Physics 2 Quiz: Lorentz Force

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Lorentz ForceQuestion 1 of 6

A singly ionized atom (charge $+e$ ) moves with speed $v$ at an angle of $30°$ to a uniform magnetic field of magnitude $B$ .

The atom undergoes helical motion. Which of the following correctly gives both the radius of the helix and the pitch (axial distance traveled per revolution)?

Radius

r = \frac{mv\sin30°}{eB} = \frac{mv}{2eB}

and pitch

p = \frac{2\pi m}{eB}\cdot v = \frac{2\pi mv}{eB}

, because the period depends on the full speed

v

rather than the component perpendicular to

\vec{B}

, so the axial advance uses the full speed times the full period

Radius

r = \frac{mv}{eB}

and pitch

p = \frac{2\pi m}{eB}\cdot v\cos30°= \frac{\pi mv\sqrt{3}}{eB}

, because the full speed determines the radius of curvature through the Lorentz force, while only the parallel component contributes to axial advance per period

Radius

r = \frac{mv\sin30°}{eB} = \frac{mv}{2eB}

and pitch

p = \frac{2\pi m}{eB}\cdot v\cos30° = \frac{\pi mv\sqrt{3}}{eB}

, because only the perpendicular velocity component drives circular motion while the parallel component advances the helix axially

Radius

r = \frac{mv\cos30°}{eB} = \frac{mv\sqrt{3}}{2eB}

and pitch

p = \frac{2\pi m}{eB}\cdot v\sin30° = \frac{\pi mv}{eB}

, because the component of velocity along the field axis drives the circular orbit while the perpendicular component advances the particle axially along the field

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Physics 2 Quiz

Physics 2 Quiz: Lorentz Force

Practice Lorentz Force 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 Lorentz Force, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

How to use this quiz

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

A singly ionized atom (charge $+e$ ) moves with speed $v$ at an angle of $30°$ to a uniform magnetic field of magnitude $B$ .

The atom undergoes helical motion. Which of the following correctly gives both the radius of the helix and the pitch (axial distance traveled per revolution)?

Radius $r = \frac{mv\sin30°}{eB} = \frac{mv}{2eB}$ and pitch $p = \frac{2\pi m}{eB}\cdot v = \frac{2\pi mv}{eB}$ , because the period depends on the full speed $v$ rather than the component perpendicular to $\vec{B}$ , so the axial advance uses the full speed times the full period
Radius $r = \frac{mv}{eB}$ and pitch $p = \frac{2\pi m}{eB}\cdot v\cos30°= \frac{\pi mv\sqrt{3}}{eB}$ , because the full speed determines the radius of curvature through the Lorentz force, while only the parallel component contributes to axial advance per period
Radius $r = \frac{mv\sin30°}{eB} = \frac{mv}{2eB}$ and pitch $p = \frac{2\pi m}{eB}\cdot v\cos30° = \frac{\pi mv\sqrt{3}}{eB}$ , because only the perpendicular velocity component drives circular motion while the parallel component advances the helix axially (correct answer)
Radius $r = \frac{mv\cos30°}{eB} = \frac{mv\sqrt{3}}{2eB}$ and pitch $p = \frac{2\pi m}{eB}\cdot v\sin30° = \frac{\pi mv}{eB}$ , because the component of velocity along the field axis drives the circular orbit while the perpendicular component advances the particle axially along the field

Explanation: When a charged particle moves at an angle to a magnetic field, its velocity splits into two independent components:

v_\perp = v\sin\theta

perpendicular to

\vec{B}

, and

v_\parallel = v\cos\theta

parallel to

\vec{B}

. The magnetic force

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

only acts on

v_\perp

, so only that component drives circular motion. The parallel component is completely unaffected by the field and causes the particle to drift along the field axis — combining both motions produces the characteristic helix. The radius comes entirely from the perpendicular component:

r = \frac{mv_\perp}{eB} = \frac{mv\sin30°}{eB} = \frac{mv}{2eB}

. The period of one circular revolution is

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

, which depends only on mass and field strength — not on speed. During that one period, the particle advances axially by

p = v_\parallel \cdot T = v\cos30° \cdot \frac{2\pi m}{eB} = \frac{\pi mv\sqrt{3}}{eB}

. This confirms C is correct. A is wrong because it incorrectly uses the full speed

v

for the pitch calculation instead of

v_\parallel

. The axial advance per revolution depends only on the parallel component. B uses the full speed for the radius, which is the core misconception — only

v_\perp

produces the magnetic force that curves the path. D swaps the roles of the two components entirely, treating the parallel component as responsible for circular motion and the perpendicular component for axial advance — the exact opposite of reality. A reliable memory anchor: perpendicular → circle, parallel → pitch. Never let the two components cross roles.

