First slide

Force on a charged particle moving in a magnetic field

Question

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

Moderate

A

$\frac{p}{\sqrt{2} π}$
B

$\sqrt{2} p$
C

$\frac{p}{2 π}$
D

$\frac{\sqrt{2} p}{π}$

Solution

$\begin{array}{l} p = \frac{2 πm}{Bq} (v \cos 45^{0}) = \frac{2 πm}{Bq} (v \sin 45^{0}) \\ ∴ \frac{mvsin 45^{0}}{Bq} = \frac{p}{2 π} \end{array}$
= radius of helix

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