Download Inