Franklin
Faraday, (9/22/1791 – 8/25/1867)
Coulomb
1700s
Ampere
Maxwell, (6/13/1831 – 11/5/1879)
1800s
Gauss
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)
1
𝑐=
𝜇0 𝜖0
Einstein, (3/14/1879 – 4/18/1955)
q0
(test charge)
r
Although it looks thick,
the cross section radius
of the wire is very
small compared to r
𝑖
Neutral current carrying conductor
Rest frame S (the lab frame in which the positive ions of the wire are at rest)
Magnetic field due
to the current
𝜇0 𝑖
𝐵=
2𝜋𝑟
q0
B
r
𝑖
Neutral current carrying conductor
Rest frame S
q0
B
Just for convenience
separate the
negative charges
and positive charges.
They are still the
same wire.
𝑖
Neutral current carrying conductor
Neutral  no electric field  no electrostatic force
𝐹𝐵 = 𝑞0 𝑣 × 𝐵  no magnetic force
Rest frame S
B
q0
−𝑞, 𝐿−
𝑞, 𝐿+
In this frame the charge q0 moves to the
right with speed 𝒗𝑫 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
Neutral  no electric field  no electrostatic force
frame S) and hence the same magnetic
field B. So 𝐹𝐵 points downwards. This
𝐹𝐵 = 𝑞0 𝑣𝐷 × 𝐵  There is a magnetic force
conclusion contradicts observations
which show that the charge q0 should
just move to right and not move towards
the wire.
Moving frame S’ (a frame moving to the left at the same speed 𝒗𝑫 as the negative charges)
B
q0
−𝑞, 𝐿−
𝑞, 𝐿+
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
Neutral  no electric field  no electrostatic force
𝐹𝐵 = 𝑞0 𝑣𝐷 × 𝐵  There is a magnetic force
Moving frame S’
q0
B
−𝑞, 𝐿−
𝑞, 𝐿+
In the rest frame, the wire is neutral. ⇒ 𝜆+ = 𝜆− . Hence, if we choose the same amount of negative (−𝑞) and
positive (+𝑞) charge then: 𝐿+ = 𝐿−
In the frame S, the negatively charged rod is moving, so the length we observe is contracted from its proper length 𝐿−0
𝐿− =
𝐿−0
𝑣𝐷2
1− 2
𝑐
(1)
𝐿+ on the other hand is the proper length of the positively charged rod since it is at rest in the frame S.
⇒ 𝐿− = 𝐿−0
𝑣𝐷2
1 − 2 = 𝐿+
𝑐
Rest frame S
(2)
B
q0
−𝑞, 𝐿′−
𝑞, 𝐿′+
In the moving frame S’ we observe the proper length of the negatively charged rod since it is at rest now.
𝐿′− = 𝐿−0
The positively charged rod contracts in this frame. Since 𝐿+ is its proper length we have:
𝐿′+ = 𝐿+
2
𝑣𝐷2
𝑣
𝐷
1 − 2 = 𝐿−0 1 − 2
𝑐
𝑐
Moving frame S’
(3) [Using (2)]
Since charge is conserved we now have,
𝜆′+ > 𝜆′− . Hence, the wire is not neutral in
the moving frame and there is an electric
field. We will now calculate the electric field.
−𝑞, 𝐿′−
q0
B
−𝑞, 𝐿−
𝑞, 𝐿′+
𝜆′
1
1
𝑞
𝑞
𝑞
1
1
𝑞
′
′
𝐸=
=
𝜆 + 𝜆− =
−
=
−
=
2𝜋𝜖0 𝑟 2𝜋𝜖0 𝑟 +
2𝜋𝜖0 𝑟 𝐿′+ 𝐿′−
2𝜋𝜖0 𝑟 𝐿′+ 𝐿′−
2𝜋𝜖0 𝑟
𝑞𝑞0
⇒ 𝐹𝐸 =
2𝜋𝜖0 𝑟𝐿−0
1
𝑣𝐷2
1− 2
𝑐
𝑞𝑞0
−1 =
2𝜋𝜖0 𝑟𝐿−0
1
Here we used: 𝑐 =
𝜇0 𝜖0
𝑣𝐷2
𝑐2
𝑣𝐷2
1− 2
𝑐
𝑞𝑞0
=
2𝜋𝜖0 𝑟𝐿−0
1
𝐿−0
𝑣𝐷2
1− 2
𝑐
𝑣𝐷2 𝜇0 𝜖0
𝑣𝐷2
1− 2
𝑐
1
− −
𝐿0
𝑞𝑞0
=
2𝜋𝑟𝐿−0
(4) [Using (3)]
𝑣𝐷2 𝜇0
𝑣𝐷2
1− 2
𝑐
𝐹𝐸 points away from the wire since the positive
charge density is higher
Moving frame S’
B
−𝑞, 𝐿′−
q0
−𝑞, 𝐿−
𝑞, 𝐿′+
𝑞𝑞0
⇒ 𝐹𝐸 =
2𝜋𝑟𝐿−0
𝑣𝐷2 𝜇0
1−
𝑣𝐷2
𝑐2
𝐹𝐸 exactly cancels out the magnetic force in the moving frame. We will show that by
calculating the magnetic force.
Moving frame S’
(5)
B
−𝑞, 𝐿′−
q0
−𝑞, 𝐿−
𝑞, 𝐿′+
Current due to negative charges = 0
𝑁
Current density due to positive charges: 𝑗 ′ = 𝑛𝑒𝑣𝐷 = ′ 𝑒𝑣𝐷
𝐴𝐿+
𝑁
𝑞𝑣𝐷
′ = 𝑗′𝐴 =
𝑖
𝑒𝑣
=
Current due to positive charges:
𝐿′+ 𝐷
𝐿′+
n = number of charges per unit volume,
N = total number of charges ,
A = cross section area of the wire
𝜇0 𝑖 ′
𝜇0 𝑞𝑣𝐷
⇒𝐵=
=
2𝜋𝑟 2𝜋𝑟 𝐿′+
𝜇0 𝑞𝑞0 𝑣𝐷2
𝜇0
⇒ 𝐹𝐵 = 𝑞0 𝑣𝐷 𝐵 =
=
2𝜋𝑟 𝐿′+
2𝜋𝑟
(6)
𝑞𝑞0 𝑣𝐷2
𝐿−0 1 −
𝑣𝐷2
𝑐2
= 𝐹𝐸
Moving frame S’
B
−𝑞, 𝐿′−
q0
−𝑞, 𝐿−
𝑞, 𝐿′+
𝜇0 𝑞𝑞0 𝑣𝐷2
𝜇0
⇒ 𝐹𝐵 = 𝑞0 𝑣𝐷 𝐵 =
=
2𝜋𝑟 𝐿′+
2𝜋𝑟
𝑞𝑞0 𝑣𝐷2
𝐿−0
𝑣𝐷2
1− 2
𝑐
= 𝐹𝐸
𝐹𝐵 points towards the wire and exactly cancels out 𝐹𝐸 . Hence in both frames S and S’ there is no force in the
vertical direction on the charge 𝑞0 consistent with observations.
Moving frame S’
z
𝐴

y
z
𝜃
𝐴


y