^2$.
Current in the loop,$I = 12 \; A$.
The magnetic moment is $\vec{m} = I \vec{A}$. The torque is $\vec{\tau} = \vec{m} \times \vec{B}$.
In a uniform magnetic field,the net force on a current loop is always zero.
$(a)$ $\vec{A}$ is along the $x-$axis. $\vec{m} = I A \hat{i}$. $\vec{\tau} = (I A \hat{i}) \times (B \hat{k}) = -I A B \hat{j} = -1.8 \times 10^{-2} \hat{j} \; Nm$.
$(b)$ Similar to $(a)$,$\vec{\tau} = -1.8 \times 10^{-2} \hat{j} \; Nm$.
$(c)$ $\vec{A}$ is along the $y-$axis. $\vec{m} = I A \hat{j}$. $\vec{\tau} = (I A \hat{j}) \times (B \hat{k}) = I A B \hat{i} = 1.8 \times 10^{-2} \hat{i} \; Nm$.
$(d)$ The angle between $\vec{m}$ and $\vec{B}$ is $60^{\circ}$. $|\vec{\tau}| = m B \sin(60^{\circ}) = 1.8 \times 10^{-2} \times \frac{\sqrt{3}}{2} \approx 1.56 \times 10^{-2} \; Nm$.
$(e)$ $\vec{m}$ is along the $z-$axis. $\vec{m} = I A \hat{k}$. $\vec{\tau} = (I A \hat{k}) \times (B \hat{k}) = 0$. This is stable equilibrium as $\vec{m} \parallel \vec{B}$.
$(f)$ $\vec{m}$ is along the $-z-$axis. $\vec{\tau} = 0$. This is unstable equilibrium as $\vec{m}$ is anti-parallel to $\vec{B}$.