Eletrodynamics problems

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Physics 2. Electromagnetism
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Problem 1. A semi-infinite cylinder of the radius R is uniformly charged on its surface so that
the charge per unit length is λ. Find the electric field in the center of the base. Answer:E = kλ/R.
Solution. Let (0, 0, 0) be the center of the base, and z be along the axis. Let σ be the surface
charge density, σ = λ/2piR. Then
E =
∫
k(r − r′)dq′
|r − r′|3
r = (0, 0, 0), r′ = (R cosϕ′, R sinϕ′, z′)
dq′ = σRdϕ′dz′
|r − r′| = (R2 + z′2)1/2
Ex = Ey = 0
Ez = −
∫ ∞
0
2pikσRz′dz′
(R2 + z′2)3/2
Ez = −2pikσ = −kλ/R
Problem 2. A hemisphere is uniformly charged on its surface with the surface charge density σ.
Find the electric field in the center. Answer:E = kpiσ.
Solution.
E =
∫
k(r − r′)dq′
|r − r′|3
r = (0, 0, 0), r′ = (R sin θ′ cosϕ′, R sin θ′ sinϕ′, R cos θ′)
dq′ = σR2 sin θ′dθ′dϕ′
|r − r′| = R
Ex = Ey = 0
Ez = −
∫ pi/2
0
2pikσR3 sin θ′ cos θ′dθ′
R3
Ez = −pikσ
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Physics 2. Electromagnetism More problems
Problem 3. A conical surface with the radius of the basis R is uniformly charged with the
surface density σ. Find the potential at the top. Answer:φ = 2pikσR.
Solution. Let us choose coordinates so that z be along the cone axis. Let H be the cone height,
0 ≤ z ≤ H, and α be the opening angle, R = H tanα. Then
φ =
∫
kdq′
|r − r′|
dq′ = 2piz tanασdl = 2piσl sinαdl
l = |r − r′| = z/ cosα
φ = 2pikσ
∫ L
0
sinαdl
L = H cosα, R = L sinα
φ = 2pikσR
Problem 4. Find the potential on the edge of a thin disk with the radius R = 20 cm and surface
charge density σ = 0.25 µC/m2. Answer:φ = 4kσR.
Solution. Polar coordinates
φ =
∫
kdq′
|r − r′|
r = (R, 0, 0), r′ = (r′ cosϕ′, r′ sinϕ′, 0)
|r − r′| = (R2 + r′2 − 2Rr′ cosϕ′)1/2
dq′ = σr′dr′dϕ′
Problem 5. Two parallel straight thin strings are uniformly charged with the linear charge
densities λ and −λ. The distance between the strings is l. Find the potential and the electric field
in the whole space. Answer:φ = (2kλl/r) cos θ, E = 2kλl/r2.
Solution. Let the strings be perpendicular to x − y plane and pass through (l/2, 0, 0) and
(−l/2, 0, 0), respectively. Then
φ = 2kλ ln(r1/r0)− 2kλ ln(r2/r0)
r1 = [(r cos θ − l/2)2 + y2], r2 = [(r cos θ + l/2)2 + y2]
φ = 2kλ ln(r1/r2)
For l� r Taylor expansion gives the answers.
Problem 6. Find the electric dipole moment of a thin rod of the length l. The linear charge
density is λ = a(2x− l), where x is the distance from one of the ends, and a is constant. Answer:p =
2
Physics 2. Electromagnetism More problems
al3/6.
Solution.
px =
∫
xdq
=
∫ l
0
λxdx
Problem 7. A system consists of a two concentric conducting spheres (radii a and b, a < b).
The inner sphere is charged with the charge q1. What should be the charge q2 of the outer sphere to
make the potential of the inner sphere φ1 = 0 ? What is then φ(r) ? Answer:q2 = −q1b/a,
φ =
kq1
r
− kq1
a
, a ≤ r ≤ b
=
kq1(a− b)
ar
, r ≥ b
Solution. Superposition.
Problem 8. The plates of a parallel plate capacitor are connected with a conducting wire. There
is a metal plate with the charge q between the plates, parallel to the plates. The metal plate is moved
by a distance l. What charge Q passed in the wire ? Answer:Q = ql/d.
