Physics 2 Quiz: Charged Particle Motion In B Field

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Charged Particle Motion In B FieldQuestion 1 of 10

Two charged particles, Particle 1 (charge $+q$ , mass $m$ ) and Particle 2 (charge $-q$ , mass $m$ ), are simultaneously injected into a region of uniform magnetic field $\vec{B} = B\hat{z}$ with identical velocities $\vec{v} = v\hat{x}$ .

Which of the following statements about the subsequent motion of the two particles is correct?

Both particles move in circles of the same radius, but the centers of their circular orbits are on opposite sides of their initial position along the

y

-axis, so the particles curve away from each other immediately.

Both particles move in circles of the same radius, and their circular orbits share the same center point, causing them to trace identical paths and reunite periodically.

Particle 1 moves in a circle while Particle 2 moves in a helix, because the negative charge of Particle 2 causes a component of force along

\hat{z}

that gives it motion out of the

xy

-plane.

The two particles move in circles of different radii because the magnetic force magnitude depends on the sign of the charge, with the negative charge experiencing a smaller net force and thus a larger orbit.

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Physics 2 Quiz

Physics 2 Quiz: Charged Particle Motion In B Field

Practice Charged Particle Motion In B Field in Physics 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Charged Particle Motion In B Field, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Which of the following statements about the subsequent motion of the two particles is correct?

Both particles move in circles of the same radius, but the centers of their circular orbits are on opposite sides of their initial position along the $y$ -axis, so the particles curve away from each other immediately. (correct answer)
Both particles move in circles of the same radius, and their circular orbits share the same center point, causing them to trace identical paths and reunite periodically.
Particle 1 moves in a circle while Particle 2 moves in a helix, because the negative charge of Particle 2 causes a component of force along $\hat{z}$ that gives it motion out of the $xy$ -plane.
The two particles move in circles of different radii because the magnetic force magnitude depends on the sign of the charge, with the negative charge experiencing a smaller net force and thus a larger orbit.

Explanation: When a charged particle moves through a magnetic field, the force it experiences is

\vec{F} = q\vec{v} \times \vec{B}

. The key insight here is that the sign of the charge determines the direction of that force, while the magnitude depends only on

|q|

v

, and

B

. For Particle 1 (

+q

), with

\vec{v} = v\hat{x}

and

\vec{B} = B\hat{z}

\vec{F} = qv\hat{x} \times B\hat{z} = qvB(\hat{x} \times \hat{z}) = -qvB\hat{y}

, so the force initially points in the

-y

direction, curving it downward. For Particle 2 (

-q

), the force is

+qvB\hat{y}

, curving it upward. Both forces have the same magnitude

|q|vB

, so both particles trace circles of equal radius

r = \frac{mv}{|q|B}

, but their centers sit on opposite sides of the initial position along

\hat{y}

. This confirms A is correct — same radius, opposite curvature, particles immediately diverge. Choice B is wrong because identical centers would require both particles to curve the same direction, which contradicts the opposite-sign forces. Choice C confuses the cross product:

\hat{x} \times \hat{z}

has no

\hat{z}

component, so neither particle ever gains a force along

\hat{z}

— both stay in the

xy

-plane. Choice D incorrectly claims the magnitude of magnetic force depends on the sign of charge; it depends only on

|q|

, so both radii are identical. Your study tip: always separate sign (determines direction) from magnitude (determines radius) when analyzing magnetic forces on opposite charges. These are two independent effects that the exam loves to conflate.

Question 2

A singly ionized atom of unknown mass $M$ enters a mass spectrometer with speed $v$ perpendicular to a uniform magnetic field $B$ . It travels a semicircle and strikes a detector. A proton (mass $m_p$ , charge $e$ ) entering with the same speed strikes the detector at a point displaced $\Delta x$ farther from the entry point than the unknown ion does.

Which of the following expressions correctly gives the mass $M$ of the unknown ion?

