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Question

A charged particle of mass m and charge q is accelerated through a potential difference of V. It enters a region of uniform magnetic field B, which is directed perpendicular to the direction of motion of the particle. The particle will move on a circular path of radius given by -

A
Vm2qB2
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B
Vm3qB2
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C
2Vmq×(1B)
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D
Vmq×(1B)
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Solution

The correct option is C 2Vmq×(1B)
When a charged particle enters perpendicular to a uniform magnetic field, then it follows a circular path.

Its radius is given by,

R=mvqB=2mKEqB ...(1)

The given charged particle q is accelerated through a potential difference of V.

KE=qV ...(2)

From equations (1) and (2),

R=2mqVqB

R=2Vmq×1B

Hence, option (C) is the correct answer.
Why this Question?

This question helps you to understand the concept of motion of a charged particle under the influence of uniform magnetic field.

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