1700s 1800s CoulombAmpereFranklin Gauss Faraday, (9/22/1791 – 8/25/1867)Maxwell, (6/13/1831 – 11/5/1879) Einstein, (3/14/1879 – 4/18/1955)

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Presentation transcript:

1700s 1800s CoulombAmpereFranklin Gauss Faraday, (9/22/1791 – 8/25/1867)Maxwell, (6/13/1831 – 11/5/1879) Einstein, (3/14/1879 – 4/18/1955)

~ 2010 (10 12 transistors)~ 1910 Electric field and temperature distributionsLaplace equation

~ 1950 (capacity ~ 1 MB)~ 2000 (capacity ~ 10 6 MB)~ 2010 (capacity ~ 10 5 MB) Magnetic fields in matter, B and H fields

“The connection between light and electricity is now established... In every ﬂame, in every luminous particle, we see an electrical process... Thus, the domain of electricity extends over the whole of nature. It even affects ourselves intimately: we perceive that we possess... an electrical organ—the eye.”

Faraday, (9/22/1791 – 8/25/1867)Maxwell, (6/13/1831 – 11/5/1879) Einstein, (3/14/1879 – 4/18/1955)

q0q0 Neutral current carrying conductor (test charge) Rest frame S (the lab frame in which the positive ions of the wire are at rest) r Although it looks thick, the cross section radius of the wire is very small compared to r

B q0q0 Neutral current carrying conductor Rest frame S Magnetic field due to the current r

B q0q0 Rest frame S Neutral current carrying conductor Neutral  no electric field  no electrostatic force  no magnetic force Just for convenience separate the negative charges and positive charges. They are still the same wire.

B q0q0 Neutral  no electric field  no electrostatic force  There is a magnetic force

B q0q0 Neutral  no electric field  no electrostatic force  There is a magnetic force Moving frame S ’ Relativity resolves this apparent paradox. First think of the moving charges as embedded in a rod with the rod moving with the same speed as the charges

B q0q0 Rest frame S (1) (2)

B q0q0 Moving frame S ’ In the moving frame S ’ we observe the proper length of the negatively charged rod since it is at rest now. (3) [Using (2)]

B q0q0 Moving frame S ’ Here we used: (4) [Using (3)]

B q0q0 Moving frame S ’ (5)

B q0q0 Moving frame S ’ Current due to negative charges = 0 Current density due to positive charges: n = number of charges per unit volume, N = total number of charges, A = cross section area of the wire Current due to positive charges: (6)

B q0q0 Moving frame S ’

y z 

y z  