$M = m_p - \frac{2eB\Delta x}{v}$ , because the proton's path has a larger radius, and accounting for both the outward and return legs of the semicircle introduces an additional factor of 2 into the displacement relationship.
$M = m_p + \frac{eB\Delta x}{2v}$ , because the unknown ion strikes closer to the entry point, meaning it has the smaller radius and therefore the smaller mass, so $M$ is obtained by subtracting the mass correction from $m_p$ — but since the ion is heavier, the correction must be added.
$M = m_p - \frac{eB\Delta x}{v}$ , because the displacement $\Delta x$ equals the difference in radii (not diameters) of the two semicircular paths, and substituting $r = Mv/eB$ directly gives a mass correction of $eB\Delta x/v$ .
$M = m_p - \frac{eB\Delta x}{2v}$ , because the proton's semicircular diameter exceeds the unknown ion's diameter by $\Delta x$ , and since diameter $= 2mv/eB$ , the mass difference is $eB\Delta x/2v$ . (correct answer)

Explanation: When a charged particle moves perpendicular to a magnetic field, the magnetic force provides centripetal acceleration:

qvB = \frac{mv^2}{r}

, giving radius

r = \frac{mv}{qB}

. In a mass spectrometer, the particle travels a semicircle, so it strikes the detector at a distance equal to the diameter (

2r

) from the entry point. This is the key geometric fact to anchor your thinking. For the proton:

2r_p = \frac{2m_p v}{eB}

. For the unknown ion:

2r_M = \frac{2Mv}{eB}

. The proton lands

\Delta x

farther, so

2r_p - 2r_M = \Delta x

. Substituting:

\frac{2m_p v}{eB} - \frac{2Mv}{eB} = \Delta x

, which gives

M = m_p - \frac{eB\Delta x}{2v}

. That's answer D, and it's correct. A is wrong because it introduces an extra factor of 2 that doesn't exist — the displacement is already the full diameter difference, so no additional doubling is needed. B is self-contradictory in its reasoning: it correctly identifies that the unknown ion has a smaller radius (smaller mass), then incorrectly adds the correction rather than subtracting it, and also uses the wrong coefficient. C uses

\Delta x = r_p - r_M

(difference in radii) instead of the difference in diameters, missing the factor of 2 from the semicircle geometry — a very common trap. Your study tip: always sketch the semicircle and label the landing point as

2r

from the entry slit. The diameter, not the radius, is the measurable displacement — confusing the two is the most frequent error on mass spectrometer problems.

Question 3

Two particles, X and Y, travel in circular orbits in the same uniform magnetic field $B$ . Particle X has charge $q$ and mass $m$ , and its orbit has radius $R_X$ . Particle Y has charge $2q$ and mass $3m$ . Both particles have the same kinetic energy $KE$ .

What is the ratio $R_Y / R_X$ ?

$\frac{R_Y}{R_X} = \sqrt{3}$ , derived by expressing the radius in terms of kinetic energy, where the ratio depends only on the mass ratio since both particles have the same kinetic energy, charge differences cancel through the momentum expression.
$\frac{R_Y}{R_X} = \frac{3}{2}$ , derived by noting that radius is proportional to mass and inversely proportional to charge, giving a factor of $3m/m = 3$ in the numerator and $2q/q = 2$ in the denominator, for a ratio of $3/2$ .
$\frac{R_Y}{R_X} = \frac{\sqrt{3}}{2}$ , derived by expressing the radius in terms of kinetic energy as $r = \frac{\sqrt{2mKE}}{qB}$ , then forming the ratio using the correct mass and charge for each particle. (correct answer)
$\frac{R_Y}{R_X} = \frac{3}{\sqrt{2}}$ , derived by applying the formula $r = mv/qB$ with speeds found from $KE = \frac{1}{2}mv^2$ , giving $v_Y = \sqrt{2KE/3m}$ and computing the ratio of momenta to charges explicitly.

