Lorentz Force - Physics 2

1. 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 $r_{NR}$?
   Answer: $r_{rel} = 2\, r_{NR}$, 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 $v$ 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.
2. 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 $r_2/r_1$?
   Answer: 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 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.
3. 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)?
   Answer: 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 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.
4. 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?
   Answer: 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 $+\hat{x}$ 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 $-\hat{x}$ to $+\hat{x}$. 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.
5. 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?
   Answer: 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 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.
6. 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?
   Answer: 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) 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.

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