An electrical and a quantum question

In summary, the conversation involved solving two problems, one involving a 31-cm-diameter coil with 20 turns of copper wire being exposed to a changing magnetic field, and the other involving finding the voltage needed to produce a specific electron wavelength. In the first problem, the formula for EMF was used to determine the current, with some confusion regarding the units and resistance of copper. In the second problem, the equations for wavelength and kinetic energy were used, but there was uncertainty about the correct equation to use and the final units of the answer. Suggestions were given for finding the correct equation and using symbols to avoid confusion.
  • #1
joej
29
0
1. a 31-cm-diameter coil consists of 20 turns of copper wire 2.6mm in diameter. A uniform megnetic field perpendicular to the plane of the coil, changes at a rate of 8.65 x 10 ^ -3 T/s

What is the current and what is the rate at which thermal energy is produced?


now I figured this would be similar to emf induced in a moving conductor, thus that would give me the emf in volts and I could calculate the current using V = IR, however that dos not seem to give me the correct answer.
I am currently trying to figure out which forumula(s) to use to get an answer to this question... however nothing that I've tried so far seems to work.


---------------------------------------------------

2. What voltage is needed to produce electron wavelength of .10nm? (assume electrons are nonrelativistic)

here I tried to use the following 2 forumulas:

wavelength = Planck's constant / (mass * velocity)

and

eV = .5 mv^2

now I used the 1st formula to to get v = ...

and inserted that into the 2nd formula to get

eV = .5m ((wavelength * m) / (Planck's constant))^2

now I convert from eV to Volts but... answer is way off... where am I going wrong here?
 
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  • #2
Post your numbers for the second problem.I'm sure u may have screwed up the numbers...

Daniel.

P.S.As for the first problem,the conductor is not moving,as,yes,u should apply Faraday's law...
 
  • #3
okay well

h = 6.63 x 10 ^ -34 Js
or
4.14 x 10 ^ -15 eVs

m = 9.11 x 10 ^ -31

wavelength = 1 x 10 ^ -10 m

I get 2.2 x 10^ -82 eV I stop here since this seems way off
 
  • #4
Though i dislike the units for "h",i find that "lambda" is reasonable.Now tell me how did u get the energy that small??

Daniel.
 
  • #5
heh if I knew that I wouldn't be here :P

erm, just plugged it all into the forumla I posted earlier, I don't know why it turned out that small.

as for h, which is Planck's constant, being that small, I just used what was given to us some time ago as the accepted value for it.

should it be different?



-------------------------------

Also tried the 1st problem using the equation:

EMF = (change in flux)/(change in time)

change in flux being BA

so if we divide it by 1 second we have B = 8.65 x 10 ^ -3 T
and surface area of cylinder is (2 * pi * r ^ 2) + (2 * pi * r * height) = 0.151 + 0.051 = 0.202m^2

BA = 1.75 x 10 ^ -3

BA / 1 = 1.75 x 10 ^ -3 V

now... resistance of copper is 1.678 (micro-ohm-cm)

actually on second thought I think I already f*cked up somewhere
 
  • #6
For the second part, I would suggest you try to work completely in symbols until the very last step.

Your answer is in electron-Volts, which is a unit of energy. The question is asking for the voltage. Even taking that into consideration, your answer is off.

Treat the electron as Newtonian (meaning that all the "usual" equations for energy and momentum apply).

Your application of De Broglie's equation is sound. And you're correct to try and find v from the kinetic energy. Use conservation of energy to equate qV (charge*voltage) to the kinetic energy 0.5*m*v^2, and get an expression (in symbols) that relates the Voltage V to h, q (charge of electron), m (mass of electron) and the wavelength. Post the equation here.

You can just plug in the values in the equation and find the answer with less chance of error or confusion.
 
  • #7
joej said:
Also tried the 1st problem using the equation:

EMF = (change in flux)/(change in time)

change in flux being BA

so if we divide it by 1 second we have B = 8.65 x 10 ^ -3 T
and surface area of cylinder is (2 * pi * r ^ 2) + (2 * pi * r * height) = 0.151 + 0.051 = 0.202m^2

BA = 1.75 x 10 ^ -3

BA / 1 = 1.75 x 10 ^ -3 V

now... resistance of copper is 1.678 (micro-ohm-cm)

actually on second thought I think I already f*cked up somewhere

The area you should be using is the cross sectional area of the coil. Just A=pi*r^2.

The flux through a single coil is BA. The flux through n coils is nBA. So, now calculate EMF.

You should be able to calculate the total resistance of the coil, and then calculate the current.
 

Related to An electrical and a quantum question

1. What is the difference between electrical and quantum phenomena?

Electrical phenomena refer to the behavior and properties of electrically charged particles, such as electrons, protons, and ions. On the other hand, quantum phenomena involve the behavior of particles on a subatomic level, where the laws of quantum mechanics govern their behavior.

2. How do electrical and quantum systems interact with each other?

Electrical systems can influence quantum systems through the use of electrical fields, such as in the case of quantum computing. Additionally, quantum systems can also generate electricity, as seen in solar cells and quantum generators.

3. Can electricity be explained using quantum mechanics?

Yes, electricity can be explained using quantum mechanics. Quantum mechanics explains the movement and behavior of electrons and other charged particles, which are responsible for the flow of electricity.

4. How do electric currents behave in quantum systems?

In quantum systems, electric currents can exhibit unique properties such as superposition, where an electron can exist in multiple states simultaneously. Electric currents can also exhibit quantum tunneling, where electrons can pass through barriers that would typically be impossible in classical electrical systems.

5. What are some practical applications of combining electrical and quantum phenomena?

Combining electrical and quantum phenomena has led to the development of technologies such as transistors, which are the foundation of modern electronics. It has also led to advancements in quantum computing, quantum sensors, and quantum communication, which have the potential to revolutionize various industries.

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