Question #51122

In the Bohr model of hydrogen atom, the electron follows a circular orbit centred on the
nucleus containing a proton. The motion of the electron along the circular orbit
constitutes a current. Calculate the magnetic field produced by the orbiting electron at the
site of the proton.

Expert's answer

Use Bohr's law


me⋅Vn⋅rn=n⋅h2⋅πm _ {e} \cdot V _ {n} \cdot r _ {n} = \frac {n \cdot h}{2 \cdot \pi}me⋅ωn⋅rn2=n⋅h2⋅πm _ {e} \cdot \omega_ {n} \cdot r _ {n} ^ {2} = \frac {n \cdot h}{2 \cdot \pi}


Acceleration of electron


an=Vn2rn=ωn2⋅rna _ {n} = \frac {V _ {n} ^ {2}}{r _ {n}} = \omega_ {n} ^ {2} \cdot r _ {n}qe24π⋅ε0⋅rn2⋅me=ωn2⋅rn\frac {q _ {e} ^ {2}}{4 \pi \cdot \varepsilon_ {0} \cdot r _ {n} ^ {2} \cdot m _ {e}} = \omega_ {n} ^ {2} \cdot r _ {n}


Solve this equations


ωn=12⋅n3⋅h3⋅meε02⋅qe4⋅π\omega_ {n} = \frac {1}{2 \cdot n ^ {3} \cdot h ^ {3}} \cdot \frac {m _ {e}}{\varepsilon_ {0} ^ {2}} \cdot q _ {e} ^ {4} \cdot \pirn=n2⋅h2⋅ε0qe2⋅me⋅πr _ {n} = n ^ {2} \cdot h ^ {2} \cdot \frac {\varepsilon_ {0}}{q _ {e} ^ {2} \cdot m _ {e} \cdot \pi}


The electron current is


I=qe⋅ωn2⋅πI = q _ {e} \cdot \frac {\omega_ {n}}{2 \cdot \pi}I=14⋅qe5n3⋅h3⋅meε02I = \frac {1}{4} \cdot \frac {q _ {e} ^ {5}}{n ^ {3} \cdot h ^ {3}} \cdot \frac {m _ {e}}{\varepsilon_ {0} ^ {2}}


Use Biot-Savart law:


B=μ0I2rnB = \mu_ {0} \frac {I}{2 r _ {n}}B=18μ0qe7n5h5me2ε03πB = \frac {1}{8} \mu_ {0} \frac {q _ {e} ^ {7}}{n ^ {5} h ^ {5}} \frac {m _ {e} ^ {2}}{\varepsilon_ {0} ^ {3}} \pi


Answer is 12.5 T

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