Question

Derive an expession for the pitch of the helix in in magnetism
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Answer 1

The magnetic force acting on the particle of charge +q is,

F= q(v×B)

where v= velocity of a charged particle perpendicular to the magnetic field.

B= Magnetic field

The magnetic force acts perpendicular to both the velocity and magnetic field, the magnitude of magnetic force is

F=qvB

Centripetal force acting is,

$f = \frac{m {v}^{2} }{r}$

where r= The radius of the circular path of the particle

Equating the two, we get,

$f = m \times \frac{ {v}^{2} }{r} = qvb \:$

$r = \frac{mv}{qb}$

The Time period to complete one oscillation, T=

$t = \frac{2\pi \: r}{v}$

The distance moved by the particle along the direction of the magnetic field in one rotation is given by its pitch.

$t = \frac{2\pi \: m}{qb}$

$p = v't$

$p = \frac{2\pi \: m}{qb}$

= velocity parallel to magnetic field.

$bts (◍•ᴗ•◍)❤ exo$