Question

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 $P_1(y=-R)$. Which of the following option(s) is/are correct?

• A. For $B=\frac{8}{13}\frac{p}{QR}$, the particle will enter region 3 through the point $P_2$ on x-axis
• B. For $B>\frac{2}{3}\frac{p}{QR}$, the particle will re-enter region 1
• C. When the particle re-enters region 1 through the longest possible path in region 2, the magnitude of the change in its linerar momentum between point $P_1$ and the farthest point from y-axis is $p/\sqrt{2}$
• D. For a fixed B, particles of same charge Q and same velocity $v$, the distance between the point $P_1$ and point of re-entry region 1 is inversely proportional to the mass of the particle

Answer 1

Answer: (A), (B)

For $B>\frac{2}{3}\frac{p}{QR}$, the particle will re-enter region 1

When $B=\frac{8p}{13QR}$

$\frac{m{v}^2}{r}=Qv\left(\frac{8p}{13QR}\right)\therefore r=\frac{13R}{8}$



Also $C{P}_2=\sqrt{C{O}^2+O{P}_2^2}$
$=\sqrt{{\left(\frac{5R}{8}\right)}^2+{\left(\frac{3R}{2}\right)}^2}$

${CP}_2=\frac{13R}{8}$

Thus the particle will enter region 3 through the point P ${}_1$ on X-axis
So, 'A' is the correct option.

For the charge $+Q$ to return region 1 , the radius of the circular path taken by charge should be $3R/2$

$\frac{m{v}^2}{\left(3R/2\right)}=QvB\therefore \frac{2p}{3R}=QB$

$\therefore B=\frac{2p}{3QR}$

i.e., B should be equal or greater than $\frac{2p}{2QR}$

'B'is the correct option.

Change in momentum $=\sqrt{2}p$.
Thus 'C' is incorrect

Further
$\frac{m{v}^2}{r}=qvB\therefore r=\frac{mv}{qB}\text{or}r\propto m$

So 'D' is incorrect.