Explanation: When a charged particle moves in a magnetic field, the magnetic force provides centripetal force:

qvB = \frac{mv^2}{r}

, giving

r = \frac{mv}{qB}

. The key insight here is that both particles share the same kinetic energy, so you should express momentum in terms of KE rather than velocity. Since

KE = \frac{p^2}{2m}

, momentum is

p = \sqrt{2mKE}

, and the radius becomes

r = \frac{\sqrt{2mKE}}{qB}

. For particle X:

R_X = \frac{\sqrt{2mKE}}{qB}

. For particle Y (mass

3m

, charge

2q

R_Y = \frac{\sqrt{2(3m)KE}}{2qB} = \frac{\sqrt{6mKE}}{2qB}

. Taking the ratio cancels

B

and

\sqrt{2mKE}

, leaving

\frac{R_Y}{R_X} = \frac{\sqrt{3}}{2}

, confirming answer C. Answer A claims the charge differences "cancel through the momentum expression," which is wrong — the charge appears explicitly in the denominator of the radius formula and does not cancel when comparing two particles with different charges. Answer B uses

r \propto m/q

, which would be correct if both particles had the same speed, but they don't — equal kinetic energy with different masses means different speeds. Answer D sets up the momentum approach correctly but makes an algebra error:

p_Y = m_Y v_Y = 3m \cdot \sqrt{\frac{2KE}{3m}} = \sqrt{6mKE}

, and dividing by

2q

gives

\frac{\sqrt{6mKE}}{2qB}

, which is

\frac{\sqrt{3}}{2}R_X

, not

\frac{3}{\sqrt{2}}R_X

. Study tip: Whenever a magnetic radius problem specifies equal kinetic energy rather than equal speed, immediately rewrite

r = \frac{\sqrt{2mKE}}{qB}

— this single substitution avoids the most common errors on this problem type.

Question 4

An electron travels in a circular orbit of radius $R$ in a uniform magnetic field $B$ directed out of the page. The electron's orbit lies entirely in the plane of the page. A student then reduces the magnetic field strength to $B/2$ while simultaneously reducing the electron's speed to $v/2$ . How does the new orbital radius compare to $R$ ?

The new radius equals $R$ , because both the speed and the magnetic field are halved, and since $r = mv/qB$ , these two reductions exactly cancel, leaving the radius unchanged. (correct answer)
The new radius equals $R/2$ , because the speed is halved, which reduces the numerator of $mv/qB$ by a factor of 2, while the field is also halved, reducing the denominator by a factor of 2, so only the numerator effect counts and the radius halves.
The new radius equals $2R$ , because halving the magnetic field halves the force on the electron, which doubles the radius, and the simultaneous halving of speed further reduces the force, compounding the radius increase.
The new radius equals $R/4$ , because the centripetal acceleration scales as $v^2/r$ , and since $v$ is halved, the centripetal acceleration decreases by a factor of 4, which must be matched by a fourfold decrease in radius at the new field strength.

Explanation: When a charged particle moves in a magnetic field, the magnetic force provides the centripetal force:

qvB = \frac{mv^2}{r}

. Solving for the orbital radius gives you the key formula:

r = \frac{mv}{qB}

. Whenever a question changes multiple quantities simultaneously, your first instinct should be to plug the new values directly into this formula rather than reasoning qualitatively about "competing effects." Here, the new speed is

v/2

and the new field is

B/2

. Substituting into the formula:

r_{new} = \frac{m(v/2)}{q(B/2)} = \frac{mv/2}{qB/2} = \frac{mv}{qB} = R

. The factors of 2 in the numerator and denominator cancel exactly, leaving the radius unchanged. Answer A is correct. Answer B claims "only the numerator effect counts," which is simply wrong algebra — when you halve both numerator and denominator of a fraction, the ratio is preserved, not halved. Answer C tries to reason about force reduction leading to radius doubling, but this qualitative chain-of-reasoning breaks down because it double-counts the effect of halving the speed (speed appears in both the magnetic force and the centripetal requirement). Answer D incorrectly imports the centripetal acceleration formula

v^2/r

and applies it as if radius must scale with acceleration alone — this ignores that the magnetic force also changes when

v

and

B

change. Study tip: When multiple quantities change at once, skip qualitative reasoning and substitute directly into

r = mv/qB

. Ratio problems on this exam are almost always easier to solve algebraically than conceptually.

