Question

A particle enters a space where exists a uniform magnetic field $\overrightarrow{B}=B_x\hat{i}+B_y\hat{j}+B_z\hat{k}$ and uniform electric field $\overrightarrow{E}=E_x\hat{i}+E_y\hat{j}+E_z\hat{k}$, with initial velocity $\overrightarrow{u}=u_x\hat{i}+u_y\hat{j}+u_z\hat{k}$. If $E=0$ and $u_x{B}_x+u_y{B}_y\ne -u_z{B}_z$, then which of the following statement regarding path of particle is correct:

• A. Straight line
• B. Helix with uniform pitch and constant radius
• C. Both $\left(a\right)$ and $\left(b\right)$
• D. Circle

Answer 1

The correct option is C Both $\left(a\right)$ and $\left(b\right)$

Given:
$\overrightarrow{B}=B_x\hat{i}+B_y\hat{j}+B_z\hat{k}$
$\overrightarrow{E}=E_x\hat{i}+E_y\hat{j}+E_z\hat{k}$
$\overrightarrow{u}=u_x\hat{i}+u_y\hat{j}+u_z\hat{k}$

Electric field, $(E=0)$

$u_x{B}_x+u_y{B}_y\ne -u_z{B}_z$

$\Rightarrow u_x{B}_x+u_y{B}_y+u_z{B}_z\ne 0$

$\Rightarrow \overrightarrow{u}\cdot \overrightarrow{B}\ne 0$

So, angle between $\overrightarrow{u}$ and $\overrightarrow{B},\theta \ne \frac{\pi }{2}$.

Thus, there is some angle $0\le \theta \le \pi$ existing between velocity and magnetic field.

Now if $\theta =0^{\circ }$ or $\theta =\pi$
then ${\overrightarrow{F}}_m=q(\overrightarrow{v}\times \overrightarrow{B})=0$

In this case particle will continue to move in straight due to absence of electric and magnetic forces.

If the angle $\left(\theta \right)$ between $\overrightarrow{v}$ & $\overrightarrow{B}$ is lies between $0^o$ & ${180}^o$, then the particle will follow a helical path with uniform pitch radius, $r=\frac{m{v}_{\perp }}{qB}=\frac{mv\sin \theta }{qB}$.


Therefore, option $\left(c\right)$ is right.

Why this question ? Concept: Whenever $\overrightarrow{u}\cdot \overrightarrow{B}\ne 0$ and if $\theta \ne 0^o\text{or}\pi$, then there must be some non-zero angle between $\overrightarrow{u}\&\overrightarrow{B}$, which will lead to two mutually perpendicular component of velocity giving rise to helical path.