Answer to Question #127839 in Field Theory for Harris

Question #127839
An electric having 450v of energy move at right angle to a unifoy magnetic field fluid density 1.50x10-3 show that the path of the electron is circle and find its rad
1
Expert's answer
2020-07-30T11:02:39-0400

As per the given question,

Applied potential (V)=450V

Charge on electron "(e)=1.6\\times 10^{-6}C"

"B=1.5 \\times 10^{-3}T"

We know that,

"qv\\times B=\\frac{mv^2}{R}"


"\\Rightarrow qv\\times B=\\frac{mv^2}{R}"


"\\Rightarrow R=\\frac{1}{qB}\\sqrt{\\frac{2m^2 v^2}{2}}"


"\\Rightarrow qv\\times B=\\frac{2eV}{R}"


"\\Rightarrow R=\\frac{\\sqrt{2 m_e V}}{\\sqrt{e}\\times B}"


"\\Rightarrow R = \\frac{\\sqrt{2\\times 9.1 \\times 10^{-31}\\times 450}}{\\sqrt{1.6\\times 10^{-19}}\\times 1.5\\times 10^{-3}}=\\frac{\\sqrt{8190\\times 10^{-31}}}{1.89\\times 10^{-3}}"


"\\Rightarrow R=\\frac{28.61\\times 10^{-15}}{1.89\\times 10^{-3}}m"

"\\Rightarrow R = 15.13\\times 10^{-12} m"


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