Topic 12.2: Magnetism and Moving Charges

What Topic 12.2 requires

Topic 12.2 sits in Unit 12, Magnetic Fields and Electromagnetism and carries two learning objectives. It prints no boundary statement, which turns out to be the most consequential fact on this page.

12.2.A, describe the magnetic field produced by moving charged objects.

• 12.2.A.1 a single moving charged object produces a magnetic field.
• 12.2.A.1.i the magnetic field at a particular point produced by a moving charged object depends on the object's velocity and the distance between the point and the object.
• 12.2.A.1.ii at a point in space, the direction of the magnetic field produced by a moving charged object is perpendicular to both the velocity of the object and the position vector from the object to that point in space and can be determined using the right-hand rule.
• 12.2.A.1.iii the magnitude of the magnetic field is a maximum when the velocity vector and the position vector from the object to that point in space are perpendicular.

12.2.B, describe the force exerted on moving charged objects by a magnetic field.

• 12.2.B.1 a magnetic field will exert a force on a charged object moving within that field, with magnitude and direction that depend on the cross-product of the charge's velocity and the magnetic field. The relevant equation is $\vec{F}_B = q(\vec{v} \times \vec{B})$.
• 12.2.B.2 in a region containing both a magnetic field and an electric field, a moving charged object will experience independent forces from each field.
• 12.2.B.3 the Hall effect describes the potential difference created in a conductor by an external magnetic field that has a component perpendicular to the direction of charges moving in the conductor.

The suggested skills are 1.B, create quantitative graphs with appropriate scales and units, including plotting data; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; 3.A, create experimental procedures that are appropriate for a given scientific question; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim.

Seven statements and one equation. The equation is the only one in the topic, and everything under 12.2.A is qualitative on purpose.

The cross product replaces the sine, and a boundary statement disappears

AP Physics 2 has a Topic 12.2 with the same title, and its objective 12.2.A is the same four statements word for word. The divergence is entirely in 12.2.B, and it is sharper than a change of notation.

The AP Physics 2 boundary statement reads in full: "Quantitative treatment of the magnitude of the magnetic force exerted by a magnetic field on a moving charge is limited to angles of 0, 90, and 180 degrees between the velocity and the magnetic field. Qualitative analysis of other angles is permitted."

AP Physics C prints no boundary statement for Topic 12.2 at all. So the fence is gone. A velocity at 53 degrees to the field is a legitimate quantitative question in this course and is not one in the algebra-based course, and the reason is not that Physics C students are expected to work harder. It is that the cross product is a single object that produces both a magnitude and a direction at any angle, whereas $qvB\sin\theta$ is a magnitude with a direction rule bolted on beside it.

The booklet backs that up. Its VECTORS table, which is printed in the exam's Table of Information appendix and sits on the same pages as the equations, gives

along with unit-vector notation and component addition. Its trigonometric-values table lists 0, 30, 37, 45, 53, 60 and 90 degrees with exact values. So the tools for an arbitrary-angle calculation are handed to you, and the absence of a boundary statement means you may be asked to use them.

Two smaller differences worth knowing. Physics C's suggested-skill list swaps 1.A for 1.B, quantitative graphs with scales and units including plotting data, and adds 3.A, create experimental procedures. Those two together are the fingerprint of a laboratory question, and this is the only topic in Unit 12 that lists 3.A.

The [AP Physics 2 Topic 12.2 page](/ap-physics-2/unit-12-magnetism-and-electromagnetism/12-2-magnetism-and-moving-charges) is for AP Physics 2 students, and this page is for AP Physics C: Electricity and Magnetism students. If you are in Physics 2, the three allowed angles are your whole quantitative world here and that page says so. If you are in Physics C, that page is still a good read on the right-hand rule and the physical picture, and then you come back for the vector form.

Objective 12.2.A prints no equation, and that is deliberate

A moving charge makes a magnetic field. That is 12.2.A.1, and the course stops there quantitatively. There is no point-charge field equation anywhere in this course description or on the equation sheet, so anything you may have seen with a $qv \times \hat{r}$ in it is outside what the exam can ask you to compute.

What the CED does give you is three pieces of functional and geometric information, and each is separately examinable.

• Depends on velocity and distance (12.2.A.1.i). More speed means more field, more distance means less. The statement gives directions of dependence and no exponents. Do not supply an inverse-square law from memory.
• Direction is doubly perpendicular (12.2.A.1.ii). At a point in space, $\vec{B}$ is perpendicular both to the object's velocity and to the position vector running from the object to that point, and the right-hand rule picks which of the two perpendicular directions it is.
• Magnitude is maximum when those two are perpendicular (12.2.A.1.iii). So a point directly ahead of the moving charge, along its line of motion, has the weakest field, and a point off to the side has the strongest.

