Charged Particle Motion in B-Field - Physics 2

1. 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?
   Answer: 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. 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.
2. Which of the following statements about the subsequent motion of the two particles is correct?
   Answer: 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. 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.
3. 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?
   Answer: 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. 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.
4. Which of the following expressions correctly gives the mass $M$ of the unknown ion?
   Answer: $M = m_p - \frac{eB\Delta x}{2v}$, because the proton's semicircular diameter exceeds the unknown ion's diameter by $\Delta x$, and since diameter $= 2mv/eB$, the mass difference is $eB\Delta x/2v$. 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.
5. 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?
   Answer: 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. 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.
6. 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 $R$?
   Answer: 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. 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.
7. 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?
   Answer: 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. 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.
8. 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?
   Answer: 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 $v$. 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 $B$ or $m$ changes.
9. What is the ratio $R_Y / R_X$?
   Answer: $\frac{R_Y}{R_X} = \frac{\sqrt{3}}{2}$, 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. 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.
10. What is the radius of the alpha particle's circular path in the same magnetic field, expressed in terms of $R$?
    Answer: $2R$, 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. 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.

Start this quiz