Question

$\sf \colorbox{pink} {Today's Question}$
A particle of charge q and mass m is moving with velocity It is subjected to a uniform magnetic field B directed perpendicular to its velocity. Show that it describes a circular path. Write the expression for its radius.


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Answer 1

$\bf \huge \colorbox {pink}{Solution :- }$

When A particle of charge q of mass m is directed to move perpendicular to the uniform magnetic field ′ B ′ with velocity V

The force on the charge

F = q ( v × B )

This magnetic force always perpendicular to the velocity of charged particle . Hence magnitude of velocity means constant but direction charges continuously.

Consequently the part of the charged particle in a perpendicular magnetic field becomes circular. The magnetic force ( qvB ) provide the necessary centripetal force to move along a circular path .Then ,

$\bf qvB \: = \frac{ {mv}^{2} }{τ} ⇒ \: r = \frac{mv}{qB}$

Here r = radius of the circular path followed by the charge .

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Answer 2

Answer:

When a particle of charge ‘q’ of mass ‘m’ is directed to move perpendicular to the uniform magnetic field ‘B’ with velocity ‘vector v' . The force on the charge This magnetic force acts always perpendicular to the velocity of charged particle. Hence magnitude of velocity remains constant but direction changes continuously. Consequently the path of the charged particle in a perpendicular magnetic field becomes circular. The magnetic force (qvB) provides the necessary centripetal force to move along a circular path. Here r = radius of the circular path followed by the charge.

See the above image for reference

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