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I need to design a procedure for an experiment to determine the concentration and sign of the charge carriers in copper. I'm given:

  1. A slab of copper 2.0 mm thick, 1.5 cm wide, and 4 cm long
  2. One very sensitive digital multimeter
  3. One standard digital multimeter
  4. A DC power supply (any voltage)
  5. Typical clips (banana clips, etc.)
  6. Super glue
  7. Scotch tape

The main confusion that I'm having is how i would be able to create a magnetic field and measure the magnitude. If i did this then to determine the concentration we could use the formula n = -$J_x$$B_y$$/$$q$$E_x$ where $J_x$ is the current density, $B_y$ is the magnetic field, $E_x$ is the electric field, and $q$ is the charge of an electron. Then to determine the sign of the charge carriers we would just look at the Hall coefficient. Which it should be negative, since the mobile charges in metals are negatively charged (electrons).

Thanks in advance for any help, I wanna stress the fact that I don't want to just be given the answer if possible. Just a step in the right direction. Thanks again! :)

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Maybe you don't have to create the magnetic field, you may just use the Earth's magnetic field. As you don't want a complete answer, maybe I should not add anything.

Another possibility: you can use a magnetic field of a current.

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  • $\begingroup$ If we just connected the copper to the DC power supply and set it to a certain amount of amps, and used the Earth's magnetic field would we have all the information needed to calculate the concentration of the charge carriers? $\endgroup$ – Linear Algebra Noob Jun 5 '16 at 3:35
  • $\begingroup$ Oh, I think we'd also need to measure the potential at the bottom of the slab, and at the top, and take the difference between them. $\endgroup$ – Linear Algebra Noob Jun 5 '16 at 3:41
  • $\begingroup$ I am not sure about your formula. I would think the electric field, the magnetic field, and the current should have three different directions. $\endgroup$ – akhmeteli Jun 5 '16 at 3:46
  • $\begingroup$ Hmm, i mean if we assumed a uniform magnetic field with magnitude 0.4 T and we run a 75 A current in the +X direction then the potential at the bottom of the slab is 8.1*10^-7 V higher than at the top. Plugging and chugging with those values makes n= 11.6 * 10^28. Which is close to the actual value of the concentration of the charge carriers in copper. Which is 8.5 * 10^28. I think understand the computation behind this problem, but now I'm just trying to figure out how to design the procedure $\endgroup$ – Linear Algebra Noob Jun 5 '16 at 3:55
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I would use a solenoid. Place two coils above and below the sheet. Something like:

Solenoid

You can simply calculate the field strength from the coil radius, number of turns and the current, and you'll get a pretty accurate result. However you need to multiply the calculated field strength by the magnetic permeability of copper.

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  • $\begingroup$ I also suggested using a magnetic field of a current, but it is not quite clear how to make a solenoid of what the OP is allowed to use. $\endgroup$ – akhmeteli Jun 5 '16 at 8:14
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You're not given a long wire and an additional current source to create a magnetic field, nor a permanent magnet, so you'll have to stick with the Earth magnetic field. You'll have to figure out where the North is and whether the North counts as a magnetic North pole or South pole. That's enough to get the sign of the charge carriers and a ballpark estimate of the concentration. For anything accurate, you'll need to know the exact magnitude and direction of the Earth field with a good accuracy.

If you do have a wire: build a Helmholtz coil, to create a uniform field.

As for the additional current source: you could put the coil in series with the copper plate.

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  • $\begingroup$ How would I be able to figure out where the North is, and from that how do I get an estimate of the concentration? I'm pretty sure the experiment needs to be done only with the equipment provided, so I'm not able to build a Helmholtz coil :/ $\endgroup$ – Linear Algebra Noob Jun 5 '16 at 12:33
  • $\begingroup$ Look out of the window around noon on a sunny day... $\endgroup$ – Han-Kwang Nienhuys Jun 5 '16 at 12:46
  • $\begingroup$ @LinearAlgebraNoob"estimate of the concentration" -- that will be the result of your measurement. $\endgroup$ – Han-Kwang Nienhuys Jun 5 '16 at 13:28
  • $\begingroup$ @LinearAlgebraNoob: you can conduct the experiment with different orientations of the slab and find the orientation where the Hall voltage is maximum. $\endgroup$ – akhmeteli Jun 5 '16 at 13:56

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