An electron is accelerated through a potential difference of 5 kV and enters a uniform magnetic field of 0.02 Wb/m${2}$ acting normal to the direction of electron motion.

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written 5.3 years ago by

vermavarsha432 • 1000

modified 2.2 years ago by

DevarenjiniP • 3.1k

engineering physics

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written 2.2 years ago by

DevarenjiniP • 3.1k

• modified 2.2 years ago

$$ \begin{array}{l} v=\frac{\sqrt{2 q V}}{m}=\frac{\sqrt{2 \times 1.6 \times 10^{-19} \times 5000}}{9.1 \times 10^{-31}}=4193 \times 10^{4} \ r=\frac{m v}{q B}=\frac{9.1 \times 10^{-31} \times 4193 \times 10^{4}}{1.6 \times 10^{-19} \times 0.02}=0.0119 m \end{array} $$ Radius of path $=0.0119 \mathrm{~m}$

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