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```				   A particle of mass m and charge q moves with a constant velocity v along the positive x direction. It enters the region containing a uniform magnetic field B directed along the negative z direction , extending from x=a to x=b. The minimum value of v(velocity) required so that particle can just enter the region x>b ?
```

6 years ago

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```										Dear Ankita Shakya,
Ans:- The force on a charged particle is F=q(v×B) Now v=in + X direction and B= -Z direction hence it will rotate in the X Y plane . Let the radious is r Hence Bqv=mv²/r
So r=mv/Bq Now if the value of r is less than (b - a) then it will continue to rotate in this region as it will not be able to reach the point x=b
Hence r>b-a
or mv/Bq>(b - a)
or v>Bq(b -a)/m  (Ans)

Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.
All the best Ankita Shakya !!!

Regards,
```
6 years ago

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