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What is the radius of the path of an electron (mass  \(9\times 10^{-31}kg\) and charge  \(1.6\times 10^{-19}C\) ) moving at a speed of  \(3\times 10^7m/s\)  in a magnetic field of  \(6\times 10^{-4}T\)  perpendicular to it? What is its frequency? Calculate its energy in \(keV\). ( \(1\mathit{eV}=1.6\times 10^{-19}\ J\)).

Answer

Radius\(=28\ cm\)
Frequency\(=18\ \mathit{MHz}\)
Energy\(=2.5\mathit{ke}V\)
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Solution

Given:
 \(m=9\times 10^{-31}kg\)
 \(q=1.6\times 10^{-19}C\)
 \(v=3\times 10^7m/s\)
 \(B=6\times 10^{-4} T\)
Radius of the path of an electron:
As we know that the radius of the circular path of a particle i.e. \(r=mv/qB\)
Using above equation we find:
 \(r=\frac{mv}{qB}\)
 \(=\frac{9\times 10^{-31}\mathit{kg}\times 3\times 10^7m/s}{1.6\times 10^{-19}C\times 6\times 10^{-4}T}\)
 \(=28\times 10^{-2}m=28\ cm\)
Frequency:
 \(\nu=\frac v{2\pi r}\)
\(\nu=\frac {3\times 10^7}{2\times 2.14 \times 28\times 10^{-2}}\)
\(=18\times 10^6Hz\)
\(=18\ \mathit{MHz}\)
Energy:
 \(E=\frac 1 2mv^2\)
 \(=\frac 1 2\times 9\times 10^{-31}\mathit{kg}\times 9\times 10^{14}m^2/s^2\)
 \(=40.5\times 10^{-17}J\)
 \(=4\times 10^{-16}J=2.5\mathit{ke}V\) 
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