Helical Motion

When the velocity is not perpendicular to the field

The circular motion of Section 10.3 assumed that the particle moves in a plane perpendicular to $\vec{B}$. If the velocity also has a component along the field, the path is no longer a circle.

Resolve $\vec{v}$ into a component $\vec{v}_\parallel$ parallel to $\vec{B}$ and a component $\vec{v}_\perp$ perpendicular to it. If $\theta$ is the angle between $\vec{v}$ and $\vec{B}$, then $v_\parallel = v\cos\theta$ and $v_\perp = v\sin\theta$.

The parallel component is untouched

For the parallel component the angle with $\vec{B}$ is zero, so

$$F_m = q\,v_\parallel B \sin(0^\circ) = 0$$

The magnetic force on $\vec{v}_\parallel$ vanishes. This component is therefore unaffected, and the particle continues to drift steadily along the direction of $\vec{B}$ at constant speed $v_\parallel$.

The perpendicular component gives circular motion

The component $\vec{v}_\perp$ lies in the plane perpendicular to $\vec{B}$, so it leads to uniform circular motion exactly as before, with

$$r = \dfrac{m v_\perp}{qB}, \qquad T = \dfrac{2\pi m}{qB}$$

Note that the period is the same as for pure circular motion — it does not depend on $v_\perp$ either.

The resulting path is a helix

Superposing a uniform circular motion in the plane perpendicular to $\vec{B}$ on a uniform translation along $\vec{B}$ gives a helix. The particle moves parallel to the field while going round a circular path perpendicular to it.

Diagram (Fig. 10.7): with $\vec{B}$ along the z axis, the trajectory is drawn as a coiled spring winding around the z axis — a circle of radius $r$ in the x-y projection, advancing steadily along z. The axis of the helix is parallel to $\vec{B}$.

The distance advanced along the field in one complete revolution is the pitch:

$$\text{pitch} = v_\parallel T = \dfrac{2\pi m v_\parallel}{qB} = \dfrac{2\pi m v\cos\theta}{qB}$$

Standard results

• $v_\parallel = v\cos\theta$ (unaffected), $v_\perp = v\sin\theta$ (gives the circle)
• Magnetic force on the parallel component is zero, since $\sin 0^\circ = 0$
• Radius of the helix $r = \dfrac{mv_\perp}{qB} = \dfrac{mv\sin\theta}{qB}$
• Period $T = \dfrac{2\pi m}{qB}$, independent of speed
• Pitch $= v_\parallel T$
• Special cases: $\theta = 90^\circ$ gives a circle (zero pitch); $\theta = 0^\circ$ gives a straight line (no force at all)
• The speed and kinetic energy are still constant, because the magnetic force does no work.