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 of magnitude 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

$\frac{q b B}{m}$

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

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$\frac{q a B}{m}$

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

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Solution

The correct option is B
$\frac{q (b - a) B}{m}$

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 $r \geq (b - a)$

Now, $r = \frac{m v}{q B} \geq (b - a)$

or $v \geq \frac{q (b - a) B}{m} \Rightarrow v_{m i n} = \frac{q (b - a) B}{m}$

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