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

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

Given,

$\overrightarrow{v}=v_0(\hat{i}+\hat{j});\overrightarrow{B}=B_0\hat{i}$ and

The horizontal component of the velocity gives drift in the horizontal direction and vertical component of the velocity is responsible for circular motion of the particle. Thus, it will undergo a helical motion.

Magnetic force on a charge particle is,

$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$

$\overrightarrow{F}=q\left[v_0\right(\hat{i}+\hat{j})\times B_0\hat{i}]$

$\overrightarrow{F}=-q{v}_0{B}_0\hat{k}$

Hence, the charge undergoes a circular motion in $\text{yz}$-plane.

Given,

$t=\frac{\pi }{B_0\alpha }=\frac{\pi }{B_0\times \frac{q}{m}}=\frac{m\pi }{q{B}_0}=\frac{T}{2}[\because T=\frac{2\pi m}{qB}]$



Where, $T$ is time period of revolution.

Let, the position of the paricle is at coordinate $P(x,y,z)$ at time $\frac{T}{2}$.

Now, for $x-$ coordinate,

$P(x,0,0)=v_{\parallel }\times t=v_0\times \frac{T}{2}$

$=\frac{v_0\pi m}{q{B}_0}=\frac{v_0\pi }{B_0\alpha }$

Similarly,

$P(0,y,0)=0$ and

$P(0,0,z)=-2\times \text{radius of the path}=-2\times \frac{m{v}_0}{q{B}_0}=-\frac{2v_0}{B_0\alpha }$

$\therefore P(x,y,z)=(\frac{v_0\pi }{B_0\alpha },0,-\frac{2v_0}{B_0\alpha })$

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

Why this Question : To understand the motion of a particle when it is projected at an angle $\theta$ with the magnetic field.