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Question

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


A
qbBm
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B
q(ba)Bm
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C
qaBm
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D
q(b+a)B2m
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Solution

The correct option is B $$\dfrac {q(b-a)B}{m}$$
Radius of curvature has to be greater than the width of magnetic field for the particle to enter the region $$x \gt B$$.

$$ \therefore \dfrac{mv}{qB}  \gt (b-a) $$

$$ \Rightarrow v \gt \dfrac{qB(b-a)}{m}$$

$$ \therefore v_{min} = \dfrac{qB(b-a)}{m}$$

Physics

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