Solution. Let the plate surface be S. Let the charge plate be at the distance x from the capacitor
plate with the charge q′ (d−x from the one with −q′, respectively). The condition that the capacitor
plates have the same potential is
4pik(q′/S)d− 2pi(q/S)x+ 2pi(q/S)(d− x) = 0
q′ = q(d− 2x)/2d
x→ x+ l⇒ q′ → q′ − ql/d
Q = ∆q′ = ql/d
Problem 9. Initially the space between the plates of a parallel plate capacitor is not filled
with anything and the electric field is E0. Then half of the space is filled with a dielectric with the
dielectric constant �, so that the surface of the dielectric is parallel to the plates. Find the electric
field inside the capacitor for the cases when a) voltage is constant, b) charge is constant. Answer:a)
E1 = 2�E0/(�+ 1), E2 = 2E0/(�+ 1); b) E1 = E0, E2 = E0/�.
Solution. Let E1 be the electric field inside the dielectric with �1, and E2 is inside �2. The
boundary conditions for the normal components gives
�1E1 = �2E2
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Physics 2. Electromagnetism More problems
The voltage is
V = E1d/2 + E2d/2
The fields are related to the charge density as �E = 4piσ, and q = σS. In the beginning V0 = E0d.
Compare all this to arrive at the answers above.
Problem 10. Same as in the previous problem but the dielectric surface is perpendicular to the
plates. Answer:a) E1 = E2 = E0; b) E1 = E2 = 2E0/(�+ 1).
Solution. See previous problem. The difference is that here the electric field in tangential, so
that E1 = E2 = E and V = Ed. Attention: charge density is different on the parts of the plate
which are in contact with different dielectrics.
Problem 11. A system consists of two concentric thin spherical conducting envelopes with the
radii R1 and R2, R1 < R2, and charges q1 and q2. Find the electric energy W1 and W2 of each
envelope, the energy of interaction of the two envelopes W12 and the total electric energy W of the
system. Answer:
W = W1 +W2 +W12 =
kq21
2R1
+
kq22
2R2
+
kq1q2
R2
Solution.
φ2 =
k(q1 + q2)
R2
φ1 = φ2 +
kq1
R1
− kq1
R2
W = 1
2
q1φ1 +
1
2
q2φ2
Problem 12. The space between two concentric conducting spheres (radii a and b, a < b),
is filled with a homogeneous weakly conducting medium. The capacity of the system is C. Find
the specific resistivity of the medium if the voltage drops during time T in η times. Answer:ρ =
4piTab/(b− a)C ln η.
Solution.
J = I/4pir2
E = ρJ = ρI/4pir2
V =
ρI
4pi
(
1
a
− 1
b
)
I = q˙ = CV˙
V˙ =
4piab
C(b− a)ρV
V = V0 exp(−t/τ)
τ =
C(b− a)ρ
4piab
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Physics 2. Electromagnetism More problems
V0/V = η ⇒ T/τ = ln(V0/V ) = ln η
Problem 13. Two long parallel wires are in a weakly conducting medium with the specific
resistivity ρ. The distance between the wires is l, the radius of each wire is a, a � l. Find a) the
current density in the point at the distance r from both wires when the potential difference is V ,
b) resistivity of the medium if the length of the wires is L, L � l. Answer:j = lV/2ρr2 ln(l/a),
R = L(ρ/pi) ln(l/a).
Solution. Each wire produces a potential φ = −2kλ ln(r/a), where λ is the linear charge density,
and r is the distance from the wire. The potential difference will then be V = 4kλ ln(l/a). The radial
electric field is Er = 2kλ/r. The total electric field in the point as in the problem E = 2Er(l/2r), and
the current density is j = E/ρ. The total current is I = 2piaLja, where ja = Ea/ρ, and Ea = 2kλ/a.
Thus,
λ = V/4k ln(l/a)
I =
piLV
ρ ln(l/a)
Problem 14. Two cylindrical resistors with the same cross-section but different specific resis-
tivities, ρ1 = 84 nΩ/m and ρ2 = 50 nΩ/m, are brought in contact at their bases. Find the charge
at the interface of the two resistors if the current I = 50 A flows from the first to the second one.
Answer:q = (ρ2 − ρ1)I/4pik.
Solution.
q = σS
=
E2 − E1
4pik
S
=
ρ2j2S − ρ1j1S
4pik
=
(ρ2 − ρ1)I
4pik
Problem 15. Between the plates 1 and 2 of a parallel plate capacitor there is an inhomogeneous
weakly conduction medium. The dielectric constant and specific resistivity vary from �1, ρ1 near
the plate 1 to �1, ρ1 near the plate 2. The capacitor is connected to a constant voltage and the
current I flows from the plate 1 to plate 2. Find the total charge inside the capacitor. Answer:q =
I(�2ρ2 − �1ρ1)/4pik.
Solution.
q =
∫
ρqdV = S
∫
ρqdx
=
S
4pik
∫
d(�Ex)
dx
dx
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Physics 2. Electromagnetism More problems
=
S
4pik
(�2E2 − �1E1)
=
S
4pik
(�2ρ2j2 − �1ρ1j1)
j1 =j2 = I/S
Problem 16. A capacitor, filled with a dielectric with the dielectric constant � = 2.1 looses half
of its charge in τ = 3 min. In the assumption that the loss is due to the conductivity of the dielectric,
find its specific resistivity. Answer:ρ = 4pikτ/� ln 2.
Solution.
E = 4pik�q/S
I = ES/ρ = (4pik�/ρ)q
I = −q˙
Problem 17. Two large parallel plates are in a vacuum. One (cathode) is the source of
electrons with the negligible initial velocity. The electron flow to the other plate results in the
charge distribution in the space between the plates, due to which the potential is φ = ax4/3, where
a = const and x is the distance from the cathode. Find the charge density and the current density.
Answer:ρ(x) = −(1/9pik)ax−2/3, j = (1/9pik)a3/2√2e/m.
Solution.
ρ = − 1
4pik
d2φ
dx2
(1)
j = ρv (2)
v =
√
2eφ/m (3)
Problem 18. Find the current density as a function of r for an axially symmetric parallel electron
flow if B = brα, where b = const > 0 and α = const > 0. Answer:j(r) = (b/4piK)(1 + α)rα−1.
Solution. If j ‖ zˆ then B = Bϕˆ, so that
4piKjz =
1
r
∂rBϕ
∂r
Problem 19. A non-conducting thin disk of the radius R, uniformly charged with the surface
density σ, rotates around its axis with the angular velocity ω. Find the magnetic field in the center
of the disk and the magnetic moment of the disk. Answer:B = 4piKσωR/2, m = piασωR4/4.
Solution. Each ring of the radius r and width dr carries current dI = ωσrdr and contributes
dBz =
2piK
r
dI
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Physics 2. Electromagnetism More problems
dmz = dI · pir2
Problem 20. A non-conducting sphere of the radius R = 50 mm, uniformly charged with the
surface charge σ = 10µK/m2, rotates around its axis with the angular velocity ω = 70 rad/s. Find
the magnetic field in the center of the sphere. Answer:B = 8KσωR/3.
Solution. Spherical coordinates: dI = (ω/2pi)σR2 sin θdθdϕ,
dBz =
2K
R
dI sin2 θ
Problem 21. A small coil K with the number of turns N = 200 is between the poles of the
magnet (see Figure). The area of the coil cross-section is S = 1 cm2, the length of the lever OA is
l = 30 cm. In the absence of the current in the coil the scales are balanced. When the current is
I = 22 mA the balance is restored with the additional mass of M = 60 mg. Find the magnetic field
of the magnet. Answer:B = mgl/NIS.
Solution. The balance is broken because of the torque N = mB on the small coil. This torque
should be balanced by Mgl. Thus
mB = Mgl
m = αNIS
Problem 22. Two long wires with parallel currents of I = 6 A each are moved apart so that
the distance between the two increases η = 2 times. What work per unit length of the wire is done
? Answer:W = 2KI2 ln η.
Solution.
W =
∫
F · dr =
∫ b
a
2KI2
r
dr = 2KI2 ln(b/a)
Problem 23. Two long parallel wires of negligible resistance are connected to a constant voltage.
The circuit is closed with a resistor. The distance between the wire axes is η = 20 times larger than
the radius of the cross-section of each wire. What should be the resistance R for the force between
the wires to vanish ? Answer:R = 4
√
Kk ln η.
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Physics 2. Electromagnetism More problems
Solution. In order to have the voltage the two wires should bear the linear charge densities λ
and −λ such that V = 4kλ ln(d/a), where d is the distance between the wires, and a is the wire
radius, d� a. The electric attractive force will be FE = λlE = 2kλ2l/d, where l is the length of the
wire. The repulsive magnetic force is FB = 2KI
2/d, where I = V/R.
Problem 24. A constant current I = 14 A flows along a long straight conductor with the
cross-section in the form of a thin semi-circle of the raduis R = 5 cm. Same current flows in the
opposite direction along the axis of the circle. Find the force of the interaction per unit length of
the conductors. Answer:F = 4KI2/piR.