Question 5

A proton undergoes uniform circular motion in a plane perpendicular to a uniform magnetic field. A student claims: 'Because the magnetic force does no work on the proton, the proton's speed must remain constant, and therefore the radius of its circular path must also remain constant regardless of any additional uniform electric field applied parallel to the magnetic field.' Which of the following best evaluates this claim?

The claim is entirely correct: the magnetic force does no work, and an electric field parallel to $\vec{B}$ cannot affect the component of velocity perpendicular to $\vec{B}$ , so the circular radius is unaffected and the orbit remains circular.
The claim is partially correct but incomplete: the magnetic force does no work, keeping perpendicular speed constant and radius unchanged, but the electric field parallel to $\vec{B}$ accelerates the proton along the field direction, converting circular motion into helical motion with a constant helical radius. (correct answer)
The claim is incorrect because any applied electric field, even if parallel to $\vec{B}$ , exerts a force on the proton that increases the total speed, which in turn increases the radius of the circular component of the motion.
The claim is incorrect because the magnetic force does perform work on the proton when the velocity has a component along the field direction, thereby changing the kinetic energy and altering the orbital radius.

Explanation: When a charged particle moves in a magnetic field, the key principle is that the magnetic force is always perpendicular to the velocity, so it does no work and cannot change the particle's speed or kinetic energy. This means the component of velocity perpendicular to

\vec{B}

stays constant, and since the circular radius is given by

r = \frac{mv_\perp}{qB}

, that radius stays constant too. So far, the student's reasoning is sound. The problem arises when an electric field is introduced parallel to

\vec{B}

. Because

\vec{E} \parallel \vec{B}

, the electric force acts entirely along the field direction — it has zero component in the plane of circular motion. This means it cannot change

v_\perp

, leaving the circular radius untouched. However, it does accelerate the proton along the field axis, adding a growing

v_\parallel

component. The result is helical motion: circular rotation in the perpendicular plane combined with linear acceleration along the field direction. The student's conclusion about the radius is correct, but the claim that "the orbit remains circular" is wrong — it becomes a helix with an ever-increasing pitch. This makes B the best answer. A is wrong because it accepts the claim entirely. While the radius argument is valid, concluding the orbit stays circular ignores the axial acceleration from the electric field. C is wrong because the electric field parallel to

\vec{B}

does not increase

v_\perp

; total speed increases, but only through the axial component, leaving the circular radius unchanged. D is wrong because the magnetic force never does work — even with a helical path,

\vec{F}_{mag} \perp \vec{v}

always holds. Study tip: Always decompose the problem into components parallel and perpendicular to

\vec{B}

— they evolve independently, and each force only affects the component it's aligned with.

Question 6

A positive charge moves in a circular orbit of radius $R$ and period $T$ in a uniform magnetic field. An experimenter triples the magnetic field strength and simultaneously triples the particle's speed. Which of the following correctly describes the new orbital radius and new period?

The radius remains $R$ and the period remains $T$ , because tripling both $v$ and $B$ leaves $r = mv/qB$ unchanged, and since both the circumference and the speed are determined by the same field and velocity conditions, the time per orbit is also unchanged.
The radius increases to $3R$ and the period remains $T$ , because tripling the speed triples the radius through $r = mv/qB$ , and since the period $T = 2\pi m/qB$ decreases by a factor of 3 due to the field increase but the faster speed means each orbit takes the same time, these effects cancel leaving $T$ unchanged.
The radius remains $R$ and the period decreases to $T/3$ , because $r = mv/qB$ is unchanged when both $v$ and $B$ are tripled by the same factor, but $T = 2\pi m/qB$ decreases by a factor of 3 since it depends on $B$ but not on $v$ . (correct answer)
The radius increases to $9R$ and the period decreases to $T/3$ , because the radius scales with $v$ and with $1/B$ independently, so tripling $v$ multiplies the radius by 3 and tripling $B$ multiplies it by another factor of 3, giving $9R$ , while the period scales as $1/B$ and decreases by 3.

