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)

 

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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