Question

A charged particle \(q\) moving with a velocity \(\vec{v}\) is subjected to a uniform magnetic field \(\vec{B}\) acting perpendicular to \(\vec{v}\). If a uniform electric field \(\vec{E}\) is also set up in the region along the direction of \(\vec{B}\), describe the path followed by the particle and draw its shape.

Hint:

Circular motion from \(B\) plus uniform acceleration along \(B\) from \(E\) gives a helix of constant radius and increasing pitch.

Answer 1

Step 1: Forces acting on the particle:
The velocity \(\vec{v}\) is perpendicular to \(\vec{B}\). The magnetic force \(q\vec{v}\times\vec{B}\) has magnitude \(qvB\) and acts at right angles to both \(\vec{v}\) and \(\vec{B}\).
The electric force \(q\vec{E}\) acts along \(\vec{B}\), since \(\vec{E}\) is set up in the direction of \(\vec{B}\).

Step 2: Motion perpendicular to B:
The magnetic force does no work and has no component along \(\vec{B}\). In the plane perpendicular to \(\vec{B}\) the particle moves on a circle of radius
\[ r = \frac{mv}{qB} \]
The time period \(T = \dfrac{2\pi m}{qB}\) does not depend on the speed.

Step 3: Motion along B:
The electric force gives a constant acceleration \(a = \dfrac{qE}{m}\) along \(\vec{B}\). The speed along \(\vec{B}\) is zero at the start and then increases as \(v_{\parallel} = at\).

Step 4: Combine the two motions:
A circular motion with fixed radius together with a uniformly accelerated motion along the axis of the circle gives a helical path.
The radius of the helix stays constant, but the distance advanced along \(\vec{B}\) in each turn (the pitch) keeps increasing, since the speed along \(\vec{B}\) keeps increasing.
Diagram: draw a spiral of constant radius wound around a line parallel to \(\vec{B}\), with the turns getting farther apart in the direction of the force (\(\vec{B}\) for a positive charge).

Final Answer:
The particle moves along a helix of constant radius \(mv/qB\) and increasing pitch, wound along the direction of \(\vec{B}\).