Question

(i) A point charge q moving with the speed $\upsilon$ enters a uniform magnetic field $\overrightarrow{B}$ that is acting into the plane of paper as shown. which is the path followed by the charge q and in which plane does it move ?
(ii) How does the path followed by the charge get affected if its velocity has a component parallel to $\overrightarrow{B}$?
(iii) If an electric field $\overrightarrow{E}$ is also applied such that the particle continues moving along the original straight line path , what should be the magnitude and direction of the electric field $\overrightarrow{E}$ ?

Answer 1

$\left(i\right)$ Let $\overrightarrow{B}=B_0(-\hat{K})$

$\overrightarrow{V}=V(-\hat{x})$

Electromagnetic force $F_m=q\overrightarrow{V}\times \overrightarrow{B}$

$=qVB(-\hat{j})$

Now we can conclude that point charge move on circular motion of radius $r=\frac{mV}{qB}$ and motion of charge in $X-Y$ plane.

$\left(ii\right)$ If point charge has velocity component parallel to $\overrightarrow{B}$ then, In this case path followed by charge particle is helical path.

$\left(iii\right)$ If electric field is also applied such that the particle continuous moving along straight line the electromagnetic force should be balanced by electrostatics force $F_e$

$F_m=qVB(-\hat{j})$

If net force on $q$ is zero.

$F_e=qV\hat{j}$

$q\overrightarrow{E}=qVB\left(\hat{j}\right)$

$\overrightarrow{E}=VB\left(\hat{j}\right)$