Download Tutorial 3 Solutions - NUS Physics Department

PC3231 Electricity and Magnetism II
Tutorial 3 Solutions
1. A long coaxial cable carries current I (the current flows down the surface of the inner
cylinder of radius a, and back along the outer cylinder of radius b). The two
conductors are maintained at a potential difference V.
(a) Find the energy stored in a section of length l.
(b) Calculate the power transported down the cable.
Solution
B
z
E
The magnetic field is due to current I. There are surface charges residing on the
surface of the inner core due to current I, giving rise to a radial electric field E. The
exact value of  is however a complicated function of how I distributes over the
material surface. Here we simply assume that the surface charge is per unit length.
E
λ 1
ŝ
2πε 0 s
B
0 I 1
φ̂
2 s







The magnetic energy stored in a cylindrical shell of length l, radius s and radial thickness
ds is
 I 2 l ds
1  0 I 

 2sdsl  0
2 0  2s 
4 s
2
Integrating from a to b for the cylindrical volume, the magnetic energy in length l is
WB 
0 I 2l b
ln
4
a
Similarly the electric energy per in length l is
WE 
2 l
b
ln
40 a
The Poynting vector is given by
S
1
0
E  B   2I
1
4  0 s 2
The power transported down the cable is
I b 1
I
b
P   S  da   S 2sds 
ds 
ln

20 a s
20 a
a
b
Comparing with Ohm’s law
 b1

b
V   E  dl 
ds 
ln

20 a s
20 a
a
b
So P = IV as in Poynting theorem, electromagnetic work is done by the fields as
Joules heating. The Poynting vector S is radially inward, and the total energy
crossing into the surface of the wire is found by integrating over the surface of the
wire:
 S  da  VI
as Ohm’s law.
For real conductor, E is nonzero inside. E 
applied).
V
zˆ
L
where V is the voltage drop (or
2. A small loop of wire (radius a) lies a distance z above the center of a large loop
(radius b), as shown.
a
θ’
z
r
b
The planes of the two loops are parallel, and perpendicular to the common axis.
(a) Suppose current I flows in the big loop. Find the flux through the little
loop. The little loop is so small that you may consider the field of the big
loop to be essentially constant.
(b) Suppose current I flows in the little loop. Find the flux through the big
loop. The little loop is so small that you may treat it as a magnetic dipole.
Solution
(a) The field (on the axis) is
B
0 I
b2
zˆ
2 (b 2  z 2 ) 3 / 2
where B is essentially constant in the small loop. The flux through the small loop
(area a2) is given by

 0Ia 2 b 2
2(b 2  z 2 ) 3 / 2
(b) The field due to dipolar loop is
B
0 m
(2 cosrˆ  sin θˆ )
3
4 r
where m = Ia2 is the magnetic dipole moment. Integrating over a spherical “cap”
bounded by the big loop and centered at the small loop, the flux is
 0 Ia 2
 0 Ia 2  '
2
   B  da 
2 cos r sin dd 
2 cos r 2 sin dd
2 

4 r
2r 0
where r  b 2  z 2 and sin  '  b

 0 Ia 2 sin 2 
r
2
'
0

r
 0Ia 2 b 2
2(b 2  z 2 ) 3 / 2
same as in Part (a)
3. A very long solenoid of radius a with N turns per unit length carries a current Is.
Coaxial with the solenoid is a circular conductor ring of radius b (b >> a) and
resistance R. When the current in the solenoid varies, a current Ir is induced in the
ring.
(a) Calculate Ir in terms of dIs/dt.
(b) Show by calculating the Poynting vector that the power Ir2R delivered to the
ring must have come from the solenoid.
Solution
(a) Assuming that the field B is uniform within the slender solenoid,
 
