AP Physics 2 · Topic 12.2

Topic 12.2: Magnetism and Moving Charges

Unit 12: Magnetism and Electromagnetism12-15% of the multiple-choice section

Topic 12.2 gives you one force equation and one direction rule. The magnitude is the charge times the speed times the field times the sine of the angle between velocity and field. The direction is perpendicular to both, from the right-hand rule for a positive charge, reversed for a negative one.

AP Physics: Unit 12 (topics 12.2 Magnetism and Moving Charges). Topic 12.2 carries two CED learning objectives: 12.2.A, describe the magnetic field produced by moving charged objects, and 12.2.B, describe the force exerted on moving charged objects by a magnetic field. It holds Unit 12's only boundary statement, which limits quantitative treatment of the magnetic force magnitude to angles of 0, 90 and 180 degrees between the velocity and the magnetic field while permitting qualitative analysis of other angles. The relevant equation, F_B = qvB sin(theta), is the first entry in the Magnetism group of the AP Physics 2 equation sheet. The CED's suggested skills for this topic are 1.A, 2.A, 2.C and 3.B. Unit 12 is weighted at 12 to 15 percent of the multiple-choice section and estimated at about 10 to 14 class periods.

What Topic 12.2 requires

Topic 12.2 is the second topic of Unit 12, which the CED weights at 12 to 15 percent of the multiple-choice section and estimates at roughly 10 to 14 class periods. Two learning objectives:

12.2.A Describe the magnetic field produced by moving charged objects.
12.2.B Describe the force exerted on moving charged objects by a magnetic field.

Read them as a pair pointing opposite ways. 12.2.A treats a moving charge as a source of magnetic field; 12.2.B treats it as a target of one. Topic 12.1 established what a magnetic field is; this is where charge and field act on each other.

The CED gives one relevant equation for the whole topic, under EK 12.2.B.2.i, printed as the first entry in the Magnetism group of the AP Physics 2 equation sheet:

F_B = qvB\sin\theta

Unit 12's only boundary statement also lands here. There is exactly one in the whole unit, and this is it, with its second sentence attached.

Boundary statement. 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.

Both sentences matter. The first caps your arithmetic at three angles, so every quantitative magnetic-force question has the velocity perpendicular to the field, along it, or against it. The second keeps the door open: a 30 degree case can still be asked about, in words rather than numbers.

The suggested skills are 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 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; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim.

A moving charge makes a magnetic field

LO 12.2.A is short, and it is the half of the topic most revision material skips. EK 12.2.A.1: a single moving charged object produces a magnetic field. Not a current, not a wire, not a magnet. One charge, moving. It is the same idea as EK 12.1.B.1 from the previous topic, which said magnetic dipoles result from the circular or rotational motion of electric charges.

Three sub-statements describe that 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.

No equation is given for this field, and none is printed on the sheet. AP Physics 2 is algebra based, and the CED asks you to describe the field of a moving charge, not to compute it. A law that builds it out of a cross product belongs to AP Physics C: Electricity and Magnetism. The requirement here stops at those three statements.

The two right-hand rules in this topic are not the same rule. 12.2.A.1.ii runs the rule on velocity and a position vector to find the direction of the field a charge produces. 12.2.B.2.ii, one objective later, runs it on velocity and the field to find the direction of the force a charge feels. Same hand, different second input. Confusing them is a tidy way to produce a confident answer at 90 degrees to the right one, so look for the word produced before you lift your hand.

EK 12.2.A.1.iii tells you where that field is strongest: off to the side of the charge's motion, where the position vector is perpendicular to the velocity. The CED states the maximum and stops. It does not say what happens directly ahead of or behind the charge, so do not put a value, or a zero, on that case from memory.

The force equation, and the only three angles you may compute with

EK 12.2.B.1 opens with a framing sentence worth pausing on: magnetic forces describe interactions between moving charged objects. Between moving charges. The magnet on a fridge door is in that sentence too, because Topic 12.1 said a magnet's dipoles are circulating charge.

EK 12.2.B.2 narrows it: a magnetic field may exert a force on a charged object moving in that field. May. Whether it does, and how hard, is EK 12.2.B.2.i, which says the magnitude is proportional to the magnitude of the charge, the magnitude of the object's velocity, and the magnitude of the field, and also depends on the angle between the velocity and magnetic field vectors. The relevant equation is the sheet's:

F_B = qvB\sin\theta

Here $q$ is the magnitude of the charge in coulombs, $v$ the speed, $B$ the field magnitude in tesla, and $\theta$ the angle between $\vec{v}$ and $\vec{B}$ . Not between the velocity and a wire, not between the velocity and a surface. Between velocity and field.

