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An electron, moving along the x-axis with an initial energy of 100 eV, enters a region of the magnetic field $\vec{B}=(1.5\times 10^{-3}T)\hat{k}$ at S(See figure). The field extends between x=0 and x=2 cm. The electron is detected at the point Q on a screen placed 8cm a way from the point S. The distance d between P and Q (on the screen) is:

(electron's charge $=1.6\times 10^{-19}C$ , mass of electron $=1.9\times 10^{-31}kg)$

Option 1)
11.65 cm

Option 2)
12.87 cm

Option 3)
1.22 cm

Option 4)
2.25 cm

Answers (1)

best_answer

Radius of charged particle -

$r=\frac{mv}{qB}=\frac{P}{qB}=\frac{\sqrt{2mK}}{qB}=\frac{1}{B}\sqrt{}\frac{2mV}{q}$

m=mass of particle

P=momentum of particle

k= kinetic energy of particle

-

$R=\frac{mv}{qB}=\frac{\sqrt{2m(kE)}}{qB}$

$R=\frac{\sqrt{2\times 9.1\times 10^{-31}\times (100\times 1.6\times 10^{-19})}}{1.6\times10^{-19}\times1.5\times10^{-3} }$

R = 2.25 cm

$\sin \Theta =\frac{2}{R}=\frac{2}{2.25}$

here $d=PA+AQ$

$AP=R(1-\cos \Theta )=2.25(1-\cos \Theta )$

$AP\approx 1.22$

Similarly $\tan \Theta =\frac{AQ}{8-2}=\frac{AQ}{6}$

$AQ=6\tan \Theta =11.7$

so $d=AQ+PA$

$d\approx 12.87cm$

Option 1)

11.65 cm

Option 2)

12.87 cm

Option 3)

1.22 cm

Option 4)

2.25 cm

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