Question

A charged particle enters a uniform magnetic field with 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)

$p=\frac{2 \pi m}{B q}\left(v \cos 45^{\circ}\right)=\frac{2 \pi m}{B q}\left(v \sin 45^{\circ}\right)$
$\therefore \frac{m v \sin 45^{\circ}}{B q}=\frac{p}{2 \pi}=$ radius of helix