That third statement is the one students skip, and it is the one a 2.C comparison question is built from. Two points at the same distance from a moving proton, one in front of it and one beside it, do not see the same field, and 12.2.A.1.iii tells you which is larger without any calculation.

Compare this with Topic 12.3, where the same three ideas reappear for a current element and this time do come with an equation, the Biot-Savart law:

Look at what each piece of that does. The $1/r^2$ is 12.2.A.1.i's distance dependence. The cross product with $\hat{r}$ is 12.2.A.1.ii's double perpendicularity. And $\left| d\vec{\ell} \times \hat{r} \right|$ is largest when the two are perpendicular, which is 12.2.A.1.iii. So 12.2.A is the Biot-Savart law with the mathematics removed, stated for a single charge instead of a current element. Reading it that way makes all three statements one idea rather than three facts to memorise.

Using the cross product without guessing

Three routes to an answer, and it is worth being able to run all three because different questions hand you different information.

1. Magnitude and direction separately. Use $\left| \vec{v} \times \vec{B} \right| = vB\sin\theta$ from the booklet's VECTORS table for the size, then the right-hand rule for the direction. State the rule once and use it identically every time: point the fingers of your right hand along $\vec{v}$, curl them toward $\vec{B}$ through the angle between them, and the thumb gives the direction of $\vec{v} \times \vec{B}$.

Then check the sign of the charge. For positive $q$ the force is along $\vec{v} \times \vec{B}$. For negative $q$, and every electron problem is one, the force is opposite to it. Reversing at the end is more reliable than trying to run the rule backwards with your left hand.

2. Components with unit vectors. When the vectors are given in $\hat{i}$, $\hat{j}$, $\hat{k}$ form this is faster and it cannot go wrong on direction. The three products you need are

with the reverse of each carrying a minus sign, and any vector crossed with itself giving zero. Cycling $i \to j \to k \to i$ forwards is positive and backwards is negative.

3. Geometry first, when the geometry is simple. If $\vec{v}$ is perpendicular to $\vec{B}$ then $\sin\theta = 1$ and the magnitude is just $qvB$. If $\vec{v}$ is parallel or antiparallel to $\vec{B}$ then $\sin\theta = 0$ and the force is exactly zero. That last case is a standing trap: a charge moving along a field line feels no magnetic force at all, however fast it moves and however strong the field.

Two habits that stop most sign errors:

• Declare your axes and your page convention before you compute. A dot in a circle is out of the page and a cross in a circle is into the page. With $\hat{i}$ to the right and $\hat{j}$ up, out of the page is $+\hat{k}$. Write that down once at the top of the working and never re-derive it mid-problem.
• Check perpendicularity at the end. The result must satisfy $\vec{F} \cdot \vec{v} = 0$ and $\vec{F} \cdot \vec{B} = 0$. In component form those are two quick dot products, and they catch a mis-signed term immediately.

One thing the vector form shows that the sine form hides. Split $\vec{v}$ into a part along $\vec{B}$ and a part perpendicular to it. The parallel part crosses with $\vec{B}$ to give zero, so it is never affected by the magnetic force and continues unchanged, while the perpendicular part is turned. A charge launched at a general angle to a uniform field therefore drifts steadily along the field while circling around it. That helical picture is a consequence of the cross product rather than an essential-knowledge statement, and no equation for it is printed, but it is the right mental image and it explains why 12.2.A.1.iii cares about perpendicularity.

A magnetic force never does work

This is not one of the topic's seven essential-knowledge statements, and it is not printed as an equation anywhere. It follows in two lines from things that are printed, and it is the single most useful consequence in the topic.

The C: E&M sheet reprints the whole C: Mechanics table, so work is available in its integral form,

and the booklet's VECTORS table gives $\vec{A} \cdot \vec{B} = AB\cos\theta$. The magnetic force is $q(\vec{v} \times \vec{B})$, and a cross product is perpendicular to both of its inputs, so $\vec{F}_B$ is perpendicular to $\vec{v}$. Since $d\vec{r}$ points along $\vec{v}$, the angle in the dot product is 90 degrees, the cosine is zero, and the integrand vanishes at every instant. The work is zero over any path, of any length, in any field.

