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  • An election emitted from a hot filament is accelerated through a potential difference of <img alt="12kV" src="h

An election emitted from a hot filament is accelerated through a potential difference of 12kV and enters a region of a uniform magnetic field of 0.5T with certain initial velocity. The trajectory of electron & radius of trajectory which make angle 30^{\circ}.

Option: 1

Circular & 0.37 mm


Option: 2

Helical & 0.37 mm


Option: 3

straight line & 40.37 mm


Option: 4

Parabolic & 0.37 mm


Answers (1)

best_answer

\begin{aligned} & V=12 \mathrm{kV}=12 \times 10^3 \mathrm{~V} \\ \end{aligned}

\begin{aligned} \frac{1}{2} m v^2=e v \Rightarrow v=\left(\frac{2 e v}{m}\right)^{1 / 2} \\ \end{aligned}

\begin{aligned} =\left(\frac{2 \times 1.6 \times 10^{-19} \times 12 \times 10^3}{9 \times 10^{-31}}\right)^{1 / 2} \\ \end{aligned}

\begin{aligned} =\left(4.26 \times 10^{31+3-19}\right)^{1 / 2} \\ \end{aligned}

\begin{aligned} =\left(4.26 \times 10^{15}\right)^{1 / 2} \\ \end{aligned}

\begin{aligned} =\left(42.6 \times 10^{14}\right)^{1 / 2} \\ \end{aligned}

\begin{aligned} =6.52 \times 10^7 \mathrm{~m} / \mathrm{sec} \\ \end{aligned}

The trajectory of the electron is helical

\begin{aligned} r & =\frac{m v}{e B} \times \sin \theta \\ \end{aligned}

\begin{aligned} =\frac{9 \times 10^{-31} \times \sin 30^{\circ} \times 6.52 \times 10^7}{1.6 \times 10^{-19} \times 0.8}=36.675 \times 10^{-5} \mathrm{~m} \\ \end{aligned}

\begin{aligned} r =0.36675 \mathrm{~mm} \end{aligned}

 

 

 

Posted by

jitender.kumar

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