Magnetic Systems

AP Physics 2· AP Physics 2 CED — Magnetism and Electromagnetic Induction· 14 min read

1. Magnetic Dipole Moment★★☆☆☆⏱ 4 min

All magnetic behavior arises from magnetic dipoles, the fundamental building blocks of magnetism with no isolated magnetic monopoles. Magnetic dipole moment is a vector quantity that describes the magnetic strength and orientation of any magnetic system, from permanent bar magnets to current-carrying coils. For a flat, N-turn current-carrying coil with current $I$ and enclosed area $A$ , the magnitude is given by:

\mu = N I A

The direction of $μ$ follows the right-hand rule for current loops: curl the fingers of your right hand along the current direction, and your thumb points in the direction of $μ$ . Dipole moment is an intrinsic property of the system—it does not depend on any external magnetic field.

📐 Worked Example

A rectangular 15-turn coil of wire with sides 1.0 cm and 4.0 cm carries a current of 2.0 A. The coil's plane makes a 45° angle with a uniform external magnetic field. What is the magnitude of the coil's magnetic dipole moment?

1
Convert all units to SI: sides are $0.010 m$ and $0.040 m$ , so area is:
2
$A = 0.010 \times 0.040 = 4.0 \times 10^{-4} \text{ m}^2$
3
Dipole moment is intrinsic, so the coil orientation and external field do not affect the calculation. Substitute values into the formula:
4
$\mu = N I A = 15 \times 2.0 \times 4.0 \times 10^{-4} = 1.2 \times 10^{-2} \text{ A·m}^2$
5
Final magnitude: $1.2 \times 1 0^{- 2} A \cdotp m^{2}$ , direction given by the right-hand rule for current.

2. Torque on a Magnetic Dipole in a Uniform Field★★★☆☆⏱ 4 min

When a magnetic dipole is placed in a uniform external magnetic field, the net force on the dipole is always zero (forces on opposite sides of the dipole cancel out). However, there is a net torque that acts to align $μ$ with the external field $B$ . The vector formula for torque is:

\vec{\tau} = \vec{\mu} \times \vec{B}

The magnitude of torque is $τ = μ B sin θ$ , where $θ$ is the angle between $μ$ and $B$ . Torque is maximum when $θ = 9 0^{\circ}$ and zero when the dipole is aligned or anti-aligned with the field. This torque is the operating principle of electric motors.

📐 Worked Example

The 15-turn coil from the previous example ( $μ = 1.2 \times 1 0^{- 2} A \cdotp m^{2}$ ) is placed in a uniform 0.50 T external magnetic field. The plane of the coil makes a 45° angle with the direction of $B$ . What is the magnitude of the torque on the coil?

1
Correct the angle: $μ$ is perpendicular to the coil plane, so the angle between $μ$ and $B$ is:
2
$\theta = 90^\circ - 45^\circ = 45^\circ$
3
Write the torque magnitude formula and substitute values:
4
$\tau = \mu B \sin\theta = (1.2 \times 10^{-2})(0.50)(\sin 45^\circ) \approx 4.2 \times 10^{-3} \text{ N·m}$
5
The torque acts to rotate the coil to align $μ$ with the external field.

3. Potential Energy and Force on Magnetic Dipoles★★★☆☆⏱ 4 min

Since torque does work to rotate a dipole into alignment with an external field, we can define a potential energy for the dipole-field system, with zero potential energy defined at $θ = 9 0^{\circ}$ . The formula is:

U = -\vec{\mu} \cdot \vec{B} = -\mu B \cos\theta

Potential energy is minimized ( $U = - μ B$ ) when $θ = 0^{\circ}$ (aligned, stable equilibrium) and maximized ( $U = + μ B$ ) when $θ = 18 0^{\circ}$ (anti-aligned, unstable equilibrium). In uniform fields net force is zero, but in non-uniform fields aligned dipoles are pulled toward regions of stronger magnetic field.

📐 Worked Example

A small bar magnet with dipole moment $μ = 0.30 A \cdotp m^{2}$ is placed in a 0.20 T uniform external magnetic field. What is the change in potential energy when the magnet is rotated from aligned with the field to 90° to the field?

1
Write the potential energy formula for both orientations: $U = - μ B cos θ$ .
2
Initial aligned orientation: $θ_{i} = 0^{\circ}$ , so:
3
$U_i = -(0.30)(0.20)\cos 0^\circ = -0.060 \text{ J}$
4
Final 90° orientation: $θ_{f} = 9 0^{\circ}$ , so:
5
$U_f = -(0.30)(0.20)\cos 90^\circ = 0 \text{ J}$
6
Calculate change in potential energy:
7
$\Delta U = U_f - U_i = 0 - (-0.060) = +0.060 \text{ J}$
8
Work must be done on the magnet to rotate it, so potential energy increases, which matches our result.

4. AP-Style Concept Check★★★★☆⏱ 6 min

✓ Quick check

Test your understanding of core concepts with these AP-style questions:

A circular current-carrying loop has a magnetic dipole moment that points along the +y axis. The loop is placed in an external magnetic field that points along the +z axis. What is the direction of the torque on the loop?
- +x
- -x
- +y
- +z
Reveal answer
+x —
Use the cross product definition $τ = μ \times B$ . The cross product of +y and +z is +x, which is the correct direction for torque that rotates $μ$ into alignment with $B$ .

📐 Worked Example

A small bar magnet with dipole moment $μ = 0.15 A \cdotp m^{2}$ is placed in a uniform 0.60 T external magnetic field. (a) What is the potential energy when aligned, and what is the orientation? (b) How much work must an external force do to rotate from aligned to anti-aligned? (c) Why is there no net force?

