Question

A uniform magnetic field $\overrightarrow{B}=(20\hat{i}-30\hat{j}+50\hat{k})\text{T}$ exists in the space. A charge particle with specific charge $\frac{{10}^3}{19}\text{C/kg}$ enters this region at time $t=0$ with a velocity $\overrightarrow{v}=(20\hat{i}+50\hat{j}+30\hat{k})\text{m/s}$. The pitch of the helical path for the motion of the charged particle will be:

• A. $\frac{\pi }{100}\text{m}$
• B. $\frac{\pi }{125}\text{m}$
• C. $\frac{\pi }{215}\text{m}$
• D. $\frac{\pi }{250}\text{m}$

Answer 1

The correct option is D $\frac{\pi }{250}\text{m}$

The pitch of the helical path is given by

$P=v_{||}\times T=\left(v\cos \theta \right)T$

The given equation can be written as:

$v\cos \theta =\frac{Bv\cos \theta }{B}=\frac{\overrightarrow{B}\cdot \overrightarrow{v}}{|\overrightarrow{B}|}$

Now substituting $v\cos \theta$, we get;

$P=\frac{\overrightarrow{B}\cdot \overrightarrow{v}}{|\overrightarrow{B}|}\times T$

Here, $T=\frac{2\pi m}{qB}$

$P=\frac{\overrightarrow{B}\cdot \overrightarrow{v}}{|\overrightarrow{B}|}\times \frac{2\pi m}{qB}$

Here,
$|\overrightarrow{B}|=B=\sqrt{400+900+2500}=\sqrt{3800}\text{T}$

and $\frac{q}{m}=\frac{{10}^3}{19}\text{C/kg}$

$P=\frac{(20\hat{i}-30\hat{j}+50\hat{k})\cdot (20\hat{i}+50\hat{j}+30\hat{k})}{\sqrt{3800}}\times \frac{38\pi }{{10}^3(\sqrt{3800}}$

$P=\frac{(400-1500+1500)}{(\sqrt{3800}{)}^2}\times \frac{38\pi }{1000}$

$\therefore P=\frac{\pi }{250}\text{m}$

Hence, option $\left(d\right)$ is the right choice.