Question

A charged particle of specific charge, $\alpha =\left(\frac{q}{m}\right)$ is released at origin at $t=0$ with velocity $v=v_0(\hat{i}+\hat{j})$ in a uniform magnetic field $B=B_0\hat{i}$. Coordinates of particle at time $t=\frac{\pi }{B_0\alpha }$ are:

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

Answer 1

The correct option is C $(\frac{v_0\pi }{B_0\alpha },0,\frac{-2v_0}{B_0\alpha })$



Time period $t=\frac{2\pi m}{qB}$

$\therefore T=\frac{2\pi }{B_0\alpha }[\because q/m=\alpha ]$

Given, $t=\frac{\pi }{B_0\alpha }\Rightarrow t=\frac{T}{2}$

The particle will undergo a circular motion in the $xz$ plane,

At time $t=\frac{T}{2}$, the particle would have reached in the$x,y,z$directions as follows,

$y$ - co-ordinate $=0$

$x$ - co-ordinate $=v_x\times \frac{T}{2}=\frac{v_{0}T}{2}=\frac{v_0\pi }{B_0\alpha }$

In the $z$ - axis, particle would have travelled $2r$ distance along negative $z$ - axis

So, $z$- coordinate $=-2r$

$=-2\left(\frac{m{v}_y}{qB}\right)=-2\left(\frac{v_0}{\alpha B_0}\right)$

$\therefore$ Coordinates of the particle at $t=\frac{T}{2},(\frac{V_0\pi }{B_{0}d},0,\frac{-2v_0}{\alpha B_0})$

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