4.
A charged particle is released from rest in a region where there is a constant electric field
and a constant magnetic field. If the two fields are parallel to each other, the path of the
particle is a
a. circle
b. straight line
c. helix
d. parabola

Charged Particle Path in Parallel Electric and Magnetic Fields

Charged particles released from rest follow a straight line when electric and magnetic fields are parallel. The electric field accelerates the particle linearly, while the magnetic field exerts no force due to zero initial velocity perpendicular to the field.​

Lorentz Force Analysis

The total force on a charged particle is $\mathbf{F}=q\mathbf{E}+q(\mathbf{v}\times \mathbf{B})$. With $\mathbf{E}\parallel \mathbf{B}$ and initial $\mathbf{v}=0$, the magnetic term $q(\mathbf{v}\times \mathbf{B})=0$ initially. Acceleration $a=qE/m$ develops velocity $\mathbf{v}$ parallel to $\mathbf{E}$ and $\mathbf{B}$, keeping $\mathbf{v}\parallel \mathbf{B}$ so $\mathbf{v}\times \mathbf{B}=0$ always. No deflection occurs; motion remains rectilinear along the field direction.​

Option Breakdown

• Circle: Requires perpendicular velocity to $\mathbf{B}$ for centripetal force; absent here.​

• Straight line: Matches exactly—pure electric acceleration with no magnetic deflection.​

• Helix: Needs parallel velocity component (circular) plus perpendicular (helical); initial rest prevents perpendicular component.​

• Parabola: Arises in electric fields with initial perpendicular velocity (like projectiles); magnetic force prevents this.​

Correct answer: b. straight line.​

The path of a charged particle in parallel electric and magnetic fields determines its trajectory under Lorentz force when released from rest. This scenario appears in competitive exams testing electromagnetic motion fundamentals.​

Physics Behind Straight Line Motion

Uniform $\mathbf{E}$ and $\mathbf{B}$ parallel exert electric force $q\mathbf{E}$ causing acceleration $q\mathbf{E}/m$ along the field. Magnetic force $q\mathbf{v}\times \mathbf{B}=0$ since $\mathbf{v}$ stays parallel to $\mathbf{B}$. Position evolves as x(t)=12(qE/m)t2x(t)=21(qE/m)t2, confirming straight line.​​

Common Exam Traps

Examinees confuse with helix (for initial perpendicular velocity) or circle (pure $\mathbf{B}$, perpendicular $\mathbf{v}$). Parallel fields with rest initial condition yield no transverse force, eliminating curves. Practice verifies straight line dominance.​