The boundary statement then hands you the entire quantitative content of the topic in three rows.

Angle between $\vec{v}$ and $\vec{B}$	$\sin\theta$	Magnitude of the force	The picture
0 degrees	0	$F_B = 0$	charge moving along the field
90 degrees	1	$F_B = qvB$ , the largest available	charge moving across the field
180 degrees	0	$F_B = 0$	charge moving against the field

Two of the three give zero, and that is the most useful line on this page. A charge fired straight along a magnetic field, or straight back against it, passes through untouched at constant velocity. There is no partial effect and no gradual turn.

Zero is also the answer whenever the charge is standing still: put $v = 0$ into the equation and the force vanishes. That is Topic 12.1's claim that a magnetic field acts on moving charges, now with an equation under it.

For every other angle the boundary statement permits qualitative analysis, so the fair question becomes comparative, which is skill 2.C. Two charges at the same speed in the same field, one at 30 degrees to it and one at 60: the 60 degree one feels the greater force, because its angle is closer to perpendicular, and you have answered without computing either.

The right-hand rule, stated once and used the same way every time

EK 12.2.B.2.ii gives the direction: the direction of the force exerted by a magnetic field on a moving charged object is perpendicular to both the direction of the magnetic field and the velocity of the charge, as defined by the right-hand rule.

Perpendicular to both. That is a free check on every answer you produce. If your force ends up lying anywhere in the plane containing the velocity and the field, it is wrong before you even think about the sign.

The rule, stated once. Point the fingers of your right hand along the velocity of the charge. Curl them toward the magnetic field. Your outstretched thumb now points along the magnetic force on a positive charge.

For a negative charge the force is opposite to your thumb. Say it every single time. The rule is built for positive charge and this course is full of electrons. Handle the reversal by running the rule as stated, taking the positive-charge answer, then flipping it as its own written step. Switching to your left hand is faster and is a dependable way to get it wrong the moment you lose track of which hand you are on. The free-response section is marked on the reasoning, and "the charge is negative, so the force is opposite to the direction the right-hand rule gives" is a clause that earns its place.

An equivalent form, if curling your fingers is awkward: point your right index finger along the velocity and your right middle finger along the field, and your thumb gives the force on a positive charge. Use one form, not both.

Page notation. Diagrams put fields perpendicular to the page constantly. The standard marks are a cross for a field going into the page, drawn as the tail feathers of an arrow going away from you, and a dot for one coming out. Skill 1.A is this topic's diagram skill, so expect to read these and draw them.

Here is the rule worked out for five arrangements. Right means across the page to your right, up means toward the top of it.

Velocity of the charge	Magnetic field	Force on a positive charge	Force on a negative charge
right	into the page	up the page	down the page
right	out of the page	down the page	up the page
up the page	into the page	left	right
right	up the page	out of the page	into the page
right	right	zero, the angle is 0	zero

Run each row with your own hand rather than reading the answers off. The fourth row catches people, because both vectors lie flat in the page and the force does not.

Why a magnetic force never does work

This consequence settles more questions than the equation does, and it is the conceptual centre of the topic. The magnetic force on a moving charge is always perpendicular to that charge's velocity. EK 12.2.B.2.ii says so in as many words. Now take the definition of work, which the AP Physics 2 sheet reprints in its Mechanics and Fluids table:

W = F_{\parallel} d = F d \cos\theta

That angle is between the force and the displacement. A moving charge is displaced along its velocity, and the magnetic force is perpendicular to that velocity, so the angle is 90 degrees, $\cos 90^\circ = 0$ , and

W_{\text{magnetic}} = F d \cos 90^\circ = 0

A magnetic force does no work on a charged particle. Not approximately and not usually: never, in any configuration, at any speed. The sheet also prints $\Delta K = \sum W_i$ , so no work means no change in kinetic energy, and at fixed mass that means no change in speed.

A magnetic field can turn a charged particle through any angle you like and cannot change how fast it goes. It steers. Be careful with the word accelerate: the particle is accelerating in the physics sense, because the direction of its velocity is changing. Constant speed is not constant velocity.