Solution.
dFy = −2KI1(I2/pi)dϕ sinϕ
Fy = −(2KI2/pi)
∫ 2pi
pi
sinϕdϕ
Problem 25. A system consists of two parallel planes with the currents which make a homoge-
neous magnetic field B between the planes. The magnetic field outside is zero. Find the force per
unit area on each planes. Answer:B2/8pik.
Solution.
dF
dV
= αJ ×B
dV = dSdx
dF
dS
=
∫
αJ ×Bdx
J =
1
4piK
rotB =
1
4piK
xˆ× ∂B
∂x
dF
dS
= −xˆ
∫
d
dx
(
B2
8piK
)dx
Problem 26. A metal disk of the radius a = 25 cm is rotated with the angular velocity ω = 130
rad/s around its axis. Find the potential difference between the center and the edge for the cases
when a) there is no magnetic field, b) magnetic field of B = 5 mT is perpendicular to the disk
surface. Answer:a) V = ω2a2me/2e; b) V = ωBa
2/2.
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Physics 2. Electromagnetism More problems
Solution. a) Centrifugal force mω2r should be balanced by electric force eE:
E = mω2r/e
V =
∫
Edr
b) Magnetic force eωrB should be balanced by electric force:
E = ωrB
V =
∫
Edr
Problem 27. A circular contour with the area S and resistivity R is rotated with the angular
velocity ω around one of its diameters, which is perpendicular to the homogeneous magnetic field B.
Neglecting the induced magnetic field find the required torque N(t). Initially the contour plane is
perpendicular to the magnetic field. Answer:N = (αωS2B2/R) sin2 ωt.
Solution.
N = |mB sin θ‖ θ = ωt
m = IS
I = −E/R = αΦ˙/R
Φ = BS cos θ = BS cosωt
Problem 28. Find the inductance per unit length of the two thin coaxial metal cylinders if the
outer cylinder radius is time η = 3.6 larger than the radius of the inner one. Answer:L1 = 2K ln η.
Solution.
B = 2KI/r
Φ = l
∫ b
a
Bdr = 2KIl ln η
Problem 29. Given two immobile contours with the mutual inductance L12. In one of the
contours the current I1 = at, where a = const and t is time. What is the current in the second
contour if its self-conductance L22 and resistivity R ? Answer:I2 = −[1− exp(−tR/L22)]aL12/R.
Solution.
E2 = −αΦ˙21 − αΦ˙22
= −L21I˙1 − L22I˙2
= −L21a− L22I˙2 = I2R
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Physics 2. Electromagnetism More problems
Problem 30. A coaxial cable consists of a central wire of the radius a and outer thin conductive
envelope of the radius b. Find the inductance per unit length if the distribution of the current inside
the central wire is homogeneous. Answer:L1 = 2K[1/4 + ln(b/a)].
Solution. Using UI = LI
2/2:
B = 2KI/r, a < r < b
B = 2KIr/a2, 0 ≤ r ≤ a
U =
∫
B2
8piK
dV
U
l
=
∫
B2
4K
rdr
= KI2[
∫ a
0
(r2/a4)rdr +
∫ b
a
dr/r]
= KI2[1/4 + ln(b/a)]
Problem 31. Two identical coils, of inductance L each, are connected in series and placed so that
the magnetic flux of one fully penetrates the other. Find the inductance of the system. Answer:4L.
Solution.
Φ11 = Φ12 = Φ21 = Φ22 = LI/α
L′I = α(Φ11 + Φ12 + Φ21 + Φ22) = 4LI
Problem 32. Two coils of the same length and almost the same cross-section are put one inside
the other. Find their mutual inductance if the self-inductances are L1 and L2. Answer:L12 =
√
L1L2.
Solution.
B1 = (4piKN1/l)I1
Φ11 = N1SB1 = (4piKN
2
1S/l)I1
Φ21 = N2SB1 = (4piKN1N2S/l)I1
L1 = L11 = 4piKαN
2
1S/l
L2 = L22 = 4piKαN
2
2S/l
L12 = L21 = 4piKαN1N2S/l
L212 = L11L22
Problem 33. Two identical coaxial circular superconducting loops, of inductance L each, are
initially at large distance one from the other. In each loop is current I initially. They are brought
together. What is the new current in each loop ? What is the change in the magnetic energy of the
system ? Answer:I ′ = I/2, W = −LI2/2.
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Physics 2. Electromagnetism More problems
Solution.
E = −αΦ˙ = IR = 0
Φ = const⇒ I ′ = I/2
W = Wnew −Wold = (4L)(I/2)2/2− 2 · LI2/2
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