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 $\frac{\pi }{2}$. Its velocity vector $\overrightarrow{v}$ returns to its initial value in a time interval of

• A. $\frac{2\pi m}{eB}$
• B. $\frac{2\times 2\pi m}{eB}$
• C. $\frac{\pi m}{eB}$
• D. depends upon angle between $\overrightarrow{v}$ and $\overrightarrow{B}$

Answer 1

The correct option is A $\frac{2\pi m}{eB}$

The velocity component perpendicular to the magnetic field will make the particle rotate in a circular path $($ uniform circular motion $)$ and the component parallel to the magnetic field make it move along the magnetic field. Thus, the particle will have a helical path.

Time period of oscillation is $T=\frac{2\pi m}{qB}$

Hence, the velocity vector $\overrightarrow{v}$ will return to its initial value after time $T$. It will not come back to the same position since it velocity has a component along the direction of the magnetic field.