Question

Consider the motion of a positive point charge in a region where there are simultaneous uniform electric and magnetic fields $E=E_0\hat{j}$ and $B=B_0\hat{j}$.At time $t=0$, this charge has velocity $v$ in the $xy$-plane making an angle $\theta$ with the $x$-axis. Which of the following option(s) is (are) correct for time $t>0$?

• A. If $\theta =0^{\circ }$, the charge moves in a circular path in the $xz$-plane
• B. If $\theta =0^{\circ }$, the charge undergoes helical motion with constant pitch along the $y$-axis
• C. If $\theta ={10}^{\circ }$, the charge undergoes helical motion with its pitch increasing with time along the $y$-axis
• D. If $\theta ={90}^{\circ }$, the charge undergoes linear but accelerated motion along the $y$-axis

Answer 1

Answer: (C), (D)

If $\theta ={90}^{\circ }$, the charge undergoes linear but accelerated motion along the $y$-axis


From Lorentz's force equation,

$\overrightarrow{F}=q(\overrightarrow{E}+(\overrightarrow{v}\times \overrightarrow{B}\left)\right)$

(or) We can say,

$\overrightarrow{F_B}=q(\overrightarrow{v}\times \overrightarrow{B})$

$\overrightarrow{F_E}=q\overrightarrow{E}$

Case- I:

When $\theta =0^{\circ }$ $v_y=0$ and $v_x=v\hat{i}$

$\overrightarrow{F_B}=q(v\hat{i}\times B_0\hat{j})=qv{B}_0\hat{k}$

$\because \overrightarrow{v}\perp \overrightarrow{B}$, Hence charge moves in a circular path in $xz$ -plane.

but there is electric force $\overrightarrow{F_E}=qE\hat{j}$ which causes charge to move in $+y$ axis with increasing velocity along $y-$ axis with respect to time.

Because of combination of electric and magnetic forces, charge will move in a helical path with increasing pitch.

$\therefore$ The charge undergoes neither pure circular nor helical path with constant pitch.

$\therefore$ Options (a) and (b) both are wrong.

Case-II:

$\theta ={10}^{\circ }$ means $\overrightarrow{v}$ has both $v_x$ and $v_y$ components. But only $v_x$ contributes in magnetic force as,
$\overrightarrow{F_B}=q(v_x\hat{i}+v_y\hat{j})\times B_0\hat{j}=q{v}_x{B}_0\hat{k}$

Due to electric force $\overrightarrow{F_E}=q\overrightarrow{E}\hat{j}$ charge accelerates along the $y$-axis.

$qE=maa=\frac{qE}{m}$

From kinematic equations of motion,

$S_y=ut+\frac{1}{2}a{t}^2$

$S_y=u_y\left(t\right)+\frac{1}{2}\left(\frac{qE}{m}\right)t^2$ and

$v_y=u_y+a_{y}t=u_y+\frac{qE}{m}t$

As $v_y$ is increasing with time.

Because of combination of electric and magnetic forces, charge will move in a helical path with increasing pitch.

$\therefore$ Option (c) is correct.

Case-III:

If $\theta ={90}^{\circ }$ then $v_y=v$ and $v_x=0$

Now, $F_B=q(\overrightarrow{v}\times \overrightarrow{B})=q(v\hat{j}\times B_0\hat{j})=0$

$\therefore$ Only electric force acts on the charge which accelerates the charge along the direction of the force.

$\overrightarrow{F_E}=qE\hat{j}$

$\Rightarrow a=\frac{qE}{m}\hat{j}$

$\therefore$ Option (d) is also correct.

Hence, options (c) and (d) are the correct answers.