Question

A charged particle moves in a magnetic field \(B\) with velocity components both along and perpendicular to \(B\). What is its path?

• A. Circular path
• B. Straight line
• C. Helical path
• D. Parabolic path

Hint:

Split the velocity into a part along B and a part perpendicular to B, and apply the magnetic force rule to each part on its own. Remember a velocity component parallel to B always gives zero force, while the perpendicular part alone causes the turning.

Answer 1

The Correct Option is C

Concept: When a charged particle enters a magnetic field, it experiences a magnetic force given by the Lorentz force law: \[ F = q(\vec{v} \times \vec{B}) \] Key cases:
• If velocity is perpendicular to \(B\) → particle moves in a circular path.
• If velocity is parallel to \(B\) → particle moves in a straight line.
• If velocity has components both parallel and perpendicular to \(B\) → particle follows a helical path.

Step 1: Identify the components of velocity. The particle has velocity components: \[ v_{\parallel} \text{ (along } B), \quad v_{\perp} \text{ (perpendicular to } B) \]

Step 2: Analyze the motion.
• The perpendicular component \(v_{\perp}\) causes circular motion.
• The parallel component \(v_{\parallel}\) causes uniform motion along the field.

Step 3: Combine the two motions. The combination of circular motion and forward motion produces a: \[ Helical path \]

Answer 2

Concept:

• The magnetic force on a moving charge is $\vec{F} = q(\vec{v} \times \vec{B})$, and the cross product of two parallel vectors is always zero.
• Splitting the velocity into a part along $B$ and a part perpendicular to $B$, and applying this cross product rule to each part separately, shows exactly what each part of the motion looks like.

Step 1: Resolve the velocity into two components.
$v_{\parallel}$, directed along $B$, and $v_{\perp}$, directed perpendicular to $B$.

Step 2: Find the force on the parallel component.
Since $v_{\parallel}$ is parallel to $B$, their cross product is zero, so this component feels no force and keeps moving at constant speed along $B$.

Step 3: Find the force on the perpendicular component.
Since $v_{\perp}$ is perpendicular to $B$, the force $q(v_{\perp} \times B)$ is nonzero and always points toward the center of a circle, giving circular motion of radius $r = \frac{mv_{\perp}}{qB}$ in the plane perpendicular to $B$.

Step 4: Combine both motions.
Uniform circular motion in one plane, combined with steady straight-line motion perpendicular to that plane, traces out a corkscrew-shaped path around the direction of $B$.

Final Answer: Helical path