SATHEE: Electrostatics 7 Question 18

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 $P_{1} (y = - R)$ . Which of the following option(s) is/are correct?

(2017 Main)

(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.

(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:

$| Δ P | = \sqrt{2} p$

(b) $r (1 - \cos θ) = R$

$r \sin θ = \frac{3 R}{2} \Rightarrow \frac{\sin θ}{1 - \cos θ} = \frac{3}{2}$

$\frac{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \sin^{2} \frac{θ}{2}} = \frac{3}{2} \Rightarrow \cot \frac{θ}{2} = \frac{3}{2}$

$\Rightarrow \tan \frac{θ}{2} = \frac{2}{3} \Rightarrow \tan θ = \frac{2 \frac{2}{3}}{1 - \frac{4}{9}} = \frac{\frac{4}{3}}{\frac{5}{9}} = \frac{12}{5}$

$\begin{aligned} \sin θ & = \frac{12}{13} \\ r \frac{12}{13} & = \frac{3 R}{2}; r = \frac{13 R}{8} = \frac{P}{Q B}; B = \frac{8 P}{13 Q R} \end{aligned}$

(d) $r = \frac{m v}{Q B}, d = 2 r = \frac{2 m v}{Q B} \Rightarrow d \propto m$

Electrostatics 7 Question 18

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Electrostatics 7 Question 18

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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