Explanation: When a charged particle moves in a magnetic field, two formulas govern its circular motion: the orbital radius

r = \frac{mv}{qB}

and the cyclotron period

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

. Notice something crucial — the period has no

v

in it at all. This is the key insight the question is testing. When both

v

and

B

are tripled simultaneously, the radius becomes

r' = \frac{m(3v)}{q(3B)} = \frac{mv}{qB} = R

. The factors of 3 cancel exactly, leaving the radius unchanged. For the period,

T' = \frac{2\pi m}{q(3B)} = \frac{T}{3}

. Because the period depends only on

B

(not

v

), tripling

B

alone drives the period down to

T/3

. This confirms C is correct. A is wrong because it correctly identifies the unchanged radius but incorrectly concludes the period is unchanged. The period does shrink — it's independent of speed, so the faster particle doesn't "compensate" for the stronger field in the period formula. B gets the period conclusion right by coincidence but for muddled reasoning, and it incorrectly states the radius triples. Tripling

v

alone would triple

r

, but tripling

B

simultaneously cancels that effect. D double-counts the radius change by treating

v

and

B

as independent multipliers, but they appear together in one ratio — you can't apply them separately to get

9R

. Study tip: Memorize that

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

is speed-independent. On any exam question involving changing

v

, the period is unaffected unless

B

m

changes.

Question 7

A proton moving with speed $v_0$ enters a region of uniform magnetic field $\vec{B}$ directed into the page. The proton travels in a semicircle of radius $R$ and exits the field region. An alpha particle (charge $+2e$ , mass $4m_p$ ) then enters the same field region with the same speed $v_0$ .

What is the radius of the alpha particle's circular path in the same magnetic field, expressed in terms of $R$ ?

$2R$ , because the alpha particle has twice the charge of the proton, which doubles the magnetic force, but four times the mass, giving a net factor of 2 increase in radius. (correct answer)
$R$ , because although the alpha particle has greater mass and greater charge than the proton, these two effects exactly cancel each other when the speed is held constant.
$\frac{R}{2}$ , because the alpha particle has twice the charge of the proton, which doubles the magnetic force and thereby halves the radius of curvature at the same speed.
$4R$ , because the alpha particle is four times more massive than the proton, and radius is directly proportional to mass, so the radius scales by a factor of 4 regardless of charge differences.

Explanation: Whenever you see a charged particle moving through a magnetic field, your anchor equation should be the condition for circular motion: the magnetic force provides the centripetal force. Setting

qvB = \frac{mv^2}{r}

and solving for radius gives

r = \frac{mv}{qB}

. This tells you that radius depends on both mass and charge — so you must account for both when comparing particles. For the proton:

R = \frac{m_p v_0}{eB}

. For the alpha particle, substitute its mass (

4m_p

) and charge (

2e

r_\alpha = \frac{4m_p \cdot v_0}{2e \cdot B} = \frac{2m_p v_0}{eB} = 2R

. The mass increases the radius by a factor of 4, the charge reduces it by a factor of 2, leaving a net factor of 2 — confirming answer A is correct. Answer B is wrong because mass and charge do not cancel. Mass scales radius up by 4, charge scales it down by 2; those are different factors, not equal ones. Answer C commits the most common error here — it focuses only on the doubled charge and concludes the radius halves, completely ignoring how the fourfold increase in mass simultaneously increases the radius. Answer D goes to the opposite extreme, correctly identifying the mass factor of 4 but ignoring the charge doubling that partially offsets it. A useful memory strategy: write out

r = \frac{mv}{qB}

and explicitly plug in the new particle's values for every variable that changes. Students lose points by mentally tracking only one change at a time.

Question 8

A particle of charge $q$ and mass $m$ moves in a uniform magnetic field $\vec{B}$ . At time $t = 0$ , its velocity is $\vec{v}_0 = v_x \hat{x} + v_z \hat{z}$ where $\vec{B} = B\hat{z}$ . Which of the following statements about the subsequent motion is correct?

