Magnetism and Moving Charges
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AP Physics 2 › Magnetism and Moving Charges
A proton moves at $2.0\times10^{5},\text{m/s}$ in a uniform magnetic field $B=0.60,\text{T}$. The velocity is perpendicular to $\vec B$. Which statement best describes the magnitude of the magnetic force on the proton?
$3.3\times10^{-25},\text{N}$
$0,\text{N}$
$1.9\times10^{-14},\text{N}$
$1.9\times10^{-9},\text{N}$
Explanation
This problem tests magnetism and moving charges. The magnetic force magnitude is F = |q|vB sin θ, where the problem states velocity is perpendicular to B, making sin θ = 1. For a proton, q = 1.60×10⁻¹⁹ C, so F = (1.60×10⁻¹⁹ C)(2.0×10⁵ m/s)(0.60 T) = 1.92×10⁻¹⁴ N ≈ 1.9×10⁻¹⁴ N. Choice C appears to use an incorrect charge value or calculation error, producing a much smaller result. To calculate magnetic force correctly, always use the fundamental charge e = 1.60×10⁻¹⁹ C for protons and electrons, and verify that perpendicular motion means sin θ = 1.
A singly charged ion ($q=+e$) moves upward at speed $v$ through a uniform magnetic field $\vec B$ directed upward. Which statement best describes the magnetic force on the ion?
The force is zero only if the ion is stationary.
The force is zero because $\vec v$ is parallel to $\vec B$.
The force is directed upward.
The force is directed sideways because the ion is charged.
Explanation
This problem tests magnetism and moving charges. The magnetic force on a charged particle is F = qv × B sin θ, where θ is the angle between velocity and magnetic field. When velocity is parallel to the magnetic field (both upward), θ = 0°, making sin θ = 0 and thus F = 0. The force is zero not because the ion is stationary, but because parallel motion produces no magnetic force. Choice B incorrectly assumes all charged particles experience sideways force regardless of their motion direction relative to the field. Remember: magnetic force is maximum when v ⊥ B and zero when v ∥ B.
A negatively charged particle moves downward at $1.0\times10^{5}\ \text{m/s}$ through a uniform magnetic field directed west. No electric fields are present, and the velocity is perpendicular to the field. Which statement best describes the direction of the magnetic force on the particle?
It is directed downward (parallel to the velocity).
It is directed south.
It is directed north.
It is directed west (parallel to the magnetic field).
Explanation
This question tests magnetism and moving charges. The magnetic force direction follows F = qv × B, perpendicular to both velocity and field. For velocity downward and field west, the right-hand rule gives force pointing south, but for a negative charge we reverse this to north. The perpendicular relationship ensures force is never parallel to velocity or field. Choice D incorrectly suggests force parallels velocity, a common misconception from everyday forces like friction. For negative charges, use the right-hand rule then flip the result.
An electron moves north at $8.0\times10^{6},\text{m/s}$ through a uniform magnetic field of $0.25,\text{T}$ directed west. Which statement best describes the magnetic force on the electron?
The force is downward.
The force is upward.
The force is zero because the electron is moving.
The force is northward.
Explanation
This problem tests understanding of magnetism and moving charges. The magnetic force on a moving charge is perpendicular to both velocity and magnetic field, following F = qv × B. Using the right-hand rule: point fingers north (velocity), curl them west (field), and your thumb points downward for a positive charge. Since this is an electron with negative charge, we reverse the direction, making the force upward. Choice D shows the misconception that electrons in motion experience no force, when actually all moving charges experience magnetic force unless v is parallel to B. To solve magnetic force problems, apply the right-hand rule consistently, then reverse the result for negative charges.
A particle with charge $q=+e$ moves upward through a uniform magnetic field of magnitude $B$. The velocity makes a $30^\circ$ angle with $\vec B$. Which statement best describes the magnitude of the magnetic force?
$F=qvB\cos 30^\circ$
$F=qvB\sin 30^\circ$
$F=qvB$
$F=0$ because the particle is not stationary.
Explanation
This problem tests magnetism and moving charges. The magnetic force magnitude is F = qvB sin θ, where θ is the angle between velocity and magnetic field vectors. With a 30° angle between v and B, we have sin 30° = 0.5, giving F = qvB sin 30°. This correctly accounts for the component of velocity perpendicular to the field. Choice B incorrectly uses cosine instead of sine, confusing the perpendicular component calculation. To find magnetic force with non-perpendicular vectors, always use F = qvB sin θ, where θ is the angle between v and B.
