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
$p=\left(vcos\theta \right)\times \frac{2\pi m}{qB}$
$\Rightarrow p=\frac{2\pi mvcos45^{\circ }}{qB}$
$\Rightarrow p=\frac{2\pi mv}{\sqrt{2}qB}$
Also radius of helix is,
$r=\frac{mvsin\theta }{qB}$
$\Rightarrow r=\frac{mv}{\sqrt{2}qB}$
$\therefore \frac{r}{p}=\frac{mv}{\sqrt{2}qB}\times \frac{\sqrt{2}qB}{2\pi mv}$
$\frac{r}{p}=\frac{1}{2\pi }$
or $r=\frac{p}{2\pi }$