Past exam of the physics 2 paper of the University of Oxford / 2014 / 4

For the intended force-balanced straight-line motion, the Lorentz force vanishes:For a nonzero uniform magnetic field, this implies $\mathbf{E}\cdot \mathbf{B}=0$ andwhile a component parallel to $\mathbf{B}$ is unrestricted. A straight trajectory by itself states only that the force has no transverse component, $\mathbf{v}\times (\mathbf{E}+\mathbf{v}\times \mathbf{B})=0$. For example, parallel electric field and motion can give straight but accelerated motion. The boxed cancellation is the usual constant-velocity interpretation of the source.

After the electric field is removed, the uniform magnetic field preserves the speed. Decompose the instantaneous initial velocity as ${\mathbf{v}}_0={\mathbf{v}}_{\parallel }+{\mathbf{v}}_{\perp }$, with $\hat{\mathbf{b}}=\mathbf{B}/B$ and signed $\Omega =qB/m$. Thenand, integrating,This is a helical motion in a uniform magnetic field of radius $m|{\mathbf{v}}_{\perp }|/(|q|B)$ and angular frequency $|q|B/m$. It is a circle if ${\mathbf{v}}_{\parallel }=0$, and a straight line if ${\mathbf{v}}_{\perp }=0$. The formula assumes a uniform, nonzero field and nonrelativistic motion.