Question

A proton is fired from with velocity $\overrightarrow{v}=v_0\hat{j}+v_0\hat{k}$ in a uniform magnetic field $\overrightarrow{B}=B_0\hat{j}$ (Take $m=$mass and $q=$ charge). $\text{X-}$coordinate of the particle after half revolution.

• A. $\frac{-m{v}_0}{q{B}_0}$
• B. $\frac{m{v}_0}{q{B}_0}$
• C. $\frac{-2m{v}_0}{q{B}_0}$
• D. $0$

Answer 1

The correct option is C $\frac{-2m{v}_0}{q{B}_0}$

Initial magnetic force acting is
${\overrightarrow{F}}_B=q(\overrightarrow{v}\times \overrightarrow{B})=q\left[\right(v_0\hat{j}+v_0\hat{k})\times B_0\hat{k}]$

$\Rightarrow {\overrightarrow{F}}_B=q{v}_0{B}_0(-\hat{i})$

So, ${\overrightarrow{F}}_B$ is acting in $-\text{ve}$ $x-$direction.

Using right-hand co- ordinate system, $y-$axis will be in upward direction.

Considering only circular motion, the circular path will be in $xz-$plane,


In half revolution, the particle will reach point $P$.

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

And, $r=\frac{mv}{qB}=\frac{m{v}_0}{q{B}_0}$

$\therefore$ $x-$ coordinate $=\frac{-2m{v}_0}{q{B}_0}$

Hence, option (c) is the correct answer.