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)=1.6×106C(e)=1.6\times 10^{-6}C

B=1.5×103TB=1.5 \times 10^{-3}T

We know that,

qv×B=mv2Rqv\times B=\frac{mv^2}{R}


qv×B=mv2R\Rightarrow qv\times B=\frac{mv^2}{R}


R=1qB2m2v22\Rightarrow R=\frac{1}{qB}\sqrt{\frac{2m^2 v^2}{2}}


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


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


R=2×9.1×1031×4501.6×1019×1.5×103=8190×10311.89×103\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}}


R=28.61×10151.89×103m\Rightarrow R=\frac{28.61\times 10^{-15}}{1.89\times 10^{-3}}m

R=15.13×1012m\Rightarrow R = 15.13\times 10^{-12} m


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