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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 ?

Askiitians Expert Soumyajit IIT-Kharagpur
28 Points
11 years ago

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)

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Regards,

Soumyajit Das IIT Kharagpur

Rishi Sharma
askIITians Faculty 646 Points
one year ago
Dear Student,
Please find below the solution to your problem.

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 isless 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

Thanks and Regards