Question

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?

Answer 1

Answer:B=12.52 T

Explanation:

Given

$r=5.30\times 10^{-11} m$

$v=2.2\times 10^{6} m/s$

$q=1.60\times 10^{-19} c$

$Magnetic field(B)=\frac{\nu _0I}{2r}$

$B=\frac{\mu _0Q}{2r\times \frac{2\pi r}{v}}$

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

B=12.52 T