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Steel in a magnetic field with a current throught it

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Andy

Thu Mar 12 2015, 06:52AM

Registered Member #4266 Joined: Fri Dec 16 2011, 03:15AM
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Posts: 874

Hi
Wonder if you have a peice of mild steel and attach some Neo/Ferrite magnetics to it, you then try and put current throught it.

What would the effects be to the voltage and current, assuming the magnetic wont move.

Dragon64

Thu Mar 12 2015, 07:10AM

Registered Member #1438 Joined: Sat Apr 12 2008, 12:57AM
Location: Canada
Posts: 218

You mean like this?

Andy

Thu Mar 12 2015, 11:08AM

Registered Member #4266 Joined: Fri Dec 16 2011, 03:15AM
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Yep exactly like that :)

Signification

Thu Mar 12 2015, 03:46PM

Registered Member #54278 Joined: Sat Jan 17 2015, 04:42AM
Location: Amite, La.
Posts: 367

From the diagram, assuming that the current flows lengthwise down the strip from the + battery connected side to the other (like from rear to front in the drawing). AND assume that the magnet's north pole is facing downward (it could be either way, but I just picked one). We now have the magnetic field directed downward from the top of the strip to the bottom (another attracting magnet on the bottom would greatly increase the B-field through the plate). Now when we consider the direction of the current flow across the plate (through the magnetic field) and the direction of the magnetic field through the plate, an interesting thing happens when we apply the right hand rule--the magnetic field causes a deflection of the positive moving charges to the right side of the plate--and negative to the left. Therefore, you have created a voltage difference across the left and right sides of the plate! Which could probably be measured with a sensitive voltmeter. Reversing the battery or the magnetic field should reverse the polarity of this "sideways" voltage.

klugesmith

Thu Mar 12 2015, 04:51PM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716

What Dragon drew and Sig explained is the standard textbook presentation of Hall effect.

The polarity of the transverse (Hall) voltage can tell us
whether the material has positive charges moving in same direction as electric current,
or negative charges moving in opposite direction as electric current.
Semiconductors, electrolytes, and plasmas could have some of each.

I think it would be practically impossible to measure in the setup as drawn.
1) Magnetic field is concentrated in a small fraction of the slab area.
2) Metals have an enormous density of mobile charge, so the drift velocity is very small at ordinary current densities. Compared to copper, steel has much higher electrical and thermal resistivity, which makes it harder to cool for a given current density.

Who wants to run the numbers, as a homework problem?
Let's say the conductor slab is "1 ounce" copper foil (0.0034 cm thick)
1 cm wide, carrying 100 amperes.
(Rule-of-thumb: sheet resistance for 1 oz Cu is 0.5 mOhm/square. So copper strip in this problem will dissipate 5 watts per cm of length.)

What's the electron drift velocity?
With normal B field of 1 tesla, what's the sideways force on a moving electron?
What strength of sideways electric field would generate an equal & opposite force on each electron, so they don't fall off the edge?

Signification

Thu Mar 12 2015, 10:52PM

Registered Member #54278 Joined: Sat Jan 17 2015, 04:42AM
Location: Amite, La.
Posts: 367

klugesmith wrote ...

Let's say the conductor slab is "1 ounce" copper foil (0.0034 cm thick)
1 cm wide, carrying 100 amperes.
(Rule-of-thumb: sheet resistance for 1 oz Cu is 0.5 mOhm/square. So copper strip in this problem will dissipate 5 watts per cm of length.)

What's the electron drift velocity?
With normal B field of 1 tesla, what's the sideways force on a moving electron?
What strength of sideways electric field would generate an equal & opposite force on each electron, so they don't fall off the edge?

I get:
Vd= 0.0216 m/s (2.16 cm/s)
Fb= 3.456EE-21 N
E= 0.0216 N/C

radiotech

Fri Mar 13 2015, 05:13PM

Registered Member #2463 Joined: Wed Nov 11 2009, 03:49AM
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Posts: 1546

By the time a given electron gets to the other end, your tea will be too cool to drink.

The current , however, will get there before the strip melts.

klugesmith

Sat Mar 14 2015, 08:24AM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716

Sig and I got the same answer, and observed a couple of interesting points.

That 1-cm ribbon of 1-oz copper foil can carry 100 A with active (but not heroic) cooling.
At current density J = 29400 A/cm^2, carrier drift velocity v is only 2.16 cm/s.

At B = 1 tesla, the transverse electric field is numerically the same as v in SI units, to make qE=qvxB.
Thus 0.0216 volts per meter, amounting to 216 microvolts across the ribbon.
Not easy to measure, considering that the ohmic voltage drop (along the direction of the current) is 51 millivolts/cm. Hats off to Dr. Hall.

Here is the same answer from another angle. Force on a current-carrying wire in a magnetic field is IlxB. In our case, 100 A x 1 T x 0.01 m = 1 newton for each centimeter of length. 3350 times the wire's weight, for projectile-oriented readers.
For each cm of length, the ribbon has 46 coulombs of mobile charge, which can be kept in line with transverse field of 1/46 volt per meter. 1/4600 volt = 216 microvolts per cm.

There's a nice presentation at

.

I spent an inordinate amount of time polishing a spreadsheet presentation, in which all numbers in non-blue cells are computed from those entered in blue cells:

Andy

Thu Mar 19 2015, 03:00AM

Registered Member #4266 Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874

Thanks Kludgesmith, sorry about the late replie my internet has been down.
Been thinking about this and salt in water as 2e22 charge carries.
Was wondering if if you parallel electrodes would the 100amps that get splitinto say one amp, would the voltage of each electrode add.

Cheers
Andy

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