Question 2

A uniform magnetic field $\vec{B} = B_0\hat{z}$ exists for $x > 0$ and is zero for $x < 0$ . A particle of charge $+q$ and mass $m$ travels in the $-x$ direction, enters the field region at the origin with speed $v$ , curves inside the field, and eventually exits back into the $x < 0$ region. Assuming the particle completes exactly a semicircle inside the field, what is the exit point's $y$ -coordinate, and in what direction does the particle travel after exiting?

Exit $y$ -coordinate: $y = +2r = \frac{2mv}{qB_0}$ , and the particle exits traveling in the $+x$ direction, because the particle enters moving in $-\hat{x}$ , the force $q(-v\hat{x})\times(B_0\hat{z}) = -qvB_0(\hat{x}\times\hat{z}) = qvB_0\hat{y}$ curves it toward $+y$ , and after a semicircle the particle is displaced $2r$ in $+y$ and its velocity has reversed to $+\hat{x}$ (correct answer)
Exit $y$ -coordinate: $y = +2r = \frac{2mv}{qB_0}$ , and the particle exits traveling in the $-x$ direction, because the semicircle brings the particle back to $x = 0$ with the same leftward velocity it had upon entering, since both entering and exiting the boundary occur at $x = 0$
Exit $y$ -coordinate: $y = -2r = -\frac{2mv}{qB_0}$ , and the particle exits traveling in the $+x$ direction, because the force $q(-v\hat{x})\times(B_0\hat{z})$ is in $-\hat{y}$ , curving the particle downward, and after a semicircle the velocity reverses to $+\hat{x}$
Exit $y$ -coordinate: $y = 0$ , and the particle exits traveling in the $+x$ direction, because the semicircle is symmetric about the $x$ -axis: the particle curves inward (in $+x$ ), reaches maximum penetration at $x = r$ , then returns to the origin and exits with velocity $+\hat{x}$

Explanation: When a charged particle enters a magnetic field, the key is to carefully apply the Lorentz force

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

to determine the initial deflection direction, then trace the resulting circular geometry. Here, the particle enters moving in

-\hat{x}

with the field

\vec{B} = B_0\hat{z}

. Computing the initial force:

\vec{F} = q(-v\hat{x}) \times (B_0\hat{z}) = -qvB_0(\hat{x} \times \hat{z}) = -qvB_0(-\hat{y}) = qvB_0\hat{y}

. The force points in

+\hat{y}

, so the particle curves upward. It traces a semicircle of radius

r = \frac{mv}{qB_0}

entirely within

x > 0

, with its center at

(0, r)

. After a semicircle, the particle has traveled a diameter in

+y

, landing at

y = 2r = \frac{2mv}{qB_0}

on the boundary

x = 0

. At that exit point, the velocity — which always stays tangent to the circle — has rotated 180° from

-\hat{x}

+\hat{x}

. Answer A captures this completely. Answer B correctly identifies the

y

-displacement but wrongly claims the exit velocity is still

-\hat{x}

. This ignores that the magnetic force continuously rotates the velocity vector; after a semicircle, it reverses direction. Answer C gets the exit direction right (

+\hat{x}

) but miscalculates the cross product, flipping the sign and sending the particle downward to

y = -2r

. Double-check:

\hat{x} \times \hat{z} = -\hat{y}

, so the negative signs cancel and the force is

+\hat{y}

. Answer D imagines the semicircle arcing into the field along the

x

-axis and returning to the origin, but this would require the initial force to point in

+\hat{x}

, which it doesn't. Study tip: Always compute the cross product explicitly before sketching the trajectory — the direction of the first force tells you which way the circle curves, and the geometry does the rest.

Question 3

Two charged particles, particle 1 (charge $+q$ , mass $m$ ) and particle 2 (charge $+2q$ , mass $4m$ ), are both launched with the same speed $v_0$ perpendicular to a uniform magnetic field $\vec{B}$ .

What is the ratio of the cyclotron period of particle 2 to that of particle 1, and what is the ratio of their orbital radii $r_2/r_1$ ?