What follows from that:

• The speed of a charged particle in a magnetic field never changes. By the work-energy theorem, $\Delta K = 0$, so $\left| \vec{v} \right|$ is constant even though $\vec{v}$ itself is turning constantly.
• A magnetic field cannot speed a particle up or slow it down. If a problem shows a particle gaining kinetic energy in a region with a magnetic field, something else is doing the work: an electric field, a contact force, or a changing field producing an induced electric field, which is Unit 13.
• A magnetic field can still change the momentum. Momentum is a vector, and turning the velocity changes it. There is no contradiction: $\vec{F}_{\text{net}} = d\vec{p}/dt$ is nonzero while $\Delta K = 0$.
• This answers the unit's own essential question. The CED asks why large-scale charged-particle accelerators, such as those at CERN, are in the shape of a circle. A magnetic force perpendicular to the velocity bends the path without spending energy, so the magnets steer while separate electric fields do the accelerating.

The place this gets tested is a justification, skill 3.B. Writing "the magnetic force does no work" is the claim. The justification is that the force is a cross product with the velocity and so is always perpendicular to it, and the work integral of a perpendicular force is zero. The unit's exam-preparation page is explicit that simply naming a law is not enough to earn credit, so include the perpendicularity step.

Circular motion, and the result the CED never prints

Put a charge into a uniform field with its velocity perpendicular to the field. The force has constant magnitude $qvB$, since the speed cannot change, and it stays perpendicular to the velocity. Constant-magnitude perpendicular force is exactly the condition for uniform circular motion.

No equation for the radius appears in the AP Physics C: Electricity and Magnetism course description, and none is printed on its equation sheet. Check the sheet's magnetism column line by line and it ends at $\omega_{LC} = 1/\sqrt{LC}$ with nothing resembling $r = mv/(qB)$ anywhere in it. What is printed, in the reprinted mechanics table on the same page, is $a_c = v^2/r = r\omega^2$ and $\vec{F}_{\text{net}} = d\vec{p}/dt$. So you assemble the radius from the force law and the centripetal condition, in one line, every time you need it. The AP Physics C: E&M formula sheet page shows both tables side by side.

That assembly is worth doing symbolically once, because the pattern it reveals is what gets examined.

and then the period, since one revolution covers a circumference at constant speed:

Read the two results against each other. The radius depends on speed and the period does not. A faster particle traces a bigger circle in exactly the same time. That is the fact worth carrying, and it is what makes a cyclotron possible: one fixed driving frequency works for a particle at every energy.

The CED's own Building the Science Practices page for Unit 12 poses precisely this as its worked illustration. It says students could describe conceptually what happens to the net force exerted on an electron moving through a magnetic field if the speed of the electron increases, and then justify what impact that change will have on the radius of the path of the electron. Notice the shape of the task: describe the force first, then justify the consequence for the radius. It is a skill 2.D functional-dependence question followed by a 3.C justification, and the CED adds that students should be comfortable making claims about the reasonableness of their claims and justifications made with functional dependence, starting with the first principles of physics.

So the answer it is looking for runs: the force magnitude is $qvB$, which increases in proportion to the speed; the required centripetal force for a fixed radius would rise as $v^2$; the force only rises as $v$, so the radius must increase, and in fact in direct proportion to $v$. Getting to "the radius increases" is half the credit. Explaining why it increases linearly rather than by some other factor is the other half.

Related patterns from the same one-line derivation, each a standard comparison question:

If the velocity is not perpendicular to the field, only the perpendicular component circles, so replace $v$ by $v\sin\theta$ in the radius while the period is untouched. For a discussion of the circular-motion side of this, the algebra-based treatment at AP Physics 1 Topic 2.9 covers the centripetal condition itself.

Both fields at once, and the Hall effect

Statement 12.2.B.2 is a superposition licence: in a region containing both a magnetic field and an electric field, a moving charged object will experience independent forces from each field. So you compute $q\vec{E}$ and $q(\vec{v} \times \vec{B})$ separately and add them as vectors. There is no combined rule to learn and no interaction term.

The classic arrangement is the velocity selector, and it is the second worked example below. Cross the two fields so their forces on a charge moving through oppose each other. The electric force $qE$ does not depend on speed and the magnetic force $qvB$ does, so exactly one speed passes through undeflected:

The charge cancels, so the selected speed is the same for every particle in the beam regardless of its charge or mass, and that is what makes the device useful. Anything faster is pushed toward the magnetic side, anything slower toward the electric side.

