The planetary model of the hydrogen atom consists of an electron in a circular orbit about a proton. The motion of the electron of charge 1.60 × 10−19 C creates an electric current. The radius of the electron orbit is 5.30 × 10−11 m and the electron's velocity is 2.20 × 106 m/s. What is the magnetic field strength at the location of the proton?

Respuesta :

Answer:B=12.52 T

Explanation:

Given

[tex]r=5.30\times 10^{-11} m[/tex]

[tex]v=2.2\times 10^{6} m/s[/tex]

[tex]q=1.60\times 10^{-19} c[/tex]

[tex]Magnetic field(B)=\frac{\nu _0I}{2r}[/tex]

[tex]B=\frac{\mu _0Q}{2r\times \frac{2\pi r}{v}}[/tex]

[tex]B=\frac{4\pi \times 10^{-7}\times 1.6\times 10^{-19}}{2\times 5.30\times 10^{-11}\times 15.138\times 10^{-17}}[/tex]

B=12.52 T

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