A rectangular loop with sides of 20 cm and 10 cm carries a current...

Area of the rectangular loop, $A = 20 \text{ cm} \times 10 \text{ cm} = 0.2 \text{ m} \times 0.1 \text{ m} = 0.02 \text{ m}^2$

Magnetic dipole moment of the loop, $M = I A = 5.0 \times 0.02 = 0.1 \text{ A m}^2$

The magnetic field $\vec{B}$ is parallel to the longer side, meaning it lies in the plane of the loop. The magnetic moment $\vec{M}$ is perpendicular to the plane of the loop. Therefore, the angle between $\vec{M}$ and $\vec{B}$ is $90^{{}^{{}^{\circ}}}$.

Magnitude of torque, $\tau = M B \sin(90^{{}^{{}^{\circ}}}) = 0.1 \times 0.20 \times 1 = 0.02 \text{ N m}$

The direction of torque is given by $\vec{\tau} = \vec{M} \times \vec{B}$. Since $\vec{M}$ is perpendicular to the loop's plane and $\vec{B}$ is parallel to the longer side, their cross product is perpendicular to both. This makes the torque vector lie in the plane of the loop and parallel to the shorter side.