Magnetic Systems — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: magnetic dipole moment calculation, torque formula $\vec{\tau }=\vec{\mu }\times \vec{B}$, potential energy $U=-\vec{\mu }\cdot \vec{B}$, force on dipoles in non-uniform fields, and right-hand rule for magnetic moment direction.

You should already know: Vector cross product and dot product rules. Magnetic force on current-carrying wires in external fields. Basic torque and potential energy concepts from mechanics.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Magnetic Systems?

Magnetic systems are collections of magnetic dipoles, the fundamental building block of all magnetic behavior from permanent bar magnets to current-carrying loops, that interact with external magnetic fields. In the AP Physics 2 Course and Exam Description (CED), this topic falls under Unit 5: Magnetism and Electromagnetic Induction, accounting for approximately 5-7% of the total AP exam score. It appears in both multiple-choice questions (MCQ) and as a component of longer free-response questions (FRQ), often combined with concepts from electric forces or electromagnetic induction. Unlike electric systems, which can have isolated positive or negative charges, magnetic systems have no isolated magnetic monopoles—all magnetic behavior arises from dipoles, which have equal-strength north and south poles separated by a small distance. Standard notation follows AP conventions: magnetic dipole moment is $\vec{\mu }$ (Greek mu), external magnetic field is $\vec{B}$, torque is $\vec{\tau }$, and potential energy is $U$. The direction of $\vec{\mu }$ is defined as pointing from the south pole to the north pole of the dipole, consistent with the right-hand rule for current loops.

2. Magnetic Dipole Moment

The magnetic dipole moment is a vector quantity that describes the magnetic strength and orientation of a magnetic system. For a flat, current-carrying coil with $N$ turns of wire, carrying constant current $I$, and enclosing area $A$, the magnitude of the dipole moment is given by: $$\mu =NIA$$ The direction of $\vec{\mu }$ follows the right-hand rule for current loops: curl the fingers of your right hand in the direction of current flow around the loop, and your extended thumb points in the direction of $\vec{\mu }$. For a permanent bar magnet, the magnitude of $\vec{\mu }$ is an intrinsic property of the magnet, with direction still pointing from the south pole to the north pole. Intuitively, a larger current or larger loop area creates a stronger magnetic dipole, just as a larger charge separation creates a stronger electric dipole. Dipole moment is an intrinsic property of the magnetic system—it does not depend on any external magnetic field the dipole is placed in.

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 is oriented so its 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 $0.010\text{ m}$ and $0.040\text{ m}$, so the area of the coil is $A=0.010\times 0.040=4.0\times 10^{-4}{\text{ m}}^2$.
2. Recall that dipole moment is an intrinsic property of the coil, so the angle of the coil relative to the external field does not affect the calculation.
3. Substitute into the dipole moment formula: $\mu =NIA=15\times 2.0\times 4.0\times 10^{-4}=1.2\times 10^{-2}{\text{ Acdotpm}}^2$.
4. The final magnitude is $1.2\times 10^{-2}{\text{ Acdotpm}}^2$, with direction given by the right-hand rule for the current direction.

Exam tip: AP questions often add distracting information about the coil's orientation or external field strength when asking for dipole moment—remember dipole moment never depends on the external field, so ignore these extra values.

3. Torque on a Magnetic Dipole in a Uniform Field

When a magnetic dipole is placed in a uniform external magnetic field, the net force on the dipole is always zero: the force on the north pole is equal and opposite to the force on the south pole, and forces on opposite sides of a current loop cancel out. However, there is a net torque that acts to align the dipole moment $\vec{\mu }$ with the external field $\vec{B}$. The vector formula for torque is: $$\vec{\tau }=\vec{\mu }\times \vec{B}$$ The magnitude of the torque is $\tau =\mu B\sin \theta$, where $\theta$ is the angle between $\vec{\mu }$ and $\vec{B}$. Torque is maximum when $\theta =90^{\circ }$ (dipole perpendicular to the field) and zero when the dipole is aligned ($\theta =0^{\circ }$) or anti-aligned ($\theta =180^{\circ }$) with the field. This torque is the operating principle of electric motors, where a rotating current-carrying coil in a permanent magnetic field converts electrical energy to mechanical work.

Worked Example

The 15-turn coil from the previous example ($\mu =1.2\times 10^{-2}{\text{ Acdotpm}}^2$) is placed in a uniform 0.50 T external magnetic field. The plane of the coil makes a 45° angle with the direction of $\vec{B}$. What is the magnitude of the torque on the coil?

1. First, correct the angle: $\vec{\mu }$ is perpendicular to the plane of the coil, so the angle between $\vec{\mu }$ and $\vec{B}$ is $\theta =90^{\circ }-45^{\circ }=45^{\circ }$.
2. Write the torque magnitude formula: $\tau =\mu B\sin \theta$.
3. Substitute values: $\tau =(1.2\times 10^{-2}{\text{ Acdotpm}}^2)(0.50\text{ T})(\sin 45^{\circ })=(6.0\times 10^{-3})(0.707)\approx 4.2\times 10^{-3}\text{ Ncdotpm}$.
4. The torque acts to rotate the coil to align $\vec{\mu }$ with the external field.

Exam tip: Always double-check what angle you are given—if the problem gives the angle between the coil plane and $\vec{B}$, you must subtract it from 90° to get the correct $\theta$ for the torque formula.