Sourcing note. The CED does not print "a magnetic force does no work" as an essential-knowledge statement in Unit 12. What it prints is EK 12.2.B.2.ii, that the force is perpendicular to the velocity. The no-work conclusion is a two-line derivation from that statement plus the definition of work, which is Topic 3.2 material carried over, and it is a good example of the chain the unit's exam advice wants written out rather than asserted.

The contrast with the electric force is the point of holding the two together. An electric force lies along the field, so it generally has a component along the displacement and does work; the sheet's $\Delta U_E = q\Delta V$ is that work kept as potential energy, covered in the electric field and potential guide. In a region with both fields, only the electric one can change a charge's speed.

Circular motion in a uniform field

Set a charge moving with its velocity perpendicular to a uniform magnetic field, the 90 degree row above, and follow what happens.

The magnitude of the force is $qvB$ and it stays $qvB$ , because $q$ and $B$ are fixed and $v$ cannot change. The direction is always perpendicular to the velocity, so as the velocity turns the force turns with it and stays perpendicular. A force of constant magnitude always perpendicular to the velocity is the definition of uniform circular motion, so the particle travels a circle at constant speed.

The CED puts this inside Topic 12.2 through its sample activities: one asks students to sketch and justify a graph of "Radius vs. speed of a circling charge in a magnetic field". Skill 2.A, derive a symbolic expression, is listed for the topic, so build the result instead of memorising it.

Two lines already on the equation sheet do the work. Circular motion gives $a_c = v^2/r$ , and Newton's second law appears as $\vec{F}_{\text{net}} = m\vec{a}$ . With the magnetic force as the only force on the particle:

qvB = m\frac{v^2}{r}

Cancel one factor of $v$ from each side and rearrange:

r = \frac{mv}{qB}

That is the radius, and the graph the CED's activity asks for is a straight line through the origin, because $r$ is proportional to $v$ when $m$ , $q$ and $B$ are held fixed. Double the speed and you double the radius.

The period follows from one circumference travelled at constant speed, $T = 2\pi r/v$ :

T = \frac{2\pi m}{qB}

The speed has cancelled. Two identical particles in the same field, one twice as fast as the other, go round circles of different sizes in the same time. Be ready to explain that rather than state it: the faster particle has proportionally further to travel and travels proportionally faster, so the two effects cancel exactly.

Neither of those two results is printed on the AP Physics 2 equation sheet. Count the Magnetism group and you find seven entries; the radius and the period are not among them. Both are derived, which is why skill 2.A hangs off this topic and not off Topic 12.1.

The free-body diagram habit from Topic 2.9 in AP Physics 1 carries over unchanged. Do not draw a centripetal force arrow next to the magnetic force arrow: the magnetic force is the inward force, one arrow pointing at the centre. The centripetal force guide drills that case by case.

Both fields at once, and the Hall effect

EK 12.2.B.3: in a region containing both a magnetic field and an electric field, a moving charged object will experience independent forces from each field.

Independent is the operative word. The electric force is along the field for a positive charge, whether or not it moves at all. The magnetic force is $qvB\sin\theta$ , perpendicular to both the velocity and the field, and exists only while the charge moves. Neither modifies the other: find each on its own, then add them as vectors.

The CED's own suggested activity for this topic asks students to "give velocity and magnetic field directions and determine the direction of an electric field that would cause balanced forces". That arrangement is the third worked example below, and its result is a good one: the speed at which the two forces balance depends on neither the size of the charge nor its sign.

EK 12.2.B.4 closes the topic. 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.

Read that back through EK 12.2.B.2 and it assembles itself. Charges moving along a conductor are moving charges in a magnetic field, so each feels $qvB\sin\theta$ . If the field has a component perpendicular to their motion, that force pushes them sideways, across the conductor rather than along it. They collect on one face and leave the opposite face short of them, and a separation of charge across the conductor is a potential difference across it.

The piece worth carrying into an exam is which face collects the charge, because that depends on the sign of the carriers: the right-hand rule reverses for negative ones. Measuring which side goes positive therefore tells you whether the moving charges are positive or negative, which the conventional current alone cannot. The CED asks you to describe the effect, and no Hall voltage equation appears on the sheet.

How Topic 12.2 is examined

Unit 12's opening page names this topic's failure mode outright. On writing justifications it says that simply referencing an equation, law, or physical principle is not sufficient, and the example it picks is this one: stating that "the force on a charged particle is to the right because of the 'right-hand rule'" is not a complete enough answer to earn points on the free-response section of the exam. Students should clearly and concisely explain the steps that lead from the equation, law or physical principle to the justification of their claim.

