Question

A proton enter in a uniform magnetic field of 2.0mT at an angle 60° with the magnetic field with speed 10m/s. Find the pitch of path

Answer 1

GIVEN :-

● A proton enters a uniform magnetic field of 2.0mT at an angle 60° with the magnetic field with speed 10m/s.

TO FIND:-

● The pitch of path.

SOLUTION:-

We know ,

$\implies \boxed{p = 2\pi \: mv \cos( \theta)/qB }$

Hence,

$Pitch = \frac{ 2\pi \: \times 1.67 \times 10 {}^{ - 27} }{1.6 \times 10 {}^{ - 19} \times 2} \\ \\ \implies \: 1.6 \times 10 {}^{ - 8} .$

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HOPE IT HELPS :)

Answer 2

A proton enter in a uniform magnetic field of 2mT at an angle 60° with the magnetic field with speed 10 m/s.

we have to find the pitch of path

solution : concept : when a charge particle enter in a uniform magnetic field at an angle with some velocity, path of charge must be helix.

because perpendicular component of velocity creates circular motion whereas parallel component of velocity creates linear motion.

now pitch of helical path is horizontal distance between two consecutive circles.

i.e., pitch of path = parallel component of velocity × time period

= (vcosθ)(T)

= (vcosθ)(2πm)/(qB)

= 2πmvcosθ/qB

= (2 × π × 1.6 × 10^-27 × 10 × cos60°)/(1.6 × 10^-19 × 2 × 10^-3)

= 50π μm

Therefore the pitch of path is 50π μm