Period ratio $T_2/T_1 = 1$ and radius ratio $r_2/r_1 = 2$ , because both particles experience the same magnetic force magnitude and therefore complete orbits in the same time, while the heavier particle curves less sharply, doubling the radius
Period ratio $T_2/T_1 = 2$ and radius ratio $r_2/r_1 = 1$ , because the period depends only on $m/q$ giving factor 2, but the radius depends on $mv/(qB)$ and since both particles have the same speed, the ratio $m/q$ cancels differently when the speed is held fixed
Period ratio $T_2/T_1 = 4$ and radius ratio $r_2/r_1 = 2$ , because the period scales with mass alone ( $T\propto m$ ) giving factor 4, while the radius scales as $m/q$ giving factor 2 when the speed is identical
Period ratio $T_2/T_1 = 2$ and radius ratio $r_2/r_1 = 2$ , because the period $T = 2\pi m/(qB)$ scales as $m/q$ , giving $(4m/2q)/(m/q)=2$ , and the radius $r = mv/(qB)$ also scales as $m/q$ , giving the same factor of 2 (correct answer)

Explanation: When a charged particle moves perpendicular to a magnetic field, the magnetic force provides centripetal acceleration, producing circular motion. Two key formulas govern this motion: the orbital radius

r = \frac{mv}{qB}

and the cyclotron period

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

. Notice that both expressions scale as $m/q$ — this is the central insight the question is testing. For particle 2 versus particle 1, compute the ratio

\frac{m_2/q_2}{m_1/q_1} = \frac{4m/2q}{m/q} = \frac{4m \cdot q}{2q \cdot m} = 2

. Since both

T

and

r

scale with

m/q

, and since both particles share the same speed

v_0

and field

B

, both ratios equal 2. This confirms answer D. A is wrong on both counts. The magnetic force magnitudes are not the same — force equals

qvB

, and particle 2 has twice the charge, so it experiences twice the force. The period-equals-1 claim is incorrect. B gets the period ratio right (2) but claims the radius ratio is 1. Since

r = mv/(qB)

and

m/q

doubles while

v

is held constant, the radius also doubles — not stays the same. C claims the period scales with mass alone (

T \propto m

), giving a factor of 4. This is the most dangerous distractor. The period formula clearly shows

T \propto m/q

, not

m

alone — the charge in the denominator matters. Study tip: Memorize that both

r

and

T

for circular motion in a magnetic field depend on the ratio

m/q

, not mass or charge separately. When you see different particles in a field, always compute

m/q

first.

Question 4

A relativistic proton (Lorentz factor $\gamma = 2$ , rest mass $m_p$ ) moves perpendicular to a uniform magnetic field $B$ .

A student calculates the radius of the proton's circular orbit using the non-relativistic formula $r = m_p v/(eB)$ and obtains $r_{NR}$ . What is the correct relativistic radius $r_{rel}$ in terms of $r_{NR}$ ?

$r_{rel} = \sqrt{2}\, r_{NR}$ , because the relativistic kinetic energy is $\gamma m_p c^2 - m_p c^2 = m_p c^2$ , and the momentum scales as $\sqrt{2}$ times the classical value when energy is doubled at the same rest mass
$r_{rel} = 2\, r_{NR}$ , because the correct formula uses the relativistic momentum $p = \gamma m_p v$ , and with $\gamma = 2$ the relativistic radius is exactly twice the non-relativistic radius computed at the same speed $v$ (correct answer)
$r_{rel} = r_{NR}/2$ , because the relativistic mass $m_{rel} = \gamma m_p = 2m_p$ appears in the denominator of the corrected force balance $eB v = m_{rel}v^2/r$ , making the radius smaller by a factor of $1/\gamma$
$r_{rel} = 4\, r_{NR}$ , because at $\gamma = 2$ the proton's speed is $v = \frac{\sqrt{3}}{2}c$ , and the relativistic momentum $\gamma m_p v = 2m_p \cdot \frac{\sqrt{3}}{2}c = \sqrt{3}m_pc$ exceeds the non-relativistic estimate by a factor of 4 when the student naively uses $v \approx c/2$ in $r_{NR}$

Explanation: When a charged particle moves through a magnetic field, the magnetic force provides centripetal acceleration. Relativistically, Newton's second law generalizes so that the magnetic force

evB

equals the rate of change of relativistic momentum, giving the orbit condition

evB = \gamma m_p v^2 / r

. Solving for the radius yields

r_{rel} = \gamma m_p v / (eB)

