# Homework Help: An electrical and a quantum question

1. Jan 18, 2005

### joej

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.

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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?

2. Jan 18, 2005

### dextercioby

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. Jan 18, 2005

### joej

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. Jan 18, 2005

### dextercioby

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. Jan 18, 2005

### joej

heh if I knew that I wouldn't be here :P

erm, just plugged it all into the forumla I posted earlier, I dunno 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?

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Also tried the 1st problem using the equation:

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

change in flux being BA

so if we devide 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. Jan 18, 2005

### Curious3141

For the second part, I would suggest you try to work completely in symbols until the very last step.

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. Jan 19, 2005

### learningphysics

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.