Question

A particle of charge $q$ , mass $m$ , moving with velocity ${v_ \circ }\left( {i + j} \right)$ enters a region of uniform magnetic field ${B_ \circ }\mathop i\limits^ \wedge$ ​. Find displacement of particle after time $t = \dfrac{{2\pi m}}{{qB}}$ ​.
$\left( A \right){\text{ }}\dfrac{{m{v_ \circ }}}{{qB}}\left( {\pi \mathop i\limits^ \wedge + 2\mathop k\limits^ \wedge } \right)$
$\left( B \right){\text{ }}\dfrac{{m{v_ \circ }}}{{qB}}\left( {\pi \mathop i\limits^ \wedge - 2\mathop k\limits^ \wedge } \right)$
$\left( C \right){\text{ 2}}\pi \dfrac{{m{v_ \circ }}}{{qB}}\left( {\mathop i\limits^ \wedge } \right)$
$\left( D \right){\text{ }}\dfrac{{2\pi m{v_ \circ }}}{{qB}}\left( {\mathop j\limits^ \wedge } \right)$

Answer 1

For solving this question we have to know first the formula for the displacement and as we know that the displacement is the product of velocity and time. So by using this formula and substituting the values, we will get to the solution.

Formula used
Displacement, $D = v \cdot t$
Here, $v$ will be the velocity
And $t$ will be the time.

Complete step by step solution:
So in this question, we have the values given as velocity is equal to ${v_ \circ }\left( {i + j} \right)$ and time given to us is $t = \dfrac{{2\pi m}}{{qB}}$ . So by substituting these values in the displacement formula, we get
$\Rightarrow D = {v_ \circ }\left( {i + j} \right) \cdot \dfrac{{2\pi m}}{{qB}}$
Since, here the magnetic field is around the $\mathop i\limits^ \wedge$ , therefore by taking this into the consideration and solving the above equation, we will get the equation as
$\Rightarrow \dfrac{{2\pi m{v_ \circ }}}{{qB}}\mathop i\limits^ \wedge$
Therefore, the displacement of particle after time $t = \dfrac{{2\pi m}}{{qB}}$ is equal to $\dfrac{{2\pi m{v_ \circ }}}{{qB}}\mathop i\limits^ \wedge$ .
Hence, the option $\left( C \right)$ is correct.

Note:
A magnetic field is created when electric charge transporters, for example, electrons travel through space or inside an electrical channel. The states of the magnetic flux lines delivered by moving charge carriers (electric flow) are like the states of the flux lines in an electrostatic field.