Question

A particle of mass m and charge q with an initial velocity v is subjected to a uniform magnetic field B along the vertical direction. The particle will

• A. follow a circular path if v is along the vertical direction
• B. make helical motion if v is along the horizontal direction
• C. make helical motion if v is neither parallel nor orthogonal to B
• D. always make circular motion
• E. always make helical motion

Hint:

The path of a charged particle in a uniform magnetic field is a straight line if \( \vec{v} \parallel \vec{B} \), a circle if \( \vec{v} \perp \vec{B} \), and a helix for any other angle between them.

Answer 1

The Correct Option is C

Step 1: Analyze the motion of a charged particle in a magnetic field.
The path of a charged particle in a uniform magnetic field \( \vec{B} \) is determined by the angle \( \theta \) between its velocity vector \( \vec{v} \) and \( \vec{B} \). The force is given by \( \vec{F} = q(\vec{v} \times \vec{B}) \).

Step 2: Analyze the different cases based on the angle \( \theta \).
Case 1: \( \theta = 0^\circ \) or \( 180^\circ \) (\(\vec{v}\) is parallel or anti-parallel to \( \vec{B} \)). The cross product \( \vec{v} \times \vec{B} \) is zero, so the force \( \vec{F} = 0 \). The particle moves in a straight line. Option (A) is incorrect because if \( \vec{v} \) is vertical (parallel to \( \vec{B} \)), the path is a straight line, not a circle.
Case 2: \( \theta = 90^\circ \) (\(\vec{v}\) is perpendicular to \( \vec{B} \)). The force \( F = qvB \) is constant in magnitude and is always perpendicular to \( \vec{v} \). This provides the necessary centripetal force for uniform circular motion. Option (B) is incorrect because if \( \vec{v} \) is horizontal (perpendicular to the vertical \( \vec{B} \)), the motion is circular, not helical.
Case 3: \( \vec{v} \) is neither parallel nor orthogonal to \( \vec{B} \). The velocity vector \( \vec{v} \) can be resolved into two components: one parallel to \( \vec{B} \) (\( v_{\parallel} = v \cos\theta \)) and one perpendicular to \( \vec{B} \) (\( v_{\perp} = v \sin\theta \)). The parallel component \( v_{\parallel} \) does not interact with the magnetic field (as \( \vec{v}_{\parallel} \times \vec{B} = 0 \)), so the particle moves with constant velocity along the field direction.
The perpendicular component \( v_{\perp} \) causes a magnetic force \( F = qv_{\perp}B \) that leads to circular motion in the plane perpendicular to \( \vec{B} \). The combination of this linear motion along the field and circular motion perpendicular to it results in a helical (spiral) path.

Step 3: Evaluate the options.
Based on the analysis in
Step 2:
- (A) is incorrect.
- (B) is incorrect.
- (C) correctly describes the condition for helical motion.
- (D) and (E) are incorrect as the motion can be a straight line, circular, or helical depending on the initial velocity's direction.


Final Answer: (C) make helical motion if v is neither parallel nor orthogonal to B