Question

A charged particle of specific charge (charge/mass) $\alpha$ is released from origin at time $t=0$ with velocity $\overrightarrow{v}=v_0(\hat{i}+\hat{j})$ in a uniform magnetic field $\overrightarrow{B}=B_0\hat{i}$. Coordinates of the particle at time $t=\frac{\pi }{B_0\alpha }$ are

• A. $(\frac{v_0}{2B_0\alpha },\frac{\sqrt{2}{v}_0}{\alpha B_0},\frac{-v_0}{B_0\alpha })$
• B. $(\frac{-v_0}{2B_0\alpha },0,0)$
• C. $(0,\frac{2v_0}{B_0\alpha },\frac{v_0\pi }{2B_0\alpha })$
• D. $(\frac{v_0\pi }{B_0\alpha },0,\frac{-2v_0}{B_0\alpha })$

Answer 1

Answer: (D)

$(\frac{v_0\pi }{B_0\alpha },0,\frac{-2v_0}{B_0\alpha })$

$\alpha =\frac{q}{m}$, path of the particle will be a helix of time period,

$T=\frac{2\pi m}{B_{0}q}=\frac{2\pi }{B_0\alpha }$

The given time $t=\frac{\pi }{B_0\alpha }=\frac{T}{2}$

$\therefore$ Coordinates of particle at time $t=T/2$ would be $(v_{x}T/2,0,-2r)$

Here, $r=\frac{m{v}_0}{B_{0}q}=\frac{v_0}{B_0\alpha }$

$\therefore$ The coordinate are $(\frac{v_0\pi }{B_0\alpha },0,\frac{-2v_0}{B_0\alpha })$