Torque on a Current Loop in a Uniform Field: 22-Elec-A7 Electromagnetics, December 2016, Q7

Question 7: 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. A multi-turn square coil lies in a vertical plane running east–west, carrying a current whose sense is specified by the view from due north, and the ambient field is vertical.

Given data
Quantity	Symbol	Value
Side of the square loop	s	10 cm = 0.10 m
Number of turns	N	10
Loop current	I	1 A
Plane of the loop	—	east–west, vertical
Current sense viewed from due north	—	clockwise
Flux density	B	0.1 T, vertically upward

Find. The magnitude of the torque on the loop and the direction of the torque vector.

Figure 7.1 — The loop seen from due north: a clockwise current puts the moment into the page (due south), and the couple m × B points due west.

Approach. Compute the magnetic dipole moment from the turns, current and area; fix its direction with the right-hand rule applied to the stated viewing direction; then take the cross product with the field, keeping a clearly declared set of axes so the final compass direction is unambiguous.

Magnitude of the magnetic moment. A flat coil of $N$ turns each enclosing area $A$ behaves as a single dipole of moment $$m=NIA=NIs^{2}=(10)(1)(0.10)^{2}=\boxed{0.100\ \text{A}\cdot\text{m}^{2}}$$
Declare a coordinate system. Take $\hat{x}$ due east, $\hat{y}$ due north and $\hat{z}$ vertically up, a right-handed set in which $\hat{x}\times\hat{y}=\hat{z}$ and $\hat{y}\times\hat{z}=\hat{x}$. The loop lies in the east–west vertical plane, that is the $x$–$z$ plane, so its normal is horizontal and lies along the north–south line. Only its sign remains to be settled.
Fix the direction of the moment with the right-hand rule. An observer standing to the north and looking south sees the current going clockwise. The right-hand rule states that the moment points toward an observer who sees the current circulating counter-clockwise, so it must point away from this observer: $$\mathbf{m}=-\,0.100\,\hat{y}\ \text{A}\cdot\text{m}^{2}\qquad\text{(due south)}$$ This is the step most worth doing slowly; the rest is mechanical.
Form the torque. With $\mathbf{B}=0.1\,\hat{z}$ T, $$\boldsymbol{\tau}=\mathbf{m}\times\mathbf{B}=\left(-0.100\,\hat{y}\right)\times\left(0.1\,\hat{z}\right)=-0.0100\left(\hat{y}\times\hat{z}\right)=-0.0100\,\hat{x}\ \ \text{N}\cdot\text{m}$$
Read off magnitude and direction. The moment is perpendicular to the field (horizontal against vertical), so the magnitude takes its largest possible value for this coil: $$|\boldsymbol{\tau}|=mB\sin90^{\circ}=(0.100)(0.1)(1)=\boxed{0.0100\ \text{N}\cdot\text{m}=10.0\ \text{mN}\cdot\text{m}}$$ and the negative $\hat{x}$ component means the torque vector points $$\boxed{\text{due west, horizontally}}$$
Interpret the sense of rotation. A torque vector pointing west drives a rotation that is right-handed about the west direction, which carries the south-pointing normal upward. Physically the couple is trying to align $\mathbf{m}$ with $\mathbf{B}$: the loop swings from its vertical east–west plane toward the horizontal plane, and it reaches equilibrium when its face is horizontal with the moment pointing up. Stating the outcome this way is a useful check, because a torque that tried to rotate the moment away from the field would indicate a sign error.
Note what the answer does not depend on. The result is independent of the loop’s position in the field (the net force on a current loop in a uniform field is zero, so there is a pure couple and no translation) and independent of the loop’s shape at fixed area — a circular coil of the same 100 cm² area, turns and current would feel the identical torque.

Question 7 — torque on the loop
Quantity	Value
Loop area, A = s²	0.0100 m²
Magnetic moment magnitude, m = NIA	0.100 A·m²
Direction of m	Horizontal, due south
Angle between m and B	90°
Torque magnitude	0.0100 N·m (10.0 mN·m)
Torque direction	Horizontal, due west
Net force on the loop	Zero (uniform field ⇒ pure couple)

Topic. Magnetic dipole moment of a current loop and the couple it experiences in a uniform field.

Key relations. $\mathbf{m}=NIA\hat{n}$ with $\hat{n}$ fixed by the right-hand rule; $\boldsymbol{\tau}=\mathbf{m}\times\mathbf{B}$; $|\boldsymbol{\tau}|=mB\sin\alpha$; potential energy $U=-\mathbf{m}\cdot\mathbf{B}$, minimised when $\mathbf{m}$ is parallel to $\mathbf{B}$.

Why this works. In a uniform field the forces $I\,d\boldsymbol{\ell}\times\mathbf{B}$ on opposite sides of the loop are equal and opposite, so they cancel in sum but not in moment: what survives is a pure couple whose axis is perpendicular to both the loop normal and the field. Collecting the geometry into the single vector $\mathbf{m}$ means the answer never requires integrating around the loop, and it also shows immediately why the torque is maximum here — the loop is oriented with its normal square across the field, the configuration of highest potential energy gradient.

Common pitfall. Getting the sense of $\mathbf{m}$ backwards by applying the right-hand rule from the wrong side; the phrase “when viewed due north” means viewed from the north looking south, and reversing it flips the final answer from west to east. A second error is to report a force rather than a torque, or to expect the loop to be pushed bodily — in a uniform field there is no net force at all.

Source. Sadiku, Elements of Electromagnetics, 7th ed., §8.6 (magnetic dipole, force and torque on a current loop); Hayt & Buck, Engineering Electromagnetics, 9th ed., ch. 8.

Question 7 of 8: Torque on a Current Loop in a Uniform Field

Question 7: Torque on a Current Loop in a Uniform Field (20 marks)