Since flux
d
(induced emf in the ring)
dt
  a 2 B , B   0 nI s ,   I r R
Therefore, the induced current I r  


dI
1
 0a 2 n s
R
dt
(b) Deduced first the electric field at the surface of the solenoid.
 E  dl  
dI
dI
d
1
 E (2a)   0a 2 n s  E    0 an s ˆ .
dt
dt
2
dt
The magnetic field at a ring of the solenoid (in blue) at the axis of thin solenoid at a
distance z above or below the plane of the ring is given by (see Eq. 5.28, p218 Griffiths):
B
0 I r
2
b
b2
2
z
2

3
zˆ
2
Assuming that B is nearly uniform across the cross-section (and same as B on the z-axis)
of the solenoid, the Poynting vector at the surface at a distance z above (a ring of the
solenoid in blue color, or below) the plane of the outer ring is:
S
1
0
E  B  
1   0 an dI s   0 I r
b2


0 
2 dt  2 b 2  z 2 3 2


ab 2 n
φˆ  zˆ    1  I dI s
rˆ .
0 r

4
dt b 2  z 2 3 2

Note in calculating S, we used the E due to Is but the radiating B was evaluated as due to
Ir, rather than due to Is. Hence the proof presented is not quite air-tight.
Total power from entire length of the solenoid is a sum of all the solenoidal rings.
P   S  da 




dI s
1
2 2
 S 2a dz   20 a b nI n dt 
z
b
1
2
 z2

E
S

Since the integral is
b
2
z b
2
2


dx
2
component of S
towards the ring
B
z
3
1  1  2
   
b2  b2  b2
dI

P   0 a 2 n s
dt


2
 I r  RI r I r  I r R

(Power delivered to the ring goes into heating)
4. A charged large parallel-plate capacitor (with uniform electric field E = E z and
plate separation d) is placed in a uniform magnetic field B = B x, as shown in the
figure below.
z
E
ey
B
y
x
Find the electromagnetic momentum in the space between the plates.
Solution
em   0 E  B    0 EByˆ
pem   0 EBAdŷ
5. A point charge q is a distance a > R from the axis of an infinite solenoid (radius
R, n turns per unit length, current I). Find the linear momentum in the fields.
(Treat the solenoid as a nonconductor so that you don’t need to worry about the
induced charges on its surface).
Magnetic field of the solenoid B   0 nIzˆ where s < R
Electric field of the point charge E 
rˆ
where r = [(x-a), y, z]
4 0 r 2
q
z
E
q
B
E
x
y
E
The momentum density is
q
1
(rˆ  zˆ )
40 r 3
P   0 ( E  B )   0 n 0 I

 0 nqI
 yxˆ  ( x  a)yˆ 
4r 3
The linear momentum is
p   Pd 

 0 nqI
 yxˆ  ( x  a)yˆ  dxdydz

4
[( x  a) 2  y 2  z 2 ]3 / 2
 0 nqI
( x  a)yˆ 
dxdydz
4  [( x  a) 2  y 2  z 2 ]3 / 2
where the term in x is odd in y and it integrates to zero.
Carry out the z integral first, where the integrand becomes
[( x  a ) 2  y 2 ]
p
z

( x  a) 2  x 2  y 2


2
[( x  a) 2  y 2 ]
 0 nqI
x  a dxdy
yˆ 
2
[( x  a) 2  y 2 ]
Now switch to polar coordinates, with
x  s cos 
y  s sin 
dxdy  sdsd
[( x  a) 2  y 2 ]  s 2  a 2  2sa cos 
p
 0 qnI
s cos  a
yˆ  2
sdsd 
2
s  a 2  2sa cos
Now using
2
cos d
 A  B cos  
0
2
d
 A  B cos  
0
2
(1 
B
2
A2  B 2
A
A2  B 2
)
here
p
A 2  B 2  (a 2  s 2 ) , and taking positive value.
 0 qnI
 a2  s2
 0 qnI R
 qnI 2
2a 2 
yˆ  1  2

sds

yˆ  sds  0
R yˆ
2
2
2
2a
2a
2a
a s 
 a s
0