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Question

A positive charge particle with charge q and mass m is moving with velocity v=v0^i in a uniform magnetic field B=B0^k. The time spent by the particle in the magnetic field if d is equal to mv2qB will be.


A
πmqB
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B
πm4qB
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C
πm3qB
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D
πm6qB
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Solution

The correct option is D πm6qB
Radius of circular path R is, R=mvqB

Given d=mv2qB=R2

d<R , width of magnetic field region is insufficient to describe a complete circle. Hence the particle will leave the magnetic field region as shown in the figure.

Thus, the angle of deflection is θ.

From the figure,

sinθ=dR

sinθ=(R2)R=12

θ=π6

Distance travelled by particle in the magnetic field region is arc SQ substending an angle θ at the center of the circular path.

Time spent in magnetic field region is;

t=(2πmqB)(θ2π)

or, t=(2πmqB)×[π6(2π)]

t=πm6qB

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

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