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Question

A charged particle of charge q and mass m is moving with velocity v (as shown in the figure) in a uniform magnetic field B along negative z – direction. Select the correct alternative(s).

A
Velocity of the particle when it comes out from the magnetic field is v=vcos60^+v sin60^j
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B
Time for which the particle was in magnetic field is πm3qB
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C
Distance travelled in magnetic field is πmv3qB
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D
None of these
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Solution

The correct option is C Distance travelled in magnetic field is πmv3qB
The charge particle q will experience some force when it moves in the region of magnetic field.
The trajectory of this particle in this region can be shown as below


The components of velocity when it exits the magnetic fields are v cos60 and v sin60 along x and y axis respectively.

v=v sin60^i+v cos60^j

Let us consider a point C on y-axis which represents the centre of circular track followed by the particle.

r is the radius of the circular track.

From the figure we can calculate ACB=600
t=θω=π3ω
mv2r=qvB
v=rω
mω=qB
ω=qBm

t=πm3qB

Distance travelled in magnetic field
l=rθ
l=mvqB×π3

l=mvπ3qB

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