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aleksandr82 [10.1K]
3 years ago
14

An electron in orbit around a (one proton) nucleus has a speed of about 106 m/s. Deduce the equivalent current due to the motion

of the electron and estimate the magnitude of the magnetic dipole moment of this one electron atom.
Physics
1 answer:
dedylja [7]3 years ago
3 0

Answer:

(a) -4.8\times 10^{-4}A (B) 4.22\times 106{-24}A/m^2

Explanation:

We have given the the speed of the electron =10^6 m/sec

The time period of the electron is given by T=\frac{2\pi r}{v} where r is the Bohr radius, which is equal to 5.29\times 10^{-11}m

So time period T=\frac{2\times \pi \times 5.29\times 10^{-11}}{10^6}=3.324\times 10^{-16}m/sec

(A) We know that current i=\frac{q}{T}=\frac{-e}{T}=\frac{-1.6\times 10^{-19}}{3.324\times 10^{-16}}=-4.8\times 10^{-4}A

(B) Magnetic moment is given by \mu =IA , where I is current and A is area

So \mu =IA=-4.8\times 10^{-4}\times 3.14\times (5.29\times 10^{-11})^2=4.22\times 10^{-24}A/m^2

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