Q.An electron and a positron are released from $(0, 0, 0)$ and $(0, 0, 1.5R)$ respectively, in a uniform magnetic field $\vec{B} = B_0\hat{i}$, each with an equal momentum of magnitude $p = eBR$. Under what conditions on the direction of momentum will the orbits be non-intersecting circles?

Concept: cyclotron motion, $r=p/(eB)$.

With $\vec B=B_0\hat i$, any momentum component along $\hat i$ makes a helix. For a circle the momentum must lie in the $y$–$z$ plane (perpendicular to $\vec B$); then both particles have

Both circles lie in the plane $x=0$. Launch the electron and positron at equal angle $\theta$ to the $y$-axis (with opposite $z$-senses of turning, so the two centres have the same $y$-coordinate). The positron starts at $(0,0,\tfrac32 R)$, so the centre-to-centre distance is

Two equal circles of radius $R$ do not intersect when $d\ge 2R$: …

1. Why the momentum must be in the $y$–$z$ plane

The field is $\vec B=B_0\hat i$. A velocity component along $\hat i$ feels no force and drifts steadily along $x$, turning the path into a helix. For a genuine circle the momentum must have no $x$-component, i.e. it lies in the $y$–$z$ plane, fully perpendicular to $\vec B$. Then the radius is

2. Locating the two centres

Each circle's centre lies a distance $R$ from the launch point, along the (centripetal) magnetic force, perpendicular to the momentum. Because $\vec p\times\hat i$ has no $x$-component, both circles lie in the plane $x=0$ and are coplanar. Write each momentum at angle $\theta$ to the $y$-axis; the electron and positron have opposite charge, hence turn in opposite senses, and we choose the geometry so their centres share the same $y$-coordinate. With the positron released at $(0,0,\tfrac32 R)$ and the electron at the origin, the centres are separated purely along $z$ by

3. Non-intersection condition

Two coplanar circles of equal radius $R$ fail to intersect when the distance between their centres exceeds the sum of radii, $2R$: …

Method: Circular vs Helical Motion in a Uniform Field, and Locating the Orbit

This method applies to any charged particle released into a uniform magnetic field, where you must decide whether the path is a circle or a helix, find its radius, and (if more than one particle is involved) work out the geometric relationship between their paths.

Steps

Step 1: Split the momentum into components parallel and perpendicular to $\vec B$

The magnetic force $\vec F=q\vec v\times\vec B$ has no component along $\vec B$, so a velocity component parallel to the field is completely unaffected — it produces steady drift, not curving. Only the perpendicular component is turned into circular motion. If the full momentum has a component along $\vec B$, the resulting path is a helix, not a closed circle; for a pure circle, the momentum must lie entirely in the plane perpendicular to $\vec B$.

Step 2: Compute the radius of the circular part

using the perpendicular momentum from Step 1 (here the full momentum, since it is chosen to be purely perpendicular). Equal-magnitude charge and momentum give equal radii regardless of the sign of the charge.

Step 3: Locate the centre of the circle …