1700s

1800s

Coulomb AmpereFranklin Gauss

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

~ 2010 (1012 transistors)~ 1910

Electric field and temperature distributions Laplace equation

~ 1950 (capacity ~ 1 MB) ~ 2000 (capacity ~ 106 MB) ~ 2010 (capacity ~ 105 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.”

𝐸⇔𝐵

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

𝑐=1

√𝜇0𝜖0

q0

Neutral current carrying conductor

𝑖

(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

Bq0

Neutral current carrying conductor

𝑖

Rest frame S

Magnetic field due to the current

𝐵=𝜇0 𝑖

2𝜋𝑟r

Bq0

Rest frame S

Neutral current carrying conductor

𝑖

Neutral no electric field no electrostatic force

�⃑�𝐵=𝑞0 �⃑�× �⃑� no magnetic force

Just for convenience separate the negative charges and positive charges. They are still the same wire.

B

−𝑞 ,𝐿−

Moving frame S’ (a frame moving to the left at the same speed as the negative charges)

q0

Neutral no electric field no electrostatic force

�⃑�𝐵=𝑞0 �⃑�𝐷× �⃑� There is a magnetic force

𝑞 ,𝐿+¿¿

In this frame the charge q0 moves to the right with speed and the positive ions move to the right with the same speed giving rise to the same current i (same magnitude and direction as the rest frame S) and hence the same magnetic field B. So points downwards. This conclusion contradicts observations which show that the charge q0 should just move to right and not move towards the wire.

B

−𝑞 ,𝐿−

𝑞 ,𝐿+¿¿

q0

Neutral no electric field no electrostatic force

�⃑�𝐵=𝑞0 �⃑�𝐷× �⃑� 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

Bq0

−𝑞 ,𝐿−

𝑞 ,𝐿+¿¿

Rest frame S

In the frame S, the negatively charged rod is moving, so the length we observe is contracted from its proper length

𝐿−=𝐿0− √1 −

𝑣𝐷2

𝑐2

on the other hand is the proper length of the positively charged rod since it is at rest in the frame S.

⇒𝐿−=𝐿0−√1−

𝑣𝐷2

𝑐2 =𝐿+¿¿

(1)

(2)

In the rest frame, the wire is neutral. . Hence, if we choose the same amount of negative () and positive () charge then:

B

−𝑞 ,𝐿−′

𝑞 ,𝐿+¿′ ¿

q0

Moving frame S’

In the moving frame S’ we observe the proper length of the negatively charged rod since it is at rest now.

𝐿−′ =𝐿0

−

The positively charged rod contracts in this frame. Since is its proper length we have:

𝐿+¿′=𝐿+¿ √1−

𝑣𝐷2

𝑐 2=𝐿0

− (1 −𝑣𝐷

2

𝑐2 )¿¿ (3) [Using (2)]

B

−𝑞 ,𝐿−

q0

Moving frame S’

−𝑞 ,𝐿−′

𝑞 ,𝐿+¿′ ¿

𝐸= 𝜆′

2𝜋 𝜖0𝑟= 1

2𝜋𝜖0𝑟¿

⇒𝐹 𝐸=𝑞𝑞0

2𝜋𝜖0𝑟 𝐿0− [ 1

(1−𝑣𝐷

2

𝑐2 )−1]= 𝑞𝑞0

2𝜋 𝜖0𝑟 𝐿0− [ 𝑣𝐷

2

𝑐2

(1 −𝑣𝐷

2

𝑐2 ) ]= 𝑞𝑞0

2𝜋 𝜖0𝑟 𝐿0− [ 𝑣𝐷

2 𝜇0𝜖0

(1−𝑣𝐷

2

𝑐2 ) ]= 𝑞𝑞0

2𝜋𝑟 𝐿0− [ 𝑣𝐷

2 𝜇0

(1 −𝑣𝐷

2

𝑐2 ) ]Here we used: 𝑐=

1

√𝜇0𝜖0

points away from the wire since the positive charge density is higher

(4) [Using (3)]

Since charge is conserved we now have, . Hence, the wire is not neutral in the moving frame and there is an electric field. We will now calculate the electric field.

B

−𝑞 ,𝐿−

q0

Moving frame S’

−𝑞 ,𝐿−′

𝑞 ,𝐿+¿′ ¿

⇒𝐹 𝐸=𝑞𝑞0

2𝜋𝑟 𝐿0− [ 𝑣𝐷

2 𝜇0

(1−𝑣𝐷

2

𝑐2 ) ] exactly cancels out the magnetic force in the moving frame. We will show that by calculating the magnetic force.

(5)

B

−𝑞 ,𝐿−

q0

Moving frame S’

−𝑞 ,𝐿−′

𝑞 ,𝐿+¿′ ¿

Current due to negative charges = 0

Current density due to positive charges: 𝑗′=𝑛𝑒𝑣𝐷=

𝑁𝐴𝐿

+¿ ′𝑒𝑣𝐷 ¿n = number of charges per unit volume, N = total number of charges , A = cross section area of the wire

Current due to positive charges: 𝑖′= 𝑗 ′ 𝐴=

𝑁

𝐿+¿ ′𝑒𝑣𝐷=𝑞𝑣𝐷

𝐿+¿ ′¿¿

⇒𝐵=𝜇0 𝑖

′

2𝜋 𝑟=

𝜇0

2𝜋 𝑟𝑞𝑣𝐷

𝐿+¿′ ¿

⇒𝐹 𝐵=𝑞0𝑣𝐷𝐵=𝜇0

2𝜋𝑟𝑞𝑞0𝑣𝐷

2

𝐿+¿ ′=𝜇0

2𝜋 𝑟𝑞𝑞0𝑣𝐷

2

𝐿0−(1−

𝑣𝐷2

𝑐2 )=𝐹𝐸¿

(6)

B

−𝑞 ,𝐿−

q0

Moving frame S’

−𝑞 ,𝐿−′

𝑞 ,𝐿+¿′ ¿

⇒𝐹 𝐵=𝑞0𝑣𝐷𝐵=𝜇0

2𝜋𝑟𝑞𝑞0𝑣𝐷

2

𝐿+¿ ′=𝜇0

2𝜋 𝑟𝑞𝑞0𝑣𝐷

2

𝐿0−(1−

𝑣𝐷2

𝑐2 )=𝐹𝐸¿

points towards the wire and exactly cancels out . Hence in both frames S and S’ there is no force in the vertical direction on the charge consistent with observations.

y

z

�⃗�

y

z

�⃗�𝜃