Question

A particle of charge per unit mass $\alpha$ is projected from origin with a velocity $\overrightarrow{v}=v_0\hat{i}$ in a uniform magnetic field $\overrightarrow{B}=-B_0\hat{k}$. If the particle passes through $(0,y,0)$ then $y$ is equal to

• A. $-\frac{2v_0}{B_0\alpha }$
• B. $\frac{v_0}{B_0\alpha }$
• C. $\frac{2v_0}{B_0\alpha }$
• D. $-\frac{v_0}{B_0\alpha }$

Answer 1

The correct option is C $\frac{2v_0}{B_0\alpha }$

The magnetic force acting on moving charge $q$ with velocity $\overrightarrow{v}$ in an uniform magnetic field $\overrightarrow{B}$

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

Given: $\overrightarrow{v}=v_0\hat{i};\overrightarrow{B}=-B_0\hat{k}$

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

$\therefore \overrightarrow{F}=q{v}_0{B}_0\hat{j}$

Thus, the force $\overrightarrow{F}$ will make the charge particle to move in circle with its radius on $y-$ axis.

Therefore, particle passes through point $(0,2R,0)$ on $y-$ axis, where $R$ will be

$R=\frac{m{v}_0}{q{B}_0}$

$\therefore y=2R=\frac{2m{v}_0}{q{B}_0}$

$\Rightarrow y=\frac{2v_0}{\alpha B_0}(\text{given,}\alpha =q/m)$

Therefore, option $\left(c\right)$ is correct.