Download 2012 Moed A - Solution

BGU Physics Dept. Physics 2 for Physicist
2012 Moed A - Solution
Exercise 1
~ = E(r)r̂, so Gauss
1. Since we have spherical symmetric charge distribution, we know that E
law tells us
Z Z
Z Z Z
1
~
~
(1)
E · dS = 4π
ρdV ⇒ E(r) = 2 Qin .
r
r<R
Z
r
Qin = 4π
ρ0 r02 dr0 =
0
4πρ0 3
4πρ0
r ⇒ E(r) =
r.
3
3
(2)
R < r < 2R
Z
R
Qin = 4π
ρ(r0 )r02 dr0 + 4πR2 σ1 = 0 ⇒ E(r) = 0 .
(3)
0
2R < r
4
R
32R
02 0
2
3 15
= 4π
32ρ0
r dr + 4π(2R) σ2 ⇒ E(r) = 4πρ0 R
− 3
.
r0
r2
r
2R
Z
Qin
r
The potential (after setting ϕ(r → ∞) = 0) is defined by ϕ(r) = −
(4)
Rr
∞ E(r)dr
2R < r
ϕ(r) = 4πρ0 R
3
15 16R
− 2
r
r
.
(5)
R < r < 2R
ϕ(r) = Const = ϕ(2R) = 14πρ0 R2 .
(6)
r<R
ϕ(r) = −
2πρ0 2
r + Const ,
3
(7)
by matching at r = R we have
ϕ(R) = −
2πρ0 2
R + Const = 14πρ0 R2 ,
3
(8)
and the solution is
ϕ(r) = −
2πρ0 2
2
r + 14 πρ0 R2 .
3
3
(9)
1
2. For the surface charge density, the Electrostatic energy is given by
1
1
4πR2 σ1 ϕ(R) + 4π(2R)2 σ2 ϕ(2R)
ΣQi ϕi =
2
2
112 2 2 5
= 28π 2 ρ0 R4 (σ1 + 4σ2 ) =
π ρ0 R .
3
Uσ =
(10)
(11)
For the volume charge density, the Electrostatic energy is given by
1
2
Z
∞
φ(r)ρ(r)dV
Z R Z ∞
1 2
1 2 2
16R
2 2
7 15
2
= 4π ρ0
− r + 7 R r dr +
64R
− 6
r dr
3
3
r5
r
0
2R
(12)
= 320.8π 2 ρ20 R5 .
(14)
Uρ =
0
(13)
The total energy is therefore
U = Uσ + Uρ ≈ 358.13π 2 ρ20 R5 .
(15)
We can also verify that result by checking
R
E 2 dV .
3. Let the charge of the new sphere be Q, then the potential for r > 1.5R will have an addition
of Qr . We demand ϕ(r = 1.5R) = 0 so
ϕ(r = 1.5R) = 14πρ0 R2 +
Q
= 0 ⇒ Q = −21π 2 ρ0 R3 .
·1.5R
(16)
The rest of the potential can be found by matching:
R < r < 1.5R ϕ(r) = 0 ,
2πρ0 2
(R − r2 ) .
r<R
ϕ(r) =
3
(17)
(18)
For the electric field we have
r < 1.5R E → E ,
r > 1.5R E → E −
(19)
84π 2 ρ
r2
0
R3
.
(20)
Exercise 2
~ = B0 ẑ. The induced emf will be created so the current will run clockwise (−ϕ̂).
1. Let B
=−
1 dΦB
d
2Bxẋ
= − B · x(t)2 = −
c dt
dt
c
The total resistance of the circuit when its length is x
Z x
4B ẋ
x2
R=
αx0 dx0 = α
⇒I=
=
2
R
cαx
0
2
(21)
(22)
2. The force on the wire due to the current for any x, ẋ is given by
FB = BIx =
4B 2 ẋ
cα
(23)
In equilibrium
0 = ΣF = F − FB ⇒ ẋ =
cαF
4B 2
(24)
3. not in equilibrium:
mẍ = F −
2
4B 2 ẋ
αF
− 4B
t
αm
+
Ce
⇒ ẋ =
cα
4B 2
(25)
Initial conditions v(t = 0) = 0:
4B 2
cαF
− cαm
t
1
−
e
v = ẋ =
4B 2
(26)
4.
