Question

A charged particle having mass $\left(m\right)$ and charge $\left(q\right)$ is projected from the origin with velocity $v=vo\hat{i}$ in a uniform magnetic field,$B=\frac{B_0}{2}\hat{i}+\frac{\sqrt{3}}{2}{B}_0\hat{j}$, where $B_0$ is a constant. The $z$-component of velocity is $\frac{\sqrt{3}}{2}{v}_0$ after time $t$. Find $t$.

• A. $\frac{2\pi m}{B_{0}q}$
• B. $\frac{\pi m}{B_{0}q}$
• C. $\frac{\pi m}{2B_{0}q}$
• D. $\frac{2\pi m}{3B_{0}q}$

Answer 1

The correct option is C $\frac{\pi m}{2B_{0}q}$

Given,

$B=\frac{B_0}{2}\hat{i}+\frac{\sqrt{3}}{2}{B}_0\hat{j}$

$\therefore \tan \theta =\frac{B_y}{B_x}=\frac{\frac{\sqrt{3}}{2}{B}_0}{\frac{B_0}{2}}=\sqrt{3}$

$\therefore \theta ={60}^{\circ }$

Now,


The component of velocity $v_0\cos {60}^{\circ }$ will be always in $xy$ plane. But $v_0\sin {60}^{\circ }$ component is responsible for the circular motion of the particle. Hence it can give $z$-component of velocity. Magnitude of this component will remain constant.

$z$-component of velocity will be equal to $\frac{\sqrt{3}{v}_0}{2}$

If particle travels perpendicular to the $xy$-plane. This will happen after time, $t=\frac{T}{4}$.

$\therefore t=\frac{2\pi m}{qB\times 4}=\frac{\pi m}{2B_{0}q}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (c) is the correct answer.