Electrostatics 7 Question 18
20. A uniform magnetic field $B$ exists in the region between $x=0$ and $x=\frac{3 R}{2}$ (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 $P _1(y=-R)$.
(2017 Main) Which of the following option(s) is/are correct?
(a) 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 $\frac{p}{\sqrt{2}}$.
(b) For $B=\frac{8}{13} \frac{p}{Q R}$, the particle will enter region 3 through the point $P _2$ on $X$-axis.
(c) For $B>\frac{2}{3} \frac{p}{Q R}$, the particle will re-enter region 1 .
(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.
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Solution:
- (a)
$|\Delta \mathbf{P}|=\sqrt{2} p$ (b) $r(1-\cos \theta)=R$
$$ r \sin \theta=\frac{3 R}{2} \Rightarrow \frac{\sin \theta}{1-\cos \theta}=\frac{3}{2} $$
$$ \frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin ^{2} \frac{\theta}{2}}=\frac{3}{2} \Rightarrow \cot \frac{\theta}{2}=\frac{3}{2} $$
$\Rightarrow \tan \frac{\theta}{2}=\frac{2}{3} \Rightarrow \tan \theta=\frac{2 \frac{2}{3}}{1-\frac{4}{9}}=\frac{\frac{4}{3}}{\frac{5}{9}}=\frac{12}{5}$
$$ \begin{aligned} \sin \theta & =\frac{12}{13} \\ r \frac{12}{13} & =\frac{3 R}{2} ; r=\frac{13 R}{8}=\frac{P}{Q B} ; B=\frac{8 P}{13 Q R} \end{aligned} $$
(c) $\frac{P}{Q B}<\frac{3 R}{2}, B>\frac{2 P}{3 Q R}$
(d) $r=\frac{m v}{Q B}, d=2 r=\frac{2 m v}{Q B} \Rightarrow d \propto m$