An electron ($q=-1.60\times10^{-19},\text{C}$) travels north at $2.0\times10^{6},\text{m/s}$ in a uniform magnetic field of $0.30,\text{T}$ directed east. Which statement best describes the direction of the magnetic force on the electron?
The force is directed north.
The force is directed upward.
The force is zero because the charge is negative.
The force is directed downward.
Explanation
This problem tests magnetism and moving charges. The magnetic force F = qv × B is always perpendicular to both the velocity vector and the magnetic field vector. Using the right-hand rule: point fingers north (velocity), curl them east (field), and the thumb points upward for a positive charge. However, since this is an electron with negative charge, we must reverse the direction, making the force downward. Choice D shows the misconception that negative charges experience no magnetic force, when actually they experience force in the opposite direction to positive charges. Always apply the right-hand rule first, then flip the direction for negative charges.
A proton ($q=+1.60\times10^{-19},\text{C}$) moves east at $3.0\times10^{6},\text{m/s}$ through a uniform magnetic field $\vec B=0.40,\text{T}$ directed upward. Which statement best describes the magnetic force on the proton?
The force is directed east.
The force is directed south.
The force is zero because the proton is moving.
The force is directed north.
Explanation
This problem tests magnetism and moving charges. The magnetic force on a moving charged particle is given by F = qv × B, where the force is perpendicular to both the velocity and magnetic field vectors. Using the right-hand rule: point your fingers east (velocity direction), curl them upward (field direction), and your thumb points north for a positive charge. Since the proton is positive, the force is indeed directed north. Choice D incorrectly assumes moving charges experience no force, missing that magnetic forces only act on moving charges. To solve these problems systematically, use the right-hand rule for positive charges, then reverse the direction for negative charges.
A particle with charge $q=+2.0,\mu\text{C}$ moves west at $500,\text{m/s}$ through a uniform magnetic field $\vec B=0.20,\text{T}$ directed south. Which statement best describes the direction of the magnetic force on the particle?
The force is directed west.
The force is zero because the field is uniform.
The force is directed downward.
The force is directed upward.
Explanation
This problem tests magnetism and moving charges. The magnetic force F = qv × B acts perpendicular to both the velocity and magnetic field vectors. Using the right-hand rule: point fingers west (velocity), curl them south (field), and the thumb points downward for a positive charge. Since the particle has positive charge, the force is directed downward. Choice D shows the misconception that uniform fields produce no force, when actually the uniformity of the field doesn't affect whether force exists. To solve magnetic force problems, always use the right-hand rule systematically: fingers along velocity, curl toward field, thumb shows force direction for positive charges.
A particle with charge $q=+3.0\times10^{-6},\text{C}$ moves north at $1.0\times10^{3},\text{m/s}$ in a uniform magnetic field $\vec B=0.80,\text{T}$ directed downward. Which statement best describes the magnitude of the magnetic force on the particle?
$3.8\times10^{-9},\text{N}$
$2.4\times10^{3},\text{N}$
$2.4\times10^{-3},\text{N}$
$0,\text{N}$
Explanation
This problem tests magnetism and moving charges. The magnetic force magnitude is F = |q|vB sin θ, where θ is the angle between velocity and field. With velocity north and field downward, they are perpendicular (θ = 90°), so sin θ = 1. Calculating: F = (3.0×10⁻⁶ C)(1.0×10³ m/s)(0.80 T)(1) = 2.4×10⁻³ N. Choice B incorrectly assumes perpendicular vectors produce zero force, when this configuration actually produces maximum force. When solving for magnetic force magnitude, always check the angle between v and B: perpendicular gives maximum force, parallel gives zero force.
An electron moves east at $4.0\times10^{6},\text{m/s}$ through a uniform magnetic field of $0.50,\text{T}$ directed upward. Which statement best describes the direction of the magnetic force on the electron?
The force is zero because the electron is moving perpendicular to $\vec B$.
The force is directed north.
The force is directed south.
The force is directed east.
Explanation
This problem tests magnetism and moving charges. The magnetic force on a moving charge is F = qv × B, perpendicular to both velocity and field vectors. Applying the right-hand rule: fingers point east (velocity), curl upward (field), giving thumb pointing north for a positive charge. Since this is an electron (negative charge), we reverse the direction, making the force point south. Choice D incorrectly states there's no force when v ⊥ B, which is actually when magnetic force is maximum. For any magnetic force problem, use the right-hand rule first, then remember to reverse the direction for negative charges like electrons.