The particle undergoes helical motion with radius $r = m\sqrt{v_x^2 + v_z^2}/qB$ , because the radius of circular motion is determined by the total speed, and both velocity components contribute to the centripetal requirement.
The particle undergoes helical motion with a time-varying radius, because the $v_z$ component produces a magnetic force along $\hat{x}$ or $\hat{y}$ that continuously transfers energy between the parallel and perpendicular components of velocity.
The particle moves in a circle in the $xz$ -plane because both $v_x$ and $v_z$ are perpendicular to $\vec{B}$ in different ways, and the magnetic force combines these to produce circular motion in the plane containing the initial velocity.
The particle undergoes helical motion: it circles in the $xy$ -plane with radius $r = mv_x/qB$ while drifting along $\hat{z}$ at constant speed $v_z$ , and the total speed remains constant because the magnetic force does no work. (correct answer)

Explanation: When a charged particle moves through a magnetic field, the key is to decompose the velocity into components parallel and perpendicular to

\vec{B}

. The magnetic force is

\vec{F} = q\vec{v} \times \vec{B}

, so only the component of velocity perpendicular to

\vec{B}

experiences a force — the parallel component is completely unaffected. Here,

\vec{B} = B\hat{z}

, so

v_z

is parallel to

\vec{B}

and contributes zero magnetic force:

v_z\hat{z} \times B\hat{z} = 0

. Meanwhile,

v_x

is perpendicular to

\vec{B}

, generating a force that curves the particle into circular motion in the

xy

-plane. The radius of that circle is

r = mv_x/qB

, using only the perpendicular speed. Since the magnetic force never does work (it's always perpendicular to velocity), the total speed — and therefore both

v_x

and

v_z

individually — remain constant. The result is helical motion: circular orbiting in

xy

combined with steady drift along

\hat{z}

. This makes D correct. A is wrong because the radius depends only on the perpendicular component

v_x

, not the total speed. Including

v_z

in the radius formula is the core misconception here. B is wrong because the magnetic force never transfers energy between components —

v_z

stays constant throughout; there is no energy exchange. C is wrong because

v_z \parallel \vec{B}

produces no force, so there's no circular motion in the

xz

-plane. Remember this rule: decompose velocity relative to $\vec{B}$ first. The perpendicular part circles; the parallel part drifts. This decomposition is the foundation of every charged-particle-in-magnetic-field problem.

Question 9

A particle of charge $q > 0$ and mass $m$ moves in a circle of radius $R$ in a uniform magnetic field of magnitude $B$ . The particle's kinetic energy is then doubled (by briefly passing through an accelerating electric field outside the magnetic field region) and it re-enters the same magnetic field. Which of the following correctly describes what happens to the radius of curvature and the period of revolution?

The radius increases by a factor of $\sqrt{2}$ and the period decreases by a factor of $\sqrt{2}$ , because both depend on speed but in different functional forms through the relations $r = mv/qB$ and $T = 2\pi m/qB$ .
The radius increases by a factor of $\sqrt{2}$ and the period remains unchanged, because the period $T = 2\pi m/qB$ depends only on mass, charge, and field strength — none of which change — while the radius scales with speed. (correct answer)
Both the radius and the period increase by a factor of $\sqrt{2}$ , because both are proportional to the particle's speed, which increases by $\sqrt{2}$ when kinetic energy is doubled.
The radius increases by a factor of 2 and the period remains unchanged, because doubling the kinetic energy doubles the speed, which doubles the radius, while the period stays constant since it is mass- and field-dependent only.

Explanation: When a charged particle moves in a magnetic field, two key formulas govern its circular motion: the radius of curvature

r = \frac{mv}{qB}

and the period

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

. Notice something important — the radius depends on speed, but the period does not. This distinction is the heart of the question. When kinetic energy doubles, you need the new speed. Since

KE = \frac{1}{2}mv^2

, doubling KE gives

2 \cdot \frac{1}{2}mv^2 = \frac{1}{2}m(v')^2

, which means

v' = \sqrt{2}\, v

. The new radius becomes

r' = \frac{m(\sqrt{2}\,v)}{qB} = \sqrt{2}\, r

— the radius increases by

\sqrt{2}

. Meanwhile, since

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

contains no velocity term, the period is completely unaffected by the speed change. Mass, charge, and field strength are all unchanged, so

