Download 2012 Moed B - Solution

BGU Physics Dept. Physics 2 for Physicist
2012 Moed B - Solution
Exercise 1
1. The charge on the capacitor is given by
− R tC
1
1
Q(t) = Q0 1 − e
≈ Q0 1 − 1 +
t
R1 C1
⇒
T
Q(T )
=
= 0.1 ,
R1 C1
Q0
(1)
so that T = 4µsec.
2. since the time constant of the system is τ = R1 C1 = 40µsec than at t = 1sec = 25, 000τ the
capacitor is fully charged and the energy is
1
U0 = C1 ε2 = 0.02J
2
(2)
3. After we move the switch, we have a closed RC circuit. at t → ∞ the potential difference on
both the capacitors should be the same, and the total charge on the capacitors should remain
the same as well (note that C1 = C2 ):
V1 =
Q1
C1
;
V2 =
Q2
C2
C1
C2 Q2 =
Q2 = Q20
⇒ Q1 =
Q1 + Q2 = Q0 ⇒ Q1 =
Q2
(3)
(4)
and the total energy on stored on the capacitors is
UT = U1 + U2 = 2 ×
1 Q20
U0
1 Q2
=
=
= 0.01J .
2 C
4 C
2
(5)
The power of the resistor is given by
P = I 2 R2 = R2 I02 e
−2 R
t
2 CT
−2 R
= R2 I02 e
t
2 CT
= R2 I02 e
−4 R t C
2
,
(6)
where CT = C/2 since the two capacitors are connected IN TOR, and I0 = V0 /R2 = Q0 /CR2
is the current right after we close the switch. The total energy wasted due to the heating of
the resistor is
Z ∞
Q0 2 R2 C
1 Q20
W =
P dt = R2
=
,
(7)
R2 C
4
4 C
0
same as before
4. since the above result does not depend on R2 , nothing changes.
Exercise 2
1. The current density is given by ~j = nqvẑ = ρvẑ = ρ0rav ẑ inside the cylinder and zero outside.
By Amper’s law:
I
ZZ
Z
4π
4π r
~
~
~
~
B · dl =
J · dS ⇒ Bϕ =
ρ(r0 )r0 dr0 ,
(8)
c
cr 0
So the magnetic field is
(
4πρ0 av
r<a
c
~ = ϕ̂
B
4πρ0 a2 v 1
c
r r >a
(9)
1
2. We will demand that the charge per unit length on the outer cylinder will be equal to minus
the charge per unit length on the inner cylinder:
ZZ
Z
a
dQin = dz
ρ0 rdrdϕ = 2πρ0 a2 dz
(10)
r
Z
Z
Z
σ(2a)dϕdz = Qout = −Qin = − dQin = 2πρ0 a2 dz
(11)
σ=−
ρ0 a
2
(12)
(13)
3. Inside the outer cylinder. The magnetic field will remain the same. Outside it will be doubled
(since the total current has been doubled.
The current density is given by j = nqv = ρv = ρ0rav inside the cylinder and zero outside. By
Amper’s law:
I
ZZ
Z
4π
4π r
~
~
~
~
B · dl =
J · dS ⇒ Bϕ =
ρ(r0 )r0 dr0 ,
(14)
c
cr 0
So the magnetic field is
 4πρ av
0

r<a

c
2v
4πρ
a
1
0
~ = ϕ̂
B
c
r a < r < 2a

 8πρ0 a2 v 1
2a < r
c
r
(15)
Exercise 3
When the rings are located far away from each other, the flux through each ring is given by φi = LI.
The EMF on each ring is given by
ε = IR .
(16)
Since the rings are made from superconducting material, with R = 0 we conclude that the flux
through each ring is constant:
0=ε=−
1
φ̇ ⇒ φ = Const .
C
(17)
After placing the rings together the flux through each ring will have contribution both from the ring
itself and from the second ring. When the rings are placed together, the mutual inductance is equal
to the self induction of each ring (they are identical and in the same place) and from symmetry
considerations, we can assume that the current in each ring is identical:
φf1 = L11 I1f + L12 I2f = 2LIf ,
(18)
comparing with the flux in the beginning we find that If = 12 I.
Energy: In the start, each ring will store energy 12 LI 2 . After the rings are moved together, each
one will hold 21 LIf2 and in addition, we have the energy of their mutual inductance LIf2 The energy
difference will be
1 2
1
1
2
∆U = Uf − Ui = 2 × LIf + LIf − 2 × LI 2 = − LI 2
(19)
2
2
2
Exercise 4
2
1. Option A:
The electric field of a dipole p1 experienced by p2 is given by
ˆ dˆ − p~1
p1 · d)
~ 1 = 3(~
E
d3
(20)
And the energy of interaction of p2 and p1 will be
~ 1 = p1 p2
U = −~
p2 · E
d3
(21)
where we simplified the above equation using the fact that p̂1 k pˆ2 ⊥ dˆ The force will be given
by
F = −∇U = 3
p1 p2
d4
(22)
Option B
we can treat this problem as 4 charges sitting on the vertices of rectangle: q1 , −q1 , q2 , −q2 .The
distance between opposite charges along the y axis is a and the distance between same sign
charges along the x axis is d. The total force that charges q1 , −q1 is acting on charges −q2 , q2
is:
q1 q2
q1 q2 d
~
ΣF = 2 2 − 2 2
x̂ ,
(23)
d
(d + a2 )( 3/2)
(the force on the y axis is zero). Expanding for d >> a

2
q1 q2
q1 q2 d
2q1 q2 
−2 2
=
1 − d2
d2
(d + a2 )(3/2)
1+

1
3p1 p2
 3q1 q2 a2
≈
=
4
(3/2) 
d
d4
2
a
(24)
d
same as above
2. If we treat the conducting sphere as a dipole p~, than its electric field is given by
pr̂)r̂ − p~
~ dip = 3(~
E
.
r3
(25)
The total electric field on the shell of the sphere will be
3(~
pr̂)r̂ − p~ ~
+ Eext .
R3
(26)
demanding that the field will be perpendicular to to shell, we set
~ ext ,
p~ = R3 E
(27)
and the total force is
2
~ ext
3p1 p2 ~ 3R6 E
d
=
d~
F~ =
d5
d5
(28)
3