A uniform magnetic field B exists in the region between x=0 and $x=\frac{3R}{2}$ (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?

$\text{ For particle to re - enter region }1,\text{ radius of circular motion should be less than }\frac{3R}{2}$

$\begin{array}{l}r<\frac{3R}{2}\\ \Rightarrow \frac{mv}{qB}<\frac{3R}{2}\\ \Rightarrow B>\frac{2mv}{3QR}\\ \Rightarrow B>\frac{2p}{3QR}\left[\therefore Option \left(3\right) is correct\right]\end{array}$

$\text{ Now, for complete semi circular path, }$

${\text{ Initial Momentum at P}}_1=p\widehat{i}$

${\text{ Final Momentum at P}}_2=p\widehat{j}$

$\therefore {\text{ Change in momentum from P}}_1\text{ \& }{P}_2=p\widehat{j}-p\widehat{i}$

$\Rightarrow \text{ Magnitude of Change in momentum }=\sqrt{p^2+p^2}=\sqrt{2}p\left[ \therefore Option \left(1\right) is wrong\right]$

$\begin{array}{l}\text{ Distance between the point }{P}_1\text{ and the point of re-entry into region }1=d=2r=\frac{2p}{qB}=\frac{2mv}{qB}\text{}\\ \Rightarrow d\propto m\left[\therefore Option \left(4\right) is wrong\right]\end{array}$

$\begin{array}{l}r\left(1-\cos \theta \right)=R----\left(1\right)\\ \text{}r\sin \theta =\frac{3R}{2}------\left(2\right)\text{}\end{array}$

$\begin{array}{l}\left(2\right) ÷\left(1\right) we get\text{}\\ \frac{r\sin \theta }{r\left(1-\cos \theta \right)}=\frac{\left(\frac{3R}{2}\right)}{R}\text{}\\ \Rightarrow \frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{\sin }^2\frac{\theta }{2}}=\frac{3}{2}\text{}\\ \Rightarrow \tan \frac{\theta }{2}=\frac{2}{3}----\left(3\right)\text{}\end{array}$

$$\begin{gathered} \tan \theta =\frac{2\tan {\displaystyle \frac{\theta }{2}}}{1-{\tan }^2\frac{\theta }{2}}\left(\because \tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }\right) \\ \tan \theta =\frac{2\left({\displaystyle \frac{2}{3}}\right)}{1-{\left(\frac{2}{3}\right)}^2}=\frac{12}{5} \end{gathered}$$

$\begin{array}{l}r\sin \theta =\frac{3R}{2}\text{}\\ \Rightarrow r\left(\frac{12}{13}\right)=\frac{3R}{2}\text{}\\ \Rightarrow r=\frac{13R}{8}\\ \Rightarrow \frac{13R}{8}=\frac{p}{QB}\\ \Rightarrow B=\frac{8p}{13QR}\\ \left[\therefore option \left(2\right) is correct\right]\text{}\end{array}$