Question

An electron is launched with velocity v in a uniform magnetic field B. The angle θ between v and B is between 0 and 90o. As a result, the electron follows a helix, its velocity vector v returning to its initial value in a time interval of:

A) 2πm/eB
B) 2πmv/eB
C) 2πmv sinθ /eB
D) 2πmv cosθ /eB
E) none of these

Answer 1

The period for an electron to return to its initial velocity vector when moving in a helical path in a magnetic field is given by the formula T = 2πm/eB, considering only the perpendicular component of velocity to the field (v sinθ). The correct answer is A) 2πm/eB.

Step-by-step explanation:

When an electron is launched with a velocity v at an angle θ with respect to a uniform magnetic field B, it follows a helical path. The period of motion for the electron to return to its initial velocity value is determined by the component of the velocity that is perpendicular to the magnetic field. This velocity component is v sinθ.

According to the formula for the period T of circular motion of a charged particle in a magnetic field, which is perpendicular to the plane of motion, the period is given by T = 2πm/qB, where m is the mass of the particle, q is the charge of the particle, and B is the magnetic field. For an electron, q can be substituted with -e (the electron charge).

Since we are interested in the helical motion, we look at the component of velocity that results in the circular motion, which is v sinθ. This means the period for a complete cycle of the helical path, returning the velocity vector to its initial value, is T = 2πm/eB, because only the perpendicular velocity component (v sinθ) is relevant for the period calculation, not the parallel component (v cosθ). Therefore, the correct answer is A) 2πm/eB.