So a full-credit direction answer has parts: the two input directions, the rule and the hand, the sign of the charge and what it does, then the direction. In one sentence: the velocity is to the right and the field is into the page, so pointing the right hand's fingers right and curling them into the page puts the thumb up the page; the particle is an electron and therefore negative, so the force is down the page.

What the CED's activities say about question style. Three of the five sample instructional activities in Unit 12 belong to Topic 12.2. One sends students to an AP Central item called "AP Physics 2 Featured Question: Charged Particle in a Magnetic Field". One has students write right-hand-rule questions for each other, and lists three shapes for them: give the charge sign, velocity, and magnetic force directions and determine the magnetic field direction; give velocity, force, and magnetic field directions and determine the charge sign; give velocity and magnetic field directions and determine the direction of an electric field that would cause balanced forces. The third asks for a sketched graph of radius against speed for a circling charge. Only one of those runs the rule forwards, so practise it backwards.

Errors that cost marks on 12.2.

Running the rule on a negative charge and forgetting to reverse. The most mechanical way there is to lose a direction mark, and the easiest to fix: write the reversal as its own step.
Measuring the angle from the wrong pair of vectors. It is the angle between the velocity and the field. Nothing else.
Computing a magnetic force at an arbitrary angle. The boundary statement limits quantitative work to 0, 90 and 180 degrees, so a question handing you 40 degrees wants a comparison or an explanation.
Saying a magnetic field speeds a particle up. It cannot. The force is perpendicular to the velocity, so it does no work and the speed is fixed.
Drawing a separate centripetal force on a circling charge. The magnetic force is the inward force.
Giving a stationary charge a magnetic force. Set the speed to zero in the equation and see what is left.
Confusing the topic's two right-hand rules. 12.2.A.1.ii finds the field a moving charge produces; 12.2.B.2.ii finds the force it feels.

A proton crossing a uniform field: force, radius, period

A proton travels at $2.0 \times 10^6 \ \mathrm{m/s}$ to the right across a page, through a uniform magnetic field of $0.50 \ \mathrm{T}$ directed into the page. Find the magnitude and direction of the magnetic force on it, the radius of its path, and the time it takes to go once around. Use the sheet values $q = 1.60 \times 10^{-19} \ \mathrm{C}$ and $m_p = 1.67 \times 10^{-27} \ \mathrm{kg}$ .

Set the convention first. Right across the page is positive $x$ , up the page positive $y$ , out of the page positive $z$ , so the field into the page lies along negative $z$ . These hold throughout.
Check the angle. The velocity lies in the page and the field is perpendicular to the page, so the angle between them is 90 degrees, one of the three the boundary statement lets you compute with, and $\sin 90^\circ = 1$ .
Magnitude, from $F_B = qvB\sin\theta$ . Take the last two factors first: $(2.0 \times 10^6)(0.50) = 1.0 \times 10^6$ . Then $(1.60 \times 10^{-19})(1.0 \times 10^6) = 1.6 \times 10^{-13}$ , so $F_B = 1.6 \times 10^{-13} \ \mathrm{N}$ .
Direction. Fingers of the right hand point right, along the velocity; curl them into the page, toward the field; the thumb points up the page. The proton is positive, so no reversal: the force is up the page, along positive $y$ . Check it against EK 12.2.B.2.ii, which requires the force perpendicular to both the velocity (along $x$ ) and the field (along $z$ ). Up the page is along $y$ , so it passes.
Radius. The force is constant in magnitude and always perpendicular to the velocity, so the path is a circle. Set the magnetic force equal to mass times centripetal acceleration, with $a_c = v^2/r$ from the sheet: $qvB = mv^2/r$ , so $r = mv/(qB)$ .
Substitute. Numerator $(1.67 \times 10^{-27})(2.0 \times 10^6) = 3.34 \times 10^{-21}$ ; denominator $(1.60 \times 10^{-19})(0.50) = 8.0 \times 10^{-20}$ . Divide: $r = 0.0418 \ \mathrm{m}$ , about 4.2 cm.
Period, from one circumference at constant speed: $T = 2\pi r/v = 2\pi(0.0418)/(2.0 \times 10^6) = 1.3 \times 10^{-7} \ \mathrm{s}$ . Symbolically the same thing is $T = 2\pi m/(qB)$ , in which the speed has cancelled. The speed and the field are each given to two significant figures, so all three answers stop at two.
Now change one input and read the structure, which is skill 2.C. Double the speed to $4.0 \times 10^6 \ \mathrm{m/s}$ : the force doubles to $3.2 \times 10^{-13} \ \mathrm{N}$ and the radius doubles to 8.4 cm, but the period does not move, because $T = 2\pi m/(qB)$ contains no $v$ .

