Question

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

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

Hint:

Remember: 1. Parallel ($\theta = 0^{\circ}$) $\rightarrow$ Straight line. 2. Perpendicular ($\theta = 90^{\circ}$) $\rightarrow$ Circle. 3. Any other angle $\rightarrow$ Helix.

Answer 1

The Correct Option is C

Concept: The motion of a charged particle in a magnetic field depends on the angle $\theta$ between its velocity $\vec{v}$ and the magnetic field $\vec{B}$. The magnetic force is $\vec{F} = q(\vec{v} \times \vec{B})$.

Step 1: {Analyze motion when velocity is parallel to the field ($\theta = 0^{\circ}$).}
If $\vec{v}$ is along the vertical direction (same as $\vec{B}$), $\vec{v} \times \vec{B} = 0$. The force is zero, so the particle continues in a straight line, not a circular path.

Step 2: {Analyze motion when velocity is perpendicular to the field ($\theta = 90^{\circ}$).}
If $\vec{v}$ is along the horizontal direction (orthogonal to vertical $\vec{B}$), the force $F = qvB$ acts as a centripetal force. The particle follows a circular path.

Step 3: {Analyze motion at an arbitrary angle.}
If $\vec{v}$ is neither parallel nor orthogonal to $\vec{B}$, it has two components:
• $v_{\parallel} = v \cos\theta$: This component remains constant (straight line motion along $\vec{B}$).
• $v_{\perp} = v \sin\theta$: This component causes circular motion in the plane perpendicular to $\vec{B}$. The superposition of linear and circular motion results in a helical path.