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). $\text{y-}$coordinate of the particle after half revolution.

• A. $0$
• B. $\frac{m\pi V_0}{q{B}_0}$
• C. $\frac{\pi m{V}_0}{2q{B}_0}$
• D. $\frac{3}{2}\frac{m{V}_0}{q{B}_0}$

Answer 1

The correct option is B $\frac{m\pi V_0}{q{B}_0}$

The diagram for the given values,


Velocity $\overrightarrow{v}$makes an angle of ${45}^{\circ }$
with the magnetic field $\overrightarrow{B}$.

So, its path will be helix due to $z-$component of velocity , particle describes circular motion and due to $y-$component of velocity it will move forward.

Due to magnetic field, there will not be any magnetic force along $y-$axis.
So it will move with velocity.

As, in $y-$direction It is moving with constant velocity $v_0$
So, after hal rerolution

$y-$coordinate$=v_0\times \frac{T}{2}$

$y=v_0\times \frac{2\pi m}{q{B}_0\times 2}$

$y=\frac{\pi m{V}_0}{q{B}_0}$

Hence, option (c) is the correct answer.