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

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$\frac{q (b + a) B}{2 m}$

$\frac{qaB}{m}$

$\frac{q (b - a) B}{m}$

$\frac{qbB}{m}$

answer is B.

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Detailed Solution

In the figure, the z-axis points out of the paper, and the magnetic field is directed into the paper, existing in the region between PQ and RS. The particle moves in a circular path of radius r in the magnetic field. It can just enter the region x > b for $r \geq (b - a)$

$\begin{aligned} Now, r = \frac{m v}{q B} \geq (b - a) \\ or v \geq \frac{q (b - a) B}{m} \Rightarrow ν_{min} = \frac{q (b - a) B}{m} \end{aligned}$

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