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Question

A particle of mass m and charge q moving with velocity v=qBl2m enters region-II normal to the boundary as shown in the figure. Region-II has a uniform magnetic field B perpendicular to the plane of paper. The particle will
1350768_05943ee1a09b498ca435929e236fd6ee.png

A
Enter the region - III
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B
Not enter the region-III
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C
Complete circular motion in region-II
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D
Move undeflected in region-II
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Solution

The correct option is A Not enter the region-III
Given :A particle of mass m and charge q moving with velocity v=qBl2mv=qBl2m enters region-II normal to the boundary as shown in the figure. Region-II has a uniform magnetic field B perpendicular to the plane of paper.

Solution :
Formula for radius of circular path when a charge enters magnetic field is R=mvqB
And we have the velocity of charge V= qBl2m
when we substitute this value in the formula of radius we get R= l2
It means particle will move in semicircular path and come out with deviation of 180
The correct opt : B

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