The force is $1.6 \times 10^{-13} \ \mathrm{N}$ directed up the page, the radius is 4.2 cm, and one full circuit takes $1.3 \times 10^{-7} \ \mathrm{s}$ , or 0.13 microseconds. The proton circles at constant speed: the magnetic force turns it and, being perpendicular to the velocity at every instant, does no work on it.

The same charge at all three allowed angles

A charge of $+3.0 \ \mu\mathrm{C}$ moves at $4.0 \times 10^5 \ \mathrm{m/s}$ in a uniform magnetic field of magnitude $0.25 \ \mathrm{T}$ . Find the magnitude of the magnetic force on it when the velocity is along the field, perpendicular to the field, and directly opposite the field. Then say what changes if the charge is swapped for $-3.0 \ \mu\mathrm{C}$ .

Convert once, at the start: $+3.0 \ \mu\mathrm{C} = 3.0 \times 10^{-6} \ \mathrm{C}$ .
Compute the product all three cases share. $(4.0 \times 10^5)(0.25) = 1.0 \times 10^5$ , then $(3.0 \times 10^{-6})(1.0 \times 10^5) = 0.30$ . So $qvB = 0.30 \ \mathrm{N}$ and every answer is $0.30\sin\theta$ newtons.
Along the field: $\theta = 0$ , $\sin 0 = 0$ , so $F_B = 0$ . As far as the magnetic field is concerned the charge sails straight on at constant velocity.
Perpendicular to the field: $\theta = 90^\circ$ , $\sin 90^\circ = 1$ , so $F_B = 0.30 \ \mathrm{N}$ , the largest magnitude available for these values.
Opposite the field: $\theta = 180^\circ$ , $\sin 180^\circ = 0$ , so $F_B = 0$ again. Reversing the direction of travel along a field line does not turn a zero into a maximum; both ends of the line give nothing.
Flip the sign of the charge. In the sheet equation $q$ is a magnitude, so all three force magnitudes are unchanged: 0, then 0.30 N, then 0. What changes is the direction in the perpendicular case, which reverses, because the right-hand rule returns the force on a positive charge.
Note what the boundary statement did and did not permit. At 45 degrees you could say the force is between zero and 0.30 N, and rank it against a 70 degree case, but not put a number on it.

Zero at 0 degrees, 0.30 N at 90 degrees, and zero again at 180 degrees. Swapping in $-3.0 \ \mu\mathrm{C}$ leaves all three magnitudes unchanged and reverses the direction of the one nonzero force, because the right-hand rule is stated for a positive charge. Those three angles are the topic's whole quantitative scope.

The electric field that balances the magnetic force

A positive charge moves to the right across the page at an unknown speed, through a uniform magnetic field of $0.060 \ \mathrm{T}$ directed into the page. A uniform electric field of magnitude $3.0 \times 10^4 \ \mathrm{N/C}$ is applied so that the two forces on the charge cancel. Find the direction of that electric field and the speed at which the forces balance.

Keep the same convention: right is positive $x$ , up the page positive $y$ , into the page negative $z$ .
Find the magnetic force direction first. Fingers right along the velocity, curl them into the page toward the field, thumb up the page. The charge is positive, so no reversal: the magnetic force is up the page.
The electric force must therefore point down the page and match it in size. EK 12.2.B.3 is what licenses handling the two separately: in a region containing both fields, a moving charged object experiences independent forces from each.
Get the field direction from the force direction. The sheet defines the field as $\vec{E} = \vec{F}_E/q$ , so for a positive charge the force lies along the field. The force has to be down the page, so the electric field points down the page.
Set the magnitudes equal. The velocity is perpendicular to the magnetic field, so $\theta = 90^\circ$ and the magnetic force is $qvB$ ; the electric force is $qE$ . Balance gives $qE = qvB$ , and the charge cancels from both sides, leaving $v = E/B$ . The balance speed depends on neither the size of the charge nor its sign: make the charge negative and both forces reverse together, so they still cancel, at the same speed.
Substitute: $v = (3.0 \times 10^4)/(0.060) = 5.0 \times 10^5 \ \mathrm{m/s}$ .
Check the units, the quickest way to catch a flipped fraction. A tesla is a newton per ampere-metre, so $\frac{\mathrm{N}/\mathrm{C}}{\mathrm{T}} = \frac{\mathrm{A} \cdot \mathrm{m}}{\mathrm{C}}$ , and an ampere is a coulomb per second, giving metres per second.
Check with a number, taking a proton as the test charge. Electric force $qE = (1.60 \times 10^{-19})(3.0 \times 10^4) = 4.8 \times 10^{-15} \ \mathrm{N}$ ; magnetic force $qvB = (1.60 \times 10^{-19})(5.0 \times 10^5)(0.060) = 4.8 \times 10^{-15} \ \mathrm{N}$ . They match.
Read the result back as physics. Only charges travelling at exactly $5.0 \times 10^5 \ \mathrm{m/s}$ cross undeflected: a slower one has too little magnetic force and is pushed down the page, a faster one too much and is pushed up.