. Compare this to the student's non-relativistic formula

r_{NR} = m_p v / (eB)

, evaluated at the same speed

v

. The only difference is the factor of

\gamma = 2

in the numerator, so

r_{rel} = \gamma \cdot r_{NR} = 2\, r_{NR}

. B is correct. Choice A is wrong because it conflates momentum scaling with energy scaling. The relativistic momentum is simply

p = \gamma m_p v

, not derived from a kinetic energy argument. Kinetic energy doubling does not mean momentum scales as

\sqrt{2}

. Choice C contains a subtle algebra error. The relativistic mass does appear in the numerator of the momentum (

p = \gamma m_p v

), not the denominator of the radius formula. Writing

evB = m_{rel}v^2/r

and solving gives

r = m_{rel}v/(eB) = \gamma m_p v/(eB)

, which is larger, not smaller — so the radius increases by

\gamma

, not decreases. Choice D invents an inconsistency: the problem states the student uses speed

v

throughout, so there is no "naive substitution" of

v \approx c/2

. The factor of 4 has no physical basis here. Study tip: Whenever a relativistic mechanics problem involves momentum, always replace

m_p v

with

\gamma m_p v

— this single substitution correctly handles most relativistic dynamics questions on this exam.

Question 5

An electron moves with velocity $\vec{v} = v_0(\hat{x} + \hat{y})/\sqrt{2}$ through a region where $\vec{B} = B_0\hat{z}$ . A student claims that the component of the Lorentz force along $\hat{x} + \hat{y}$ is zero, and that the force is purely in the direction $\hat{x} - \hat{y}$ . Which of the following best evaluates this claim?

The student is incorrect on both counts. The magnetic force on an electron can have a component along $\vec{v}$ because the electron's negative charge reverses the usual right-hand rule, effectively rotating the force vector by 180°, which can produce a component parallel to $\vec{v}$
The student is correct that the force has no component along $\hat{x}+\hat{y}$ , but incorrect about the direction: the force is along $+\hat{z}$ because the cross product of a vector in the $xy$ -plane with $\hat{z}$ always produces a vector along $\hat{z}$
The student is correct on both counts. The magnetic force is always perpendicular to $\vec{v}$ , so it has no component along $\hat{x}+\hat{y}$ . Computing $\vec{F} = (-e)\frac{v_0}{\sqrt{2}}(\hat{x}+\hat{y})\times B_0\hat{z}$ gives a result proportional to $\hat{y}-\hat{x}$ , confirming the force lies entirely along $\hat{x}-\hat{y}$ (up to a sign) (correct answer)
The student is correct that the force is perpendicular to $\vec{v}$ , but incorrect about the direction: the force on the electron is along $+(\hat{x}+\hat{y})/\sqrt{2}$ , not along $\hat{x}-\hat{y}$ , because the negative charge of the electron reverses the sign and thereby rotates the force into the same direction as $\vec{v}$

Explanation: When a charged particle moves through a magnetic field, the Lorentz force law states

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

. Two key principles govern this: (1) the magnetic force is always perpendicular to the velocity, meaning it can never do work or have a component along

\vec{v}

, and (2) for an electron,

q = -e

, which flips the sign of the force compared to a positive charge. Let's work through the calculation directly. First, compute the cross product for a positive charge:

(\hat{x} + \hat{y}) \times \hat{z} = \hat{x}\times\hat{z} + \hat{y}\times\hat{z} = -\hat{y} + \hat{x} = \hat{x} - \hat{y}

Now apply the full force for the electron:

\vec{F} = (-e)\frac{v_0}{\sqrt{2}}B_0(\hat{x}-\hat{y}) \propto -(\hat{x}-\hat{y}) = \hat{y}-\hat{x}

This result is indeed perpendicular to

\hat{x}+\hat{y}

(their dot product is zero), and lies entirely along

\hat{x}-\hat{y}

up to a sign — exactly what the student claimed. Answer C is correct. Answer A is wrong because the magnetic force is never parallel to

\vec{v}

, regardless of charge sign — this is a fundamental property of the cross product. Answer B incorrectly claims the cross product of an in-plane vector with

\hat{z}

produces a

\hat{z}

component, which is geometrically impossible. Answer D confuses "reversing the sign" with "rotating into the same direction as

\vec{v}

" — negating

\hat{x}-\hat{y}

gives

\hat{y}-\hat{x}

, which is still perpendicular to

\vec{v}

, not parallel. Study tip: Always do the cross product first for a positive charge, then apply the sign of

q

. This two-step habit prevents sign errors and keeps the geometry clear.

