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Question

A charged particle of specific charge, α=(qm) is released at origin at t=0 with velocity v=v0(^i+^j) in a uniform magnetic field B=B0^i. Coordinates of particle at time t=πB0α are:

A
(v0π2B0α,0,2v0B0α)
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B
(v0π2B0α,0,0)
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C
(v0πB0α,0,2v0B0α)
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D
(v0πB0α,0,2v0B0α)
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Solution

The correct option is C (v0πB0α,0,2v0B0α)


Time period t=2πmqB

T=2πB0α [q/m=α]

Given, t=πB0αt=T2

The particle will undergo a circular motion in the xz plane,

At time t=T2, the particle would have reached in the x, y, z directions as follows,

y - co-ordinate =0

x - co-ordinate =vx×T2=v0T2=v0πB0α

In the z - axis, particle would have travelled 2r distance along negative z - axis

So, z- coordinate =2r

=2(mvyqB)=2(v0αB0)

Coordinates of the particle at t=T2,(V0πB0d,0,2v0αB0)

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (c) is the correct answer.

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