Answer to Question #157135 in Electric Circuits for Ivo

Question #157135
In an experiment to determine the specific charge (e/m) of an electron, electrons where accelerated by an electric field strength of 10N/C and the magnetic field strength of 0.1H is used to deflect them. Find the speed of the accelerated electrons. What is the radius of the path described by the electron?
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Expert's answer
2021-02-04T02:34:27-0500

a) Using the Newton’s Second Law of Motion we can write:


evB=mv2r,evB=\dfrac{mv^2}{r},r=mveB.r=\dfrac{mv}{eB}.

Let's assume that both fields are adjusted so that remains undeflected and arrives at the point at the center of the screen. Then, we can write:


Fe=Fm,F_e=F_m,Bev=eE,Bev=eE,v=EB=Eμ0H=10 NC4π107 Hm0.1 H=7.96107 ms.v=\dfrac{E}{B}=\dfrac{E}{\mu_0H}=\dfrac{10\ \dfrac{N}{C}}{4\pi\cdot10^{-7}\ \dfrac{H}{m}\cdot0.1\ H}=7.96\cdot10^7\ \dfrac{m}{s}.

Then, we can find the radius of the path described by the electron:


r=9.11031 kg7.96107 ms1.61019 C4π107 Hm0.1 H=3602 m.r=\dfrac{9.1\cdot10^{-31}\ kg\cdot7.96\cdot10^7\ \dfrac{m}{s}}{1.6\cdot10^{-19}\ C\cdot4\pi\cdot10^{-7}\ \dfrac{H}{m}\cdot0.1\ H}=3602\ m.

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