P = I 2R =
Z
U=
0
T
8B 2 v 2
cαF 2
=
cα
2B 2
mc2 α2 F 2
P dt =
16B 4
4B 2
1 − e− cαm t
2
4B 2
8B 2
8B 2 T
− 3 + 4e− cαm T − e− cαm T
cαm
(27)
Exercise 3
~ =
If there is a current I in ẑ direction, that is due to an electric field E
µ
I
0
~ =
the wire is B
2πr ϕ̂. The Poynting vector is given by
~= 1E
~ ×B
~ = −r̂ IV ,
S
µ0
2πaL
(28)
V
L ẑ.
The magnetic field of
(29)
where a is the radii of the wire. The total energy per unit time floating into the wire (due to
heating) is
Z
~ · dA
~ = S2πaL = IV ,
P = S
(30)
which is the known expression for the power of a resistor.
Exercise 4
1. Option A:
Due to the rotation ω
~ = ω φ̂, the free charges (electrons) will experience centrifugal force
(outward). the electrons the will flow in the radial direction until equilibrium achieved:
2
~ in = me ω r r̂ .
ΣF~ = me ω 2 rr̂ − eEin r̂ = 0 ⇒ E
e
(31)
Note that we have neglected the influence of the magnetic field, since we know that B ∼
~ out = 0 since the total charge is zero.
The electric field outside is E
3
E
c.
The charge density is given by
ρ=
1 ~
1
me ω 2
∇E = d(rEr )dr =
4π
r
2πe
(32)
The total charge remains zero, which means that in addition to positive volume density, there
is a negative surface density on the surface which is given by
Z Z Z
Z Z
ρR
me ω 2 R
ρdV +
σdA = 0 ⇒ σ = −
=−
(33)
2
4πe
since the charges are rotating with angular velocity ω the current per unit length is given by
I=
∆Q
2πRσ
ω 3 me R 2
=
=
∆t
2π/ω
4πe
(34)
for the surface density, where the direction of the current is counter-clockwise. Using Amper’s
law, we can find the associated magnetic field:
3
2
~ σ = ω me R ẑ .
B
ce
(35)
For the volume density, we will divide each disc of unit length into rings, each ring carrying
the charge 2πρrdr. The magnetic field due to each ring is
~ =−
dB
2me ω 3 rdr
ẑ .
ce
(36)
Note that the direction of the current is clockwise, so the magnetic field is in the opposite
direction. each ring contributes only to the magnetic field inside (since the field outside of a
coil is zero):
Z R
3
2
2
~
~ = − me ω (R − r ) ẑ .
Bρ =
dB
(37)
ce
r
So, the total magnetic field is
3 2
~ =B
~ρ + B
~ σ = µ0 me ω r ẑ ,
B
4πe
(38)
and the magnetic field outside is zero
Option B:
The velocity of the charges is given by v = rω. The current density is related to the charge
density:
J = nqv = ρ(r)v = ρ(r)rω .
(39)
~ = Er r̂ where
Since the problem have cylindrical symmetry, Gauss law imply that E
Z
4π r
Er =
ρ(r0 )r0 dr0 .
r 0
(40)
~ = Bz (r)ẑ we have
On the other hand, from Amper’s law, and recalling that J~ = J ϕ̂, B
h
i
~ = ∂Bz = 4π J = 4π ρrω
∇×B
∂r
c
c
ϕ
(41)
4
which leads us to
4π
Bz (r) =
ω
c
Z
r
ρ(r0 )r0 dr0 =
0
ωr
E(r)
c
(42)
The equilibrium condition is
ωr 2 −1 m ω 2 r
e
me ω 2 r
e
~
~
1+
me~a = me ω rr̂ = eE + ~v × B ⇒ E(r) =
≈
c
e
c
e
B(r) ≈
2
(43)
me ω 3 r 2
ce
(44)
2.
ρ=
me ω 2
2πe
σ=−
(45)
me ω 2 R
4πe
(46)
3. The potential difference is given by
Z
∆ϕ = −
R
E(r)dr =
0
mω 2 R2
2e
(47)
4. If E(r) = 0 the equilibrium equation (with external magnetic field) take the form
e ~ ext → Bext = mωc
mω 2 r = ~v × B
c
e
5
(48)