Statement 12.2.B.3, the Hall effect, is the same balance happening inside a conductor without anyone arranging it. The statement reads: the Hall effect describes the potential difference created in a conductor by an external magnetic field that has a component perpendicular to the direction of charges moving in the conductor.

Work through the mechanism, because the CED prints no Hall equation and the derivation is the content.

1. Charge carriers drift along a conducting strip with some drift speed $v$, and an external field $\vec{B}$ has a component perpendicular to that drift.
2. Each carrier feels $q(\vec{v} \times \vec{B})$, which pushes it sideways, across the strip rather than along it.
3. Carriers pile up on one edge, leaving the opposite edge with the opposite sign. That separation is itself a charge distribution, so it creates an electric field across the strip.
4. That transverse electric field pushes back on the drifting carriers. It grows until the two forces balance, $qE = qvB$, and then no more carriers cross.
5. The steady transverse field means a steady potential difference across the width of the strip. With a uniform field across a width $w$, and using $E_x = -dV/dx$ from the sheet, the size of it is $\Delta V = Ew = vBw$.

Two points examiners like:

• The sign of the Hall voltage tells you the sign of the carriers. Positive carriers drifting one way and negative carriers drifting the other way constitute the same conventional current, but the magnetic force pushes them to the same edge, so the edge that goes positive differs. The sheet's exam conventions state that the direction of current is the direction in which positive charges would drift, which is the convention to reason with, and the Hall effect is the experiment that can tell the difference.
• The field only needs a perpendicular component. The statement says "a component perpendicular to the direction of charges moving in the conductor". A field partly along the drift still produces a Hall voltage, just a smaller one, because only the perpendicular part contributes to the cross product.

How Topic 12.2 is tested

None of the fifteen sample multiple-choice questions and none of the four sample free-response questions in the CED aligns to 12.2.A or 12.2.B. Two of the five optional sample instructional activities for Unit 12 do, and both are quantitative.

The first has students research a model of old CRT television, find the potential difference through which the electrons are accelerated to determine their speed entering the magnetic field region, and then find the strength of the magnetic field needed to deflect them to different points on the screen. That is a two-stage problem: energy from Unit 9 to get the speed, then this topic to get the deflection. The second is a graph-and-switch pairing on Topic 12.3 rather than this one.

The suggested-skill list tells you the rest.

1. Skill 2.A, derive a symbolic expression. Produce $r = mv/(qB)$, or the velocity-selector condition, or a Hall voltage, from the printed force law plus one other printed relationship. This is the most likely free-response shape in this topic, and skill 2.A carries 25 to 30 percent of the multiple-choice weighting, more than any other single skill.
2. Skill 2.C, compare between scenarios. Two particles, or one particle at two speeds, or the same particle in two fields. The table in the circular-motion section above is the shape of the answer.
3. Skill 3.B, apply a law to make a claim. Claim the force is zero for a charge moving along a field line, or that the speed cannot change, and back it with the cross product.
4. Skill 1.B, quantitative graphs with scales, units and plotted data. Radius against speed is a straight line through the origin of slope $m/(qB)$. Force against speed is a straight line. Force against the sine of the angle is a straight line. Any of those is a plot-the-data-and-take-the-slope task.
5. Skill 3.A, create experimental procedures. This is the only Unit 12 topic that lists 3.A, and the natural question is how you would measure a field strength, or a charge-to-mass ratio, from deflection data. Pairing 3.A with 1.B is the signature of a full experimental-design question.

Traps worth rehearsing:

• Forgetting the sign of the charge. Run the right-hand rule to get $\vec{v} \times \vec{B}$, then reverse if $q$ is negative. Doing it in that order is what makes it reliable.
• Using $qvB$ when the angle is not 90 degrees. Physics C has no boundary statement fencing off other angles, so the sine factor is live. If a question gives you 37 or 53 degrees it is telling you to use the trigonometric-values table in the booklet.
• Assuming a magnetic field changes the speed. It cannot, ever.
• Treating the two fields in 12.2.B.2 as combining into one rule. They do not; add the two force vectors.
• Confusing the field a charge makes with the force a field puts on it. Objective 12.2.A is the first and has no equation; 12.2.B is the second and has the only one. A question about the field at a point near a moving proton is qualitative; a question about the force on that proton is not.

Where this goes next. Objective 12.2.A becomes the Biot-Savart law once the moving charges are organised into a current, which is Topic 12.3, and objective 12.2.B becomes the force on a wire in the same topic. Topic 12.4 then supplies the fields those forces act in, and Unit 13 puts the whole apparatus in motion.