T

stays the same. This confirms B. A is wrong because it correctly identifies the radius change but incorrectly claims the period decreases by

\sqrt{2}

. The period formula has no

v

dependence — that's a critical misconception to avoid. C is wrong because it assumes both quantities scale with speed. The period does not scale with speed at all. D is wrong about the speed change itself. Doubling kinetic energy multiplies speed by

\sqrt{2}

, not 2, since KE depends on

v^2

. The period conclusion is correct, but the radius factor is wrong. A useful habit: always write out both formulas and explicitly check which variables each one contains before reasoning about what changes.

Question 10

A cyclotron accelerates protons using a magnetic field of magnitude $B$ and a pair of D-shaped electrodes (dees) with an alternating voltage. Each time a proton crosses the gap between the dees, it gains kinetic energy $\Delta KE$ . The proton starts essentially at rest at the center.

After the proton has crossed the gap $n$ times (gaining total kinetic energy $n\Delta KE$ ), which of the following correctly describes the radius of the proton's orbit and the time it takes to complete one full semicircle?

The orbital radius is $r_n = \frac{n\Delta KE}{eBv_n}$ and the semicircle transit time is $t = \frac{\pi m_p}{eB}$ , independent of the number of gap crossings.
The orbital radius is $r_n = \frac{\sqrt{2n m_p \Delta KE}}{eB}$ and the semicircle transit time is $t = \frac{\pi m_p}{eB}$ , which is independent of $n$ and confirms why the dee voltage frequency need not change as the proton spirals outward. (correct answer)
The orbital radius is $r_n = \frac{n\Delta KE}{eBv_n}$ and the semicircle transit time increases with each gap crossing because the proton travels a longer arc at higher speed, and these two effects do not cancel.
The orbital radius is $r_n = \frac{\sqrt{2n m_p \Delta KE}}{eB}$ and the semicircle transit time increases proportionally to $\sqrt{n}$ , because although the proton moves faster, the larger circumference of each successive semicircle more than compensates for the speed increase.

Explanation: Cyclotron problems test two key relationships: how a charged particle's orbital radius depends on its speed in a magnetic field, and why the cyclotron frequency is independent of that speed. Start with the orbital radius. A proton moving at speed

v_n

in magnetic field

B

follows a circular path where the magnetic force provides centripetal acceleration:

eBv_n = \frac{m_p v_n^2}{r_n}

, giving

r_n = \frac{m_p v_n}{eB}

. After

n

gap crossings, the proton's total kinetic energy is

\frac{1}{2}m_p v_n^2 = n\Delta KE

, so

v_n = \sqrt{\frac{2n\Delta KE}{m_p}}

. Substituting into the radius formula gives

r_n = \frac{\sqrt{2nm_p\Delta KE}}{eB}

. Now for the semicircle transit time:

t = \frac{\pi r_n}{v_n} = \frac{\pi m_p v_n}{eBv_n} = \frac{\pi m_p}{eB}

. The

v_n

cancels entirely — the time is constant regardless of

n

. This is exactly what answer B states, and it's the brilliant insight behind the cyclotron: because faster protons curve in larger arcs but travel those arcs proportionally faster, each semicircle takes the same time, so the alternating voltage frequency never needs adjustment. Answer A gets the transit time right but expresses the radius as

\frac{n\Delta KE}{eBv_n}

, which is dimensionally inconsistent and doesn't simplify correctly — it conflates energy with momentum. Answer C compounds that radius error with the false claim that transit time increases, misunderstanding the speed-versus-arc-length cancellation. Answer D has the correct radius formula but incorrectly concludes the transit time grows as

\sqrt{n}

— this would only be true if speed didn't increase, but the cancellation shows both effects are perfectly balanced. Remember: whenever you see a cyclotron question, immediately write

t_{semi} = \frac{\pi m_p}{eB}

— the fact that this is constant is the entire operating principle of the device.