What Is the Current in a Rotating Electron Model of a Hydrogen Atom?

[coffeem]

Hi I have tried the question below. However I am failing at the first hurdle part a! Some help and advice would certaintly be appretiated. Thanks

Taking a simple model of the hydrogen atom as an electron rotating around the nucleus in a circle or radius 0.53ee-10m at a frequency of 0.66ee16Hz, estimate:

a) The look current,

b) The magnitude of the dipole-moment of the loop,

c) The magnitude and ridection of the magnetic field at a distance of 2ee-9m, from the center of the loop.


For part (a) I tried using the formula: I = NevA, however this gave me an answer which was many orders of magnitude wrong lol.

 

[Dick]

The loop current is just the rate at which charge is a passing a point in the loop. The electron passes every point 0.66E16 times per second. How much current is that? I don't know what the formula I=NevA is referring to.

 

[thomate1]

I'm not sure what I'm doing wrong, any advice would be appreciated.

For part (a), you are on the right track by using the formula I = NevA. However, it seems like you may have made a mistake in your calculations. Let's break down the formula and how to use it in this scenario.

I = current (in Amperes)
N = number of turns in the loop (1 in this case)
e = elementary charge (1.6 x 10^-19 Coulombs)
v = velocity of the electron (in meters per second)
A = area of the loop (in square meters)

To calculate the current, we need to find the velocity and area of the loop. The velocity can be calculated using the formula v = ωr, where ω is the angular frequency (0.66 x 10^16 radians per second) and r is the radius of the loop (0.53 x 10^-10 meters). This gives us a velocity of approximately 3.5 x 10^6 meters per second.

Next, we need to calculate the area of the loop. Since the loop is circular, we can use the formula A = πr^2, where r is the radius of the loop. This gives us an area of approximately 8.8 x 10^-20 square meters.

Plugging these values into the formula I = NevA, we get a current of approximately 5.5 x 10^-13 Amperes. This may seem like a small number, but keep in mind that we are dealing with a single electron in a very small loop.

For part (b), we can use the formula μ = NIπr^2, where μ is the dipole moment, N is the number of turns (1 in this case), and r is the radius of the loop. This gives us a dipole moment of approximately 1.5 x 10^-28 Coulomb-meters.

For part (c), we can use the formula B = μ0I/2πr^3, where μ0 is the permeability of free space (4π x 10^-7 Tesla-meters per Ampere), I is the current, and r is the distance from the center of the loop. Plugging in the values, we get a magnetic field of approximately 3.4 x 10^-8 Tesla, directed along the axis of the loop.

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