JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Magnetic Effects of Current

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?

1. (A)For $B>\frac{2}{3}\frac{p}{QR}$, the particle will re-enter region 1✓ Correct
2. (B)For $B=\frac{8}{13}\frac{p}{QR}$, the particle will enter region 3 through the point $P_2$ on $x$-axis✓ Correct
3. (C)When 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 the farthest point from $y$-axis is $p/\sqrt{2}$
4. (D)For 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

Answer: (A), (B)

Solution

The radius of circular path is $R^{\prime }=\frac{p}{QB}$. For the particle to reach $P_2$ on the x-axis from $P_1(y=-R)$, using geometry: $(R^{\prime }-R{)}^2+R^2=R^{\prime 2}$, which gives $R^{\prime }=\frac{13R}{8}$. Therefore $B=\frac{8p}{13QR}$. For the particle to re-enter region 1, $R^{\prime }<\frac{3R}{2}$, which requires $B>\frac{2p}{3QR}$. Options (C) and (D) are incorrect based on momentum change analysis and mass dependence.

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · One or more correct options