Question

A charge particle projected with a velocity $v$ in a uniform magnetic field moves along a helical path of radius $3\text{cm}$ and pitch $6\pi \text{cm}$. Find the velocity of projection [Take, $B=0.3\text{T},q=0.02\text{C},m=3\times {10}^{-3}\text{kg}]$

• A. $6\sqrt{2}\text{cm/s}$
• B. $6\text{cm/s}$
• C. $\frac{3}{\sqrt{2}}\text{cm/s}$
• D. $3\sqrt{2}\text{cm/s}$

Answer 1

The correct option is A $6\sqrt{2}\text{cm/s}$

Given,

$\text{Radius},r=3\text{cm}\text{Pitch},P=6\pi \text{cm}B=0.3\text{T}q=0.02\text{C}m=3\times {10}^{-3}\text{kg}$

$\because \text{pitch}\left(P\right)=v_{\parallel }\times T=v\cos \theta \times \left(\frac{2\pi m}{qB}\right).........\left(1\right)$
Radius, $r=\frac{m{v}_{\perp }}{qB}=\frac{mv\sin \theta }{qB}.........\left(2\right)$

On dividing $\left(1\right)$ and $\left(2\right)$ we get,

$\frac{P}{r}=\frac{2\pi }{\tan \theta }$

$\Rightarrow \frac{6\pi }{3}=\frac{2\pi }{\tan \theta }$

$\Rightarrow \tan \theta =1$

$\therefore \theta ={45}^{\circ }$

Now using $\left(1\right)$

$P=v\cos \theta \times \frac{2\pi m}{qB}$

$6\pi =\frac{v\times \cos {45}^{\circ }\times 2\pi \times 3\times {10}^{-3}}{0.02\times 0.3}$

$\Rightarrow v=6\sqrt{2}\text{cm}/\text{s}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (a) is the correct answer.