The electric field must point down the page, opposite the magnetic force on the positive charge, and the forces balance at 500,000 metres per second. The condition reduces to $v = E/B$ once the charge cancels, so the same speed works for a charge of any size and either sign: the arrangement selects particles by speed alone.

Frequently asked questions

What is the formula for the magnetic force on a moving charge?

F sub B equals q v B sine theta, printed as the first entry in the Magnetism group of the AP Physics 2 equation sheet. The q is the magnitude of the charge, v the speed, B the magnitude of the magnetic field, and theta the angle between the velocity vector and the magnetic field vector, not between the velocity and anything else. Because the equation contains v, a charge at rest feels no magnetic force. The CED limits quantitative treatment to 0, 90 and 180 degrees, so in practice the force is either zero, when the velocity lies along or against the field, or its maximum q v B, when the velocity is perpendicular to it.

How do you use the right-hand rule for magnetic force?

Point the fingers of your right hand along the velocity of the charge, curl them toward the magnetic field, and your thumb points along the force on a positive charge. For a negative charge, an electron included, the force is in the exact opposite direction to your thumb. Check the answer against EK 12.2.B.2.ii, which says the force is perpendicular to both the field and the velocity: if your answer lies in the plane containing those two vectors, it is wrong. On the free-response section, naming the rule is not enough. Write down the two directions you used and what the sign of the charge did.

Why does a magnetic force do no work on a charged particle?

Because it is always perpendicular to the particle's velocity, which is what EK 12.2.B.2.ii states, and work is force times displacement times the cosine of the angle between them. The displacement is along the velocity, so that angle is 90 degrees, its cosine is zero, and the work is zero. No work means no change in kinetic energy, and at constant mass no change in speed. A magnetic field can turn a charged particle through any angle and still cannot make it go faster or slower. An electric force acts along the field whether or not the charge moves, so it generally does work.

Why does a charged particle move in a circle in a magnetic field?

When the velocity is perpendicular to a uniform magnetic field, the force has constant magnitude q v B, because the speed cannot change, and it stays perpendicular to the velocity as the particle turns. A force of constant size always perpendicular to the velocity is exactly the condition for uniform circular motion. Setting q v B equal to m v squared over r gives a radius of m v divided by q B, and a period of 2 pi m divided by q B, which contains no speed at all. Neither is printed on the AP Physics 2 equation sheet; both are derived, which is why the CED lists skill 2.A here.

Does the right-hand rule change for an electron or a negative charge?

The rule itself does not change, but the answer flips. The right-hand rule as normally stated gives the direction of the force on a positive charge, so for a negative charge the force points in exactly the opposite direction. The reliable habit is to run the rule as written, take the positive-charge answer, then reverse it as a separate written step, rather than switching to your left hand and hoping you remember which hand you are on. On a free-response question that reversal is worth writing out, because the mark is on the reasoning.

What angles can you use for magnetic force calculations in AP Physics 2?

Only 0, 90 and 180 degrees. The boundary statement for Topic 12.2 says 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, and that qualitative analysis of other angles is permitted. A question set at 35 degrees is legitimate, but it will ask you to compare, rank or explain rather than produce a number. At 0 and 180 degrees the force is zero; at 90 degrees it takes its largest value, q v B.

What is the Hall effect in AP Physics 2?

EK 12.2.B.4 defines it as 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 mechanism is the topic's own force law: charges moving along the conductor are moving charges in a magnetic field, so they are pushed sideways, collect on one face, and that separation of charge is a potential difference across the conductor. Which face collects them depends on the sign of the moving charges, because the right-hand rule reverses for negative ones. No Hall voltage equation is printed on the sheet.