Question

A proton is fired from origin with velocity $\overrightarrow{v}=v_0\hat{j}+v_0\hat{k}$ in a uniform magnetic field $\overrightarrow{B}=B_0\hat{j}$. In the subsequent motion of the proton

• A. its $y-$coordinate will be proportional to its time of flight
  
• B. its $x-$coordinate can never by positive
• C. its $x-$ and $z-$coordinate cannot be zero at the same time
• D. None of the above

Answer 1

Answer: (A), (B)

The correct options are
A its $y-$coordinate will be proportional to its time of flight
B its $x-$coordinate can never by positive

$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})=q\left(\right(v_o\hat{j}+v_o\hat{k})\times B_0\hat{j})=-v_0{B}_0\hat{i}$
The force will be always along -ve x-axis.
$\therefore$, the particle will never have $x-$cordinate positive since it starts from the origin.
The velocity component along $y-$axis will make the path helical.
$Y$ co-ordinate will be $y=v_{0}t$ i.e $y\propto t$