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Current
The problem:
Electrons (charge −e, mass m) and protons (charge e, mass M ) with equal number densities n are
accelerated by the homogeneous constant electric field from the rest. There is a friction force of
form
F~e = −νe m~ve
F~i = −νi M~vi
(1)
(2)
where ~νi , ~νe are friction coefficients. Find the current density when the velocities do not change any
longer .
The solution:
The second law:
~ − νe m~ve
m~ae = −eE
~ − vi M~vi
M~ai = eE
(3)
(4)
If the velocities do not change the accelerations are zero.
e ~
E
νe m
e ~
E
νi M
~ve = −
(5)
~vi =
(6)
The current density is
J~ = en~vi + (−e)n~ve
1
1
~
J~ = ne2 E
+
vi M
ve m
(7)
(8)
1
Current
Submitted by: I.D. 065944332
The problem:
The space between two concentric spheres, r1 > r2 is filled with a conductor with the resistivity ρ.
What is the resistance between the inner and outer surfaces.
The solution:
The solution is given also in e 38 2 167.
Define the conductivity σ = 1/ρ.
First way
Suppose that there is a potential V . It gives rise to a radial current I, and then R = V /I by the
Ohm’s law.
Z R2
~ · d~r
V = −
E
(1)
R1
~ =
E
1~
J
σ
(2)
Then remembering that J = I/A for homogeneous current we get
Z
1 R2
Jdr
V =
σ R1
Z
1 R2 I
I R2 − R1
=
dr =
σ R1 4πr2
4πσ R1 R2
(3)
(4)
Therefore,
R=
V
1 R1 − R2
=
I
4πσ R1 R2
(5)
Second way
L
For a wire of a length L, area A and conductivity σ the resistance is R = σA
.
Since the current in our case is radial, we have many infinitecimal thin spheres in series.
dr
dr
=
σA
σ4πr2
Z
Z R2
R =
dR =
dR =
R1
(6)
dr
1 R1 − R2
=
2
σ4πr
4πσ R1 R2
1
(7)
Current density
The problem:
t
Current density is given by J~ = J cos2 θe− τ rb (spherical coordinates).
Find the charge inside the sphere with the radius R as a function of time if the initial charge was
Q0 .
The solution:
Z2π
I=
Z2π Zπ
t
4πJR2 − t
dϕ R sin θdθJ (θ, ϕ) = dϕ R2 sin θdθJ cos2 θe− τ =
e τ
3
0
Zπ
2
0
0
(1)
0
The current is directed along the positive rb axis (away from the charge inside the sphere).
Therefore,
dQ
= −I
dt
(2)
From here
Zt
Q (t) = Q0 −
0
Zt Idt = Q0 +
0
4πJR2 − t0
e τ
−
3
dt0 = Q0 −
t
4πJR2 τ 1 − e− τ
3
(3)
0
Finally,
Q (t) = Q0 −
t
4πJR2 τ 1 − e− τ
3
(4)
1