1
(a) Aligned orientation means $θ = 0^{\circ}$ . Substitute into the potential energy formula:
2
$U = -\mu B \cos 0^\circ = -(0.15)(0.60)(1) = -0.090 \text{ J}$
3
The dipole moment $μ$ is parallel to the external field $B$ in this state.
4
(b) Work done by external force equals the change in potential energy. For anti-aligned, $θ = 18 0^{\circ}$ :
5
$U_{anti} = -\mu B \cos 180^\circ = +0.090 \text{ J}, W = \Delta U = 0.090 - (-0.090) = 0.18 \text{ J}$
6
(c) The field is uniform, so the force on the north pole is equal and opposite to the force on the south pole, giving zero net force. The separated forces produce a net torque.

📐 Worked Example

An electric motor has a 50-turn rectangular coil 2.0 cm × 3.0 cm carrying 10 A, placed in a uniform 0.80 T magnetic field. What is the torque when the coil plane is parallel to the field? What is the torque's role in the motor?

1
First calculate dipole moment, convert units to SI:
2
$A = 0.020 \times 0.030 = 6.0 \times 10^{-4} \text{ m}^2, \mu = N I A = 50 \times 10 \times 6.0 \times 10^{-4} = 0.30 \text{ A·m}^2$
3
When coil plane is parallel to $B$ , $μ$ is perpendicular to $B$ , so $θ = 9 0^{\circ}$ :
4
$\tau = \mu B \sin 90^\circ = (0.30)(0.80)(1) = 0.24 \text{ N·m}$
5
This torque is the driving force that rotates the motor shaft, converting electrical energy from the current into mechanical rotational energy.

5. Common Pitfalls

Wrong move:

Using the angle between the coil plane and $B$ directly as $θ$ in torque or potential energy formulas.

Why:

AP questions intentionally give the plane angle to test student understanding of what $θ$ describes, and many students misinterpret the definition.

Correct move:

Always confirm $θ$ is the angle between $μ$ and $B$ ; for a flat coil, $μ$ is perpendicular to the coil plane, so subtract the given plane angle from $9 0^{\circ}$ to get $θ$ .

Wrong move:

Calculating a non-zero net force on a magnetic dipole in a uniform external magnetic field.

Why:

Students confuse torque and force, and incorrectly generalize the non-uniform field force rule to all cases.

Correct move:

Always check if the field is uniform before calculating force—if uniform, net force on any dipole is always zero, only torque can be non-zero.

Wrong move:

Forgetting the negative sign in $U = - μ \cdot B$ and concluding aligned dipoles have higher potential energy than anti-aligned dipoles.

Why:

The negative sign is easy to drop when memorizing, and students mix up magnetic potential energy with other forms of potential energy.

Correct move:

Always check your result against the rule that aligned dipoles are in stable equilibrium, so they must have lower potential energy than anti-aligned dipoles.

Wrong move:

Claiming the magnetic dipole moment of a coil depends on the strength of the external magnetic field it is placed in.

Why:

Students mix up intrinsic properties of the dipole with interaction properties between the dipole and the field.

Correct move:

Dipole moment depends only on the coil's current, number of turns, and area—ignore any extra field or angle information when calculating $μ$ .

Wrong move:

Using the right-hand rule for magnetic force on a moving charge to find the direction of $μ$ for a current loop.

Why:

Students confuse the multiple right-hand rules used in magnetism.

Correct move:

For $μ$ direction, always use the current curl rule: curl your right fingers along the current direction, thumb points to $μ$ .

6. Quick Reference Cheatsheet

Category	Formula	Notes
Magnetic dipole moment (flat coil)	$μ = N I A$	Intrinsic property, independent of external field. Direction: right-hand rule, curl fingers along current, thumb = $μ$ .
Torque on magnetic dipole	$τ = μ \times B$ , $τ = μ B sin θ$	$θ$ = angle between $μ$ and $B$ . Net force = 0 in uniform fields.
Potential energy of dipole	$U = - μ \cdot B = - μ B cos θ$	Zero potential at $θ = 9 0^{\circ}$ . Minimum $U = - μ B$ (stable aligned), maximum $U = + μ B$ (unstable anti-aligned).
Force on dipole in non-uniform field	$F \propto μ \frac{d B}{d x}$	Aligned dipoles are pulled toward regions of stronger magnetic field.
Dipole direction (permanent magnet)	Points from S pole to N pole	Matches the direction of the magnetic field produced by the dipole outside the magnet.
Work to rotate dipole	$W = Δ U$	Work done by external force equals the change in potential energy of the dipole-field system.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · AP Physics 2
MCQ: Dipole moment and torque
2022 · AP Physics 2
FRQ: Potential energy change

Going deeper

official syllabusAP Physics 2 Course and Exam DescriptionUnit 5 learning objectives

What's Next

This sub-topic on magnetic systems is the foundation for all further study of magnetic interactions in AP Physics 2. Immediately next, you will apply your understanding of magnetic dipoles and torque to analyze electromagnetic induction, specifically the behavior of generators and electric motors, which are common free-response question topics on the AP exam. Without mastering the relationship between dipole moment, torque, and potential energy, you will struggle to connect microscopic magnetic behavior of dipoles to the macroscopic behavior of these devices. Magnetic systems also connect to the broader study of electromagnetic interactions, unifying electric and magnetic dipole behavior as parallel phenomena in classical physics.