A positive charge moves with velocity $\vec{v}$ PARALLEL to a uniform magnetic field $\vec{B}$. The magnetic force on it is
A$qvB$ (maximum)
B$\dfrac{1}{2}\,qvB$
C$qvB / \sqrt{2}$
Dzero
Answer & Solution
Correct answer: D. zero
1. The magnetic force magnitude is $F = qvB\sin\theta$, where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$.
2. When $\vec{v}\parallel\vec{B}$, $\theta = 0$ and $\sin 0 = 0$.
3. So $F = 0$. The charge experiences no magnetic force when its velocity is parallel (or antiparallel) to the field.
4. Consequence: a charged particle moving along magnetic field lines is unaffected by the field — explains why charged particles streaming along Earth's magnetic field lines can reach the poles (aurora).
5. Option A is for $\theta = 90^\circ$. Options B and C correspond to specific intermediate angles ($30^\circ$, $45^\circ$).
_Source: syllabus Physics Part 1, Ch 4, §4.2 (Magnetic Force — perpendicular dependence), p. 2–3._
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