Question

In Niels Bohr's 1913 model of the hydrogen atom, an electron circles the proton at a distance of 5.29 ✕ 10-11 m with a speed of 2.19 ✕ 106 m/s. Compute the magnitude of the magnetic field that this motion produces at the location of the proton.

Answer 1

Radius r = 5.29 x10 ^-11 m

Speed v = 2.19 x10 ⁶ m/s

Orbital period of the electron T = 2r / v

T = 2(3.1415)(5.29 x10 ^-11 ) / (2.19 x10 ⁶ )

     = 1.5177 x10 -16 s

Current setup by electron i = q/ T

i = (1.6 x10 ^-19 ) /(1.5177 x10 ^-16 )

    = 1.054 x10 ^-3 A

the magnitude of the magnetic field that this motion produces at the location of the proton i.e., magnetic field at the center of the loop B = i / 2r

                              = (4x10 ^-7 ) (1.054 x10 ^-3 ) /[2( 5.29 x10 ^-11)]

                              = 12.52 T