Question 6

A particle with charge $q > 0$ and mass $m$ travels in a straight line through a region containing both a uniform electric field $\vec{E} = E_0\hat{y}$ and a uniform magnetic field $\vec{B} = B_0\hat{z}$ . The particle's velocity is $\vec{v} = v_0\hat{x}$ . If the electric field is then turned off while $\vec{B}$ remains, the particle immediately begins to curve. In which direction does the particle initially curve, and what was the equilibrium speed $v_0$ when both fields were present?

The particle curves in the $+\hat{y}$ direction once $\vec{E}$ is removed, and the equilibrium speed was $v_0 = E_0/B_0$ , because with $\vec{E}$ in $+\hat{y}$ and $\vec{B}$ in $+\hat{z}$ , the magnetic force $q\vec{v}\times\vec{B}$ is in $+\hat{y}$ , requiring the electric force to oppose it in $-\hat{y}$ , and removing $\vec{E}$ leaves the $+\hat{y}$ magnetic force unbalanced
The particle curves in the $-\hat{y}$ direction once $\vec{E}$ is removed, and the equilibrium speed was $v_0 = E_0/B_0$ , because in the velocity selector condition $qE_0 = qv_0B_0$ with $\vec{E}$ in $+\hat{y}$ and $\vec{B}$ in $+\hat{z}$ , the magnetic force on the positive charge moving in $+\hat{x}$ is $q\vec{v}\times\vec{B} = qv_0B_0(-\hat{y})$ , which balances $+\hat{y}$ electric force (correct answer)
The particle curves in the $-\hat{y}$ direction once $\vec{E}$ is removed, and the equilibrium speed was $v_0 = B_0/E_0$ , since the velocity selector condition requires $v_0 = B_0/E_0$ from the force balance $qB_0 = qv_0E_0$ , and the magnetic force was directed in $-\hat{y}$ while present
The particle curves in the $+\hat{z}$ direction once $\vec{E}$ is removed, and the equilibrium speed was $v_0 = E_0/B_0$ , because the magnetic force $q(v_0\hat{x})\times(B_0\hat{z}) = qv_0B_0(\hat{x}\times\hat{z})$ has a component along $\hat{z}$ that was previously masked by the electric field, and once $\vec{E}$ is off, this $\hat{z}$ -component deflects the particle

Explanation: Whenever you see a charged particle moving in straight line through crossed electric and magnetic fields, you're looking at a velocity selector — a configuration where the electric and magnetic forces exactly cancel. Your job is to correctly compute each force's direction using the right-hand rule before drawing any conclusions. Here's the key calculation: with

\vec{v} = v_0\hat{x}

and

\vec{B} = B_0\hat{z}

, the magnetic force on the positive charge is

q\vec{v}\times\vec{B} = qv_0B_0(\hat{x}\times\hat{z}) = qv_0B_0(-\hat{y})

, pointing in the

-\hat{y}

direction. For the particle to travel in a straight line, the electric force must cancel this — so

\vec{E} = E_0\hat{y}

provides an upward

+\hat{y}

force that balances the downward magnetic force. Setting

qE_0 = qv_0B_0

gives the equilibrium speed

v_0 = E_0/B_0

. When

\vec{E}

is removed, the uncanceled magnetic force

(-\hat{y})

curves the particle downward. This is exactly what B describes — making it correct. A gets the equilibrium speed right but inverts the force directions entirely, incorrectly claiming the magnetic force points in

+\hat{y}

. This reversal is a classic right-hand rule error — double-check

\hat{x}\times\hat{z} = -\hat{y}

, not

+\hat{y}

. C flips the velocity selector formula upside down, writing

v_0 = B_0/E_0

instead of

E_0/B_0

. Dimensional analysis alone exposes this error. D invents a

\hat{z}

-component of the magnetic force that doesn't exist. Since

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

lies entirely in the

\hat{y}

-direction, there is no

\hat{z}

deflection. Study tip: Always compute

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

explicitly using the right-hand rule or the cyclic identity — this single cross product determines everything in velocity selector problems.