Question

An electron is launched with velocity $\overrightarrow v$ in a uniform magnetic field $\overrightarrow B .$ The angle $\theta$ between $\overrightarrow v$ and $\overrightarrow B$ lies between 0 and $\dfrac{\pi }{2}.$ Its velocity vector $\overrightarrow v$ returns to its initial value in a time interval of
(A). $\dfrac{{2\pi m}}{{eB}}$
(B). $\dfrac{{2 \times 2\pi m}}{{eB}}$
(C). $\dfrac{{\pi m}}{{eB}}$
(D). Depends upon angle between $\overrightarrow v$ and $\overrightarrow B$

Answer 1

Hint: In order to solve this question, we will use the formula of Lorentz force and If we change the speed vector into two components i.e. v parallel and v perpendicular (one along the magnetic field $\overrightarrow B$ and other perpendicular to the magnetic field $\overrightarrow B .$ Then we will solve the problem separately step by step.

Formula used- $F = q(\overrightarrow v \times \overrightarrow B )$
Where F is force, q is the charge, v is the velocity and B is the magnetic field.

Complete step-by-step solution -
We will resolve the speed vector into two components as mentioned above in the hint.
Thus, for the V parallel component we will have
Force F parallel, since speed is parallel to the magnetic field the angle between them is zero.
$F = q(\overrightarrow v \times \overrightarrow B ) \\\ = evB\sin {0^0} \\\ = 0 \\\$
And for the perpendicular component of the speed
Force F perpendicular, and the angle between speed and magnetic field is 90 degree.
$F = q(\overrightarrow v \times \overrightarrow B ) \\\ = evB\sin {90^0} \\\ = evB \\\$
For perpendicular component charge will perform circular motion and for parallel component of speed it will have constant speed and overall will be the superposition of circular and rectilinear motion.
The time period is given by
Since the charge performs circular motion for perpendicular component
F perpendicular = $\dfrac{{m{v^2}}}{r}{\text{ [where m is the mass, v is the velocity and r is the radius of circle ]}}$
Substituting the value of F perpendicular from above
$\Rightarrow evB = \dfrac{{m{v^2}}}{r} \\\ \Rightarrow r = \dfrac{{mv}}{{Be}} \\\$
Time period of revolution is given by
$T = \dfrac{{2\pi r}}{v}$
Substituting the value of radius in the time period equation from above
$\Rightarrow T = \dfrac{{2\pi mv}}{{vBe}} \\\ \Rightarrow T = \dfrac{{2\pi m}}{{Be}} \\\$
Thus the correct answer is A.

Note: As we know when a charge particle moves in a uniform magnetic field it always experiences a net force perpendicular to its motion. The trajectory of the particle in a uniform magnetic field is helical and the particle will reach the same point after the time period of $T = \dfrac{{2\pi m}}{{Be}}$ and will have the same speed. Also the work done by the parallel component of the particle is zero.