Torque on a Current Loop in a Uniform Field: 04-BS-9, May 2013, Q4

Question 4: Torque on a Current Loop in a Uniform Field (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Loop area $A$	100 cm² $=1.0\times10^{-2}$ m²
Turns $N$	10
Current $I$	3 A
Loop plane	vertical, east–west (normal is horizontal, north–south)
Current sense	clockwise viewed from due south, looking north
Field $B$	0.2 T, due north and 45° above horizontal

Find. The magnitude and rotational sense of the torque on the loop.

Vertical North–Up plane containing both $\vec m$ (horizontal, due north) and $\vec B$ (45° above north); the torque $\vec m\times\vec B$ is horizontal, along the East–West axis.

Approach. Find the loop's magnetic-moment vector $\vec m=NIA\,\hat n$ from the stated current sense via the right-hand rule, then compute $\vec\tau=\vec m\times\vec B$ using the angle between the (horizontal) moment and the (45°-elevated) field.

Magnetic moment direction and magnitude. Viewed from due south looking north, the current is clockwise; curling the right hand in that sense points the thumb away from the viewer, i.e. due north. So $\vec m$ is horizontal, pointing due north, with $$m=NIA=(10)(3)(0.01)=0.3\ \text{A}\cdot\text{m}^2.$$
Angle between $\vec m$ and $\vec B$. $\vec m$ is horizontal (due north); $\vec B$ shares the same north azimuth but is tilted 45° above horizontal, so the angle between them is exactly $\theta=45^\circ$.
Torque magnitude. $$\tau = mB\sin\theta = (0.3)(0.2)\sin45^\circ$$ $$\tau=\boxed{0.0424\ \text{N}\cdot\text{m}}$$
Torque sense. With North $=\hat x$, East $=\hat y$, Up $=\hat z$ (right-handed, $\hat x\times\hat y=\hat z$), $\vec m=m\hat x$ and $\vec B=B\cos45^\circ\,\hat x+B\sin45^\circ\,\hat z$, so $$\vec\tau=\vec m\times\vec B = -mB\sin45^\circ\,\hat y,$$ i.e. $\vec\tau$ points due West, about a horizontal East–West axis. This is exactly the rotation that tips the loop's north-pointing moment upward toward alignment with $\vec B$ (the sense that reduces $U=-\vec m\cdot\vec B$), consistent with $\hat x\times\hat z=-\hat y$.

Quantity	Result
Magnetic moment $m$	0.3 A·m²
Angle between $\vec m,\vec B$	45°
Torque magnitude	0.0424 N·m
Torque sense/axis	due West, tipping the loop's north face upward toward $\vec B$

Concept & Theory

Topic. Torque on a magnetic dipole (current loop) in a uniform field.

Key relations. Magnetic moment $\vec m=NIA\,\hat n$ (right-hand rule); torque $\vec\tau=\vec m\times\vec B$; potential energy $U=-\vec m\cdot\vec B$.

Why this works. Every side of the loop experiences a force $I\vec L\times\vec B$; summing these forces about the loop's centre reduces exactly to the single-dipole formula $\vec\tau=\vec m\times\vec B$, valid for any planar loop regardless of shape. The torque always acts to rotate $\vec m$ toward $\vec B$, i.e. to minimize the orientational energy.

Common pitfall. Getting the moment's direction backwards from the stated viewing direction — "viewed due north" means the observer looks northward (stands south of the loop), and applying the right-hand rule to a clockwise sense as seen by that specific observer is what fixes $\hat n$; reversing the assumed viewpoint flips the final compass direction of $\vec\tau$.

Source. Sadiku, Elements of Electromagnetics, Ch. 8 (Magnetic Forces, Materials and Devices — Force on a Current Loop).

Question 4 of 8: Torque on a Current Loop in a Uniform Field

Question 4: Torque on a Current Loop in a Uniform Field (20 marks)