4. Potential Energy and Force on Magnetic Dipoles

Since torque does work to rotate a magnetic dipole into alignment with an external field, we can define a potential energy for the dipole-field system. The standard formula for potential energy (with zero potential defined at $\theta =90^{\circ }$) is: $$U=-\vec{\mu }\cdot \vec{B}=-\mu B\cos \theta$$ Potential energy is minimized ($U=-\mu B$) when $\theta =0^{\circ }$ (dipole aligned with the field), which is the stable equilibrium position. Potential energy is maximized ($U=+\mu B$) when $\theta =180^{\circ }$ (dipole anti-aligned), which is unstable equilibrium. In a uniform field, net force is zero, but in a non-uniform field, a net force arises: an aligned dipole is always pulled toward the region of stronger magnetic field. This effect explains why paramagnetic and ferromagnetic materials are attracted to permanent magnets: induced dipoles align with the external field and are pulled into the stronger field near the magnet.

Worked Example

A small bar magnet with dipole moment $\mu =0.30{\text{ Acdotpm}}^2$ 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: $U=-\mu B\cos \theta$.
2. Initial aligned orientation: ${\theta }_i=0^{\circ }$, so $U_i=-\mu B\cos 0^{\circ }=-(0.30)(0.20)(1)=-0.060\text{ J}$.
3. Final 90° orientation: ${\theta }_f=90^{\circ }$, so $U_f=-\mu B\cos 90^{\circ }=-(0.30)(0.20)(0)=0\text{ J}$.
4. Change in potential energy: $\Delta U=U_f-U_i=0-(-0.060)=+0.060\text{ J}$. Work must be done on the magnet to produce this rotation, so potential energy increases, which matches our result.

Exam tip: Always remember the negative sign in the potential energy formula—if you drop it, you will get the sign of the potential energy change wrong, which is a common error on sign-focused MCQs.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using the angle between the coil plane and $\vec{B}$ directly as $\theta$ in torque or potential energy formulas. Why: AP questions intentionally give the plane angle to test student understanding of what $\theta$ describes, and many students misinterpret the definition. Correct move: Always confirm $\theta$ is the angle between $\vec{\mu }$ and $\vec{B}$; for a flat coil, $\vec{\mu }$ is perpendicular to the coil plane, so subtract the given plane angle from $90^{\circ }$ to get $\theta$.
• 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 $U=-\vec{\mu }\cdot \vec{B}$ 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 $\mu$.
• Wrong move: Using the right-hand rule for magnetic force on a moving charge to find the direction of $\vec{\mu }$ for a current loop. Why: Students confuse the multiple right-hand rules used in magnetism. Correct move: For $\vec{\mu }$ direction, always use the current curl rule: curl your right fingers along the current direction, thumb points to $\vec{\mu }$.

6. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

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? A) +x B) -x C) +y D) +z

Worked Solution: Use the cross product definition of torque: $\vec{\tau }=\vec{\mu }\times \vec{B}$. Here $\vec{\mu }$ is +y, $\vec{B}$ is +z. By the right-hand rule for cross products, the cross product of +y and +z is +x. The torque acts to rotate $\vec{\mu }$ into alignment with $\vec{B}$, which matches a +x direction torque for this rotation. The correct answer is A.

Question 2 (Free Response)

A small bar magnet with dipole moment $\mu =0.15{\text{ Acdotpm}}^2$ is placed in a uniform 0.60 T external magnetic field. (a) What is the potential energy of the magnet when it is aligned with the field? What is the orientation of the magnet for this state? (b) How much work must an external force do to rotate the magnet from aligned to anti-aligned with the field? (c) Explain why there is no net force on the magnet in this setup, even though there is a torque.

Worked Solution: (a) Aligned orientation means $\theta =0^{\circ }$. Substitute into the potential energy formula: $U=-\mu B\cos 0^{\circ }=-(0.15)(0.60)(1)=-0.090\text{ J}$. The dipole moment $\vec{\mu }$ is parallel to the external field $\vec{B}$ in this state. (b) Work done by an external force equals the change in potential energy of the system. For anti-aligned orientation, $\theta =180^{\circ }$, so $U_{anti}=-\mu B\cos 180^{\circ }=-(0.15)(0.60)(-1)=+0.090\text{ J}$. Work $W=\Delta U=0.090-(-0.090)=0.18\text{ J}$. (c) The magnetic field is uniform in this setup. The force on the north pole of the bar magnet is equal in magnitude and opposite in direction to the force on the south pole, so the net force adds to zero. The forces are separated by a distance, however, so they produce a net torque.

Question 3 (Application / Real-World Style)

An electric motor has a 50-turn rectangular coil with dimensions 2.0 cm × 3.0 cm that carries a current of 10 A. The coil is placed in a uniform 0.80 T magnetic field. When the plane of the coil is parallel to the magnetic field, what is the magnitude of the torque on the coil? What does this torque do for the motor's operation?

Worked Solution: First calculate the dipole moment of the coil: $A=0.020\text{ m}\times 0.030\text{ m}=6.0\times 10^{-4}{\text{ m}}^2$, so $\mu =NIA=50\times 10\times 6.0\times 10^{-4}=0.30{\text{ Acdotpm}}^2$. When the plane of the coil is parallel to $\vec{B}$, $\vec{\mu }$ is perpendicular to $\vec{B}$, so $\theta =90^{\circ }$. Torque magnitude is $\tau =\mu B\sin 90^{\circ }=(0.30)(0.80)(1)=0.24\text{ Ncdotpm}$. This torque is the driving torque that rotates the motor's shaft, converting electrical energy from the current into mechanical rotational energy.

7. Quick Reference Cheatsheet

8. What's Next

This chapter 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 FRQ topics on the AP exam. Without mastering the relationship between dipole moment, torque, and potential energy, you will struggle to connect the 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.

Magnetic Force on Currents Electromagnetic Induction Faraday's Law of Induction