Asked by Stryker
An electron moves on a circular orbit in a uniform magnetic field. The strength of the field is 1.00×10-5 tesla. The kinetic energy of the electron is 178.0 eV. (1 eV = 1 electron volt = 1.6 ×10−19 joules.) Calculate the radius of the orbit.
(electron parameters: the charge is -e where e = 1.602×10-19 C; the mass is m = 9.11×10-31 kg.)
(electron parameters: the charge is -e where e = 1.602×10-19 C; the mass is m = 9.11×10-31 kg.)
Answers
Answered by
Elena
KE =m•v²/2,
v= sqrt(2•Ke/m) = sqrt(2•178•1.6•10^-19/9.11•10^-31) =7.9•10^6.
Lorentz force
F=q•v•B•sinα,
Since the circular orbit, sinα = 1.
mv²/R = q•v•B.
R= m•v/q•B = m•v/e•B =
=9.11•10^-31•7.9•10^6/1.6•10^-19•1•10^5 =4.5•10^-10 m
v= sqrt(2•Ke/m) = sqrt(2•178•1.6•10^-19/9.11•10^-31) =7.9•10^6.
Lorentz force
F=q•v•B•sinα,
Since the circular orbit, sinα = 1.
mv²/R = q•v•B.
R= m•v/q•B = m•v/e•B =
=9.11•10^-31•7.9•10^6/1.6•10^-19•1•10^5 =4.5•10^-10 m
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