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. 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
• D. 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 $p/\sqrt{2}$

Answer 1

Answer: (A), (B)

The correct options are
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

If particle enters regions through $P_2$

Radius = $\sqrt{(\frac{3R}{2}{)}^2+x^2}=R+x$

$\frac{9R^2}{4}+x^2=R^2+x^2+2Rx$

$x=5R/8$

Total radius = $R+x=13R/8$

Radius is given by $\frac{P}{qB}$

Hence $13R/8=\frac{p}{QB}$

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

Particle re-enters region 1 if radius is less than

Hence $\frac{p}{qB}<\frac{3R}{2}$

$B>\frac{2p}{3qR}$