This question tests the understanding of the motion of charged particles in magnetic fields. The magnetic force on a charged particle is given by \(\vec{F} = q(\vec{v} \times \vec{B})\), which is perpendicular to both velocity and magnetic field, causing deflection without changing the particle's speed. Hypothetically changing electron charge to +e reverses force direction, curving oppositely with same radius (same m, |q|, v). The path flips direction but keeps size. A distractor like choice A fails by keeping direction same, ignoring sign change. Verify reversed q flips F. This, with right-hand rule, illustrates charge effect on trajectory.

This question tests the understanding of the motion of charged particles in magnetic fields. The magnetic force on a charged particle is given by \(\vec{F} = q(\vec{v} \times \vec{B})\), which is perpendicular to both velocity and magnetic field, causing deflection without changing the particle's speed. Electrons move right with field downward in the plane, neglecting electric fields. For negative charge, the force is out of the page, causing initial deflection outward. A distractor like choice B fails by choosing the opposite, often from not reversing for negative q. Verify with v right (+x), B down (-y), q<0 yielding +z force. This method, incorporating right-hand rule with sign, applies to beam deflection experiments.

This question tests the understanding of the motion of charged particles in magnetic fields. The magnetic force on a charged particle is given by \(\vec{F} = q(\vec{v} \times \vec{B})\), which is perpendicular to both velocity and magnetic field, causing deflection without changing the particle's speed. For straight path with unchanged speed, despite being called charged, the observation implies neutral q=0, as force requires q ≠ 0 and sinθ ≠ 0. This condition prevents deflection, consistent with zero force. A distractor like choice B fails by suggesting perpendicular velocity causes straight motion, when it maximizes curving. Verify q=0 gives F=0 always. This reasoning, with right-hand rule absent for q=0, distinguishes neutral from charged paths.

This question tests the understanding of the motion of charged particles in magnetic fields. The magnetic force on a charged particle is given by \(\vec{F} = q(\vec{v} \times \vec{B})\), which is perpendicular to both velocity and magnetic field, causing deflection without changing the particle's speed. A positive ion follows a circular arc, and increasing speed with v ⊥ B unchanged increases radius. The radius grows as r = mv / qB is proportional to v, reducing curvature. A distractor like choice B fails by predicting tighter path, confusing with field strength effects. Calculate r with higher v, confirming increase. This reinforces the formula and right-hand rule for consistent path analysis.

This question tests the understanding of the motion of charged particles in magnetic fields. The magnetic force on a charged particle is given by \(\vec{F} = q(\vec{v} \times \vec{B})\), which is perpendicular to both velocity and magnetic field, causing deflection without changing the particle's speed. A negative ion enters with upward velocity into a field into the page. Adjusting the right-hand rule for negative charge, the force points to the right, causing initial deflection right. A distractor like choice A fails by applying the positive charge direction, ignoring sign reversal. Verify via q<0 reversing v × B direction to right. This check, combined with right-hand rule, aids in analyzing charged beams in various field orientations.

This question tests the understanding of the motion of charged particles in magnetic fields. The magnetic force on a charged particle is given by \(\vec{F} = q(\vec{v} \times \vec{B})\), which is perpendicular to both velocity and magnetic field, causing deflection without changing the particle's speed. Proton and electron with identical v ⊥ B curve in opposite directions due to opposite charges reversing force. Their paths differ in direction but may have different radii based on mass. A distractor like choice A fails by ignoring charge sign effect on direction. Verify opposite q reverses F direction. This, using right-hand rule per charge, predicts opposing curvatures.

This question tests the understanding of the motion of charged particles in magnetic fields. The magnetic force on a charged particle is given by \(\vec{F} = q(\vec{v} \times \vec{B})\), which is perpendicular to both velocity and magnetic field, causing deflection without changing the particle's speed. Positive ions enter with v right and B up, both in-plane, yielding force out of page. The path curves out of the page for positive charges. A distractor like choice A fails by choosing into, often rule misapplication. Verify v right (+x), B up (+y), +q gives +z curve. This check, using right-hand rule, predicts steering in setups.

This question tests the understanding of the motion of charged particles in magnetic fields. The magnetic force on a charged particle is given by \(\vec{F} = q(\vec{v} \times \vec{B})\), which is perpendicular to both velocity and magnetic field, causing deflection without changing the particle's speed. For protons in circular motion within the field, the force is centripetal, always perpendicular to velocity. The speed remains constant because magnetic forces do no work, preserving kinetic energy. A distractor like choice A fails by claiming positive work, misapplying that perpendicular forces change speed like friction. Confirm with work = F · ds = 0 since F ⊥ v. This principle, tied to the right-hand rule, explains constant speed in magnetic deflections universally.

This question tests the understanding of the motion of charged particles in magnetic fields. The magnetic force on a charged particle is given by \(\vec{F} = q(\vec{v} \times \vec{B})\), which is perpendicular to both velocity and magnetic field, causing deflection without changing the particle's speed. Here, an electron with negative charge is injected upward into a uniform magnetic field to the right, with velocity perpendicular to the field and no electric field present. The perpendicular force results in a circular path in the plane perpendicular to B, as the magnetic force provides the centripetal force for uniform circular motion. A distractor like choice D fails because magnetic fields do no work and cannot increase kinetic energy, a common misapplication thinking forces along velocity. As a transferable check, confirm the path is circular when v is perpendicular to B, with radius r = mv / |q|B. Applying the right-hand rule adjusted for negative charge (reversing the force direction) helps predict the sense of rotation in such systems.

This question tests the understanding of the motion of charged particles in magnetic fields. The magnetic force on a charged particle is given by \(\vec{F} = q(\vec{v} \times \vec{B})\), which is perpendicular to both velocity and magnetic field, causing deflection without changing the particle's speed. In this lab demonstration, a positive ion enters a uniform magnetic field with velocity components parallel and perpendicular to B, and no electric field. The parallel component remains unaffected (zero force), while the perpendicular component causes circular motion, combining into helical motion with constant speed along the field. A distractor like choice D fails because the magnetic force is not constant in direction; it varies, unlike gravity causing parabolas, a common confusion with other fields. For verification, decompose velocity: the parallel part gives linear motion, perpendicular gives circling. This approach transfers to analyzing complex paths, using the right-hand rule to determine the helix handedness based on charge sign.