Question

A charged particle enters a uniform magnetic field with a velocity vector at an angle of $45^\circ $ with the magnetic field. The pitch of the helical path followed by the particle is $p$ . The radius of the helix will be

• A. $\frac{p}{\sqrt{2} \pi }$
• B. $\sqrt{2}p$
• C. $\frac{p}{2 \pi }$
• D. $\frac{\sqrt{2} p}{\pi }$

Answer 1

Answer: (C)

Pitch of helical path is
$p=\left(v cos \theta \right)\times \frac{2 \pi m}{q B}$
$\Rightarrow p=\frac{2 \pi m v cos 45 ^\circ }{q B}$
$\Rightarrow p=\frac{2 \pi m v}{\sqrt{2} q B}$
Also radius of helix is,
$r=\frac{m v sin \theta }{q B}$
$\Rightarrow r=\frac{m v}{\sqrt{2} q B}$
$\therefore \, \, \, \frac{r}{p}=\frac{m v}{\sqrt{2} q B}\times \frac{\sqrt{2} q B}{2 \pi m v}$
$\frac{r}{p}=\frac{1}{2 \pi }$
or $r=\frac{p}{2 \pi }$