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Question

A charged particle of specific charge α is released from origin at time t=0 with velocity v=v0(^i+^j) in a uniform magnetic field B=B0^i. Find the coordinates of the particle at time t=πB0α are [α=(q/m)]

A
(v02B0α, 2v0αB0, v0B0α)
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B
(v02B0α, 0, 0)
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C
(0, 2v0B0α, v0π2B0α)
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D
(v0πB0α, 0, 2v0B0α)
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Solution

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

Given,

v=v0(^i+^j);B=B0^i and

The horizontal component of the velocity gives drift in the horizontal direction and vertical component of the velocity is responsible for circular motion of the particle. Thus, it will undergo a helical motion.

Magnetic force on a charge particle is,

F=q(v×B)

F=q[v0(^i+^j)×B0^i]

F=qv0B0^k

Hence, the charge undergoes a circular motion in yz-plane.

Given,

t=πB0α=πB0×qm=mπqB0=T2 [ T=2πmqB]



Where, T is time period of revolution.

Let, the position of the paricle is at coordinate P(x,y,z) at time T2.

Now, for x coordinate,

P(x,0,0)=v×t=v0×T2

=v0πmqB0=v0πB0α

Similarly,

P(0,y,0)=0 and

P(0,0,z)=2×radius of the path=2×mv0qB0=2v0B0α

P(x,y,z)=(v0πB0α , 0 , 2v0B0α)

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (D) is the correct answer.
Why this Question :
To understand the motion of a particle when it is projected at an angle θ with the magnetic field.



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