Question

For a positively charged particle moving in a $x-yplane$ initially along the $x-axis$, there is a sudden change in its path due to the presence of electric &/or magnetic field beyond $P$. The curve path is shown $\varphi$ in the $x-yplane$ & is found to be non-circular. which of the following combination is

• A. $\overrightarrow{E}=0;\overrightarrow{B}=b\hat{i}+c\hat{k}$
• B. $\overrightarrow{E}=a\hat{i};\overrightarrow{B}=c\hat{k}+a\hat{i}$
• C. $\overrightarrow{E}=0;\overrightarrow{B}=c\hat{j}+b\hat{k}$
• D. $\overrightarrow{E}=a\hat{i};\overrightarrow{B}=c\hat{k}+b\hat{j}$

Answer 1

The correct option is A $\overrightarrow{E}=0;\overrightarrow{B}=b\hat{i}+c\hat{k}$

The velocity at P is in the x direction (given). Let $\overrightarrow{V}=m\hat{i}$.

After P, the positively charged particle gets deflected in the x-y plane toward-y direction and the path is non-circular.

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

$=q[m\hat{i}\times (c\hat{k}\times a\hat{i}\left)\right]\left[foroption\right(b\left)\right]$

$=q[mc\hat{i}\times \hat{k}+ma\hat{i}\times \hat{i}]=mcq(-\hat{i})$

Since in option (b) electric field is present, i.e., $\overrightarrow{E}=a\overrightarrow{i}$ therefore, it will also exert a force in the +x direction. The net result of the two forces will be a non-circular path.

Only option (b) fits for the above logic. For other options, we get some other results.