A uniform magnetic field B exists in the region between x = 0 and x...

A uniform magnetic field B exists in the region between $x = 0$ and $x = \dfrac{3R}{2}$ (region 2 in the figure) pointing normally into the plane of the paper. A particle with charge $+ Q$ and momentum p directed along $x$ -axis enters region 2 from region 1 at point ${P}_{1}\left(y = - R\right) .$ Which of the following option(s) is/are correct?
A uniform magnetic field B exists in the region between x = 0 and x = (3R)/(2) (region 2 in the figure) pointing

Options

AFor $B = \dfrac{8}{13}\dfrac{p}{QR}$ , the particle will enter region 3 through the point ${P}_{2}$ on $x$ -axis
BFor $B > \dfrac{2}{3}\dfrac{p}{QR}$ , the particle will re-enter region 1
CFor a fixed B, particles of same charge Q and same velocity $v$ , the distance between the point ${P}_{1}$ and the point of re-entry into region 1 is inversely proportional to the mass of the particle
DWhen the particle re-enters region 1 through the longest possible path in region 2, the magnitude of the change in its linear momentum between point ${P}_{1}$ and farthest point from $y$ -axis is $\dfrac{p}{\sqrt{2}}$

Correct answer

A. For $B = \dfrac{8}{13}\dfrac{p}{QR}$ , the particle will enter region 3 through the point ${P}_{2}$ on $x$ -axis

Step-by-step solution

A uniform magnetic field B exists in the region between x = 0 and x = (3R)/(2) (region 2 in the figure) pointing

For $B = \dfrac{8}{13}\dfrac{p}{QR},$ radius of path

${R}^{'} = \dfrac{p}{O . B} = \dfrac{p \times 13QR}{Q8p} = \dfrac{13}{8}R$

Using pythagorous theorem, $y = \dfrac{5R}{8}$

$∴$ particle will enter region 3 through point ${P}_{2}$ for $B > \dfrac{2}{3}\dfrac{p}{QR}$

Radius of path $< \dfrac{3PQR}{Q2p} = \dfrac{3}{2}R$

$∴$ Particle will not enter in region 3 and will re-enter region 1

Charge on momentum $= \sqrt{2}p .$ When particle enters region 1 between entry point and farthest point from y-axis.

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