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Question

A particle of mass m and charge q enters with velocity v0 perpendicular to a magnetic field B (coming out of the plane of the paper) as shown in the figure. It moves in the magnetic field for time t=πm4qB, and then enters into a constant electric field region. The electric and magnetic fields are present only in a rectangular region of thickness d. The length of rectangular region is l. The particle enters parallel to and grazing the side RQ and leaves the region at P.Given : v0=(2+1)ms1.
EB=8 ms1,l=425v0 metres and mqB=45

Time taken by the particle to cross the electric field region is

A
45s
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B
35s
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C
15s
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D
None of these
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Solution

The correct option is A 45s
The trajectory of particle before it enters the region of electric field can be shown as
Consider a point C to be centre of the circular path followed by the particle.


Displacement of the particle in x direction = r cosθ
We know that t=πm4qB-----(1)
and mv2r=qvB
mω=qB
m×2πT=qB
T=2πmqB------(2)

On comparing (1) and (2) we get
t=T8
T3600
t36008=450

The angle made by velocity when it exits the magnetic region with horizontal =450

Horizontal component of velocity
=v cos450

To calculate time we can say
t×v cos45=lx
t=lr cos450v02
l=452v0-----(Given)

We know that r=mv0qB
and x=r cosθ
x=mv0qB2

Further it is given that v0=(2+1) ms1 and mqB=45
x=45×(2+1)×12
x=45(1+12) m
r cosθ=45(1+12)=452+12=45v02

Now, t=(lrcosθv0)2

t=[452v045v02v0]2

t=45 s

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