Final answer:
The radius of the electron's trajectory when moving through a magnetic field after being accelerated through a potential difference is given by A) √(2eV/m)/B. This is found through the relationship between kinetic energy, the work done by the electric field, and the magnetic force acting as the centripetal force.
Step-by-step explanation:
The question is asking to determine the radius of the electron trajectory when it's deflected by a magnetic field after being accelerated from rest through a potential difference. The key to solving this problem is to use the relationship between the force exerted by the magnetic field and the centripetal force required for the electron's circular motion.
When an electron with charge −e is accelerated through a potential difference V, it gains kinetic energy equal to the work done by the electric field, which is eV. The kinetic energy of the electron can be represented as ½mv², with m being the electron's mass and v its velocity. Hence, from the work-energy principle, we get eV = ½mv². Solving for v gives v = √(2eV/m).
Once in the magnetic field B, the electron experiences a magnetic force (F = evB) perpendicular to its velocity, which acts as a centripetal force (F = mv²/r) necessary to keep it in a circular path. Equating the two forces gives evB = mv²/r, which simplifies to r = mv/(eB). Substituting the expression for v that we found earlier, the radius r of the electron's trajectory is r = (√(2eV/m)) / B. Therefore, the correct answer is A) √(2eV/m)/B.