# 14.3.3 Helical Motion in Magnetic Field

What happens if a charged particle is moving at an angle (neither parallel nor perpendicular) to the magnetic field B?

For example, in the diagram above, we have a uniform B-field directed in the +x direction. A positive charge q is moving at velocity v in the xy plane, making an angle q  with the x-axis.

The analysis becomes much easier if we make an astute decision to resolve the velocity into

• the component parallel to B, $\displaystyle {{v}_{\parallel }}=v\cos \theta$ and
• the component perpendicular to B, $\displaystyle {{v}_{\bot }}=v\sin \theta$.

If the charge had only velocity $\displaystyle \displaystyle {{v}_{\parallel }}$, it would have experienced zero magnetic force, since its velocity is completely parallel to the magnetic field. This charge would continue travelling forward (in the +x direction) at a constant velocity.

If the charge had only velocity $\displaystyle {{v}_{\bot }}$, it would have experienced the centripetal magnetic force $\displaystyle {{F}_{b}}=Bq{{v}_{\bot }}$, since its velocity is perpendicular to the magnetic field. Its motion would have been circular motion of radius $\displaystyle r=\frac{{m{{v}_{\bot }}}}{{Bq}}$ in the y-z plane.

Since the charge had both $\displaystyle \displaystyle {{v}_{\parallel }}$ and $\displaystyle {{v}_{\bot }}$, we have to superpose the two motions together. What we get is a circular motion of radius $\displaystyle r=\frac{{m{{v}_{\bot }}}}{{Bq}}$ that moves forward at a constant speed $\displaystyle {{v}_{\parallel }}$. The resultant motion is a helical path, like a screw with a pitch of $\displaystyle v\cos \theta \times \frac{{2\pi m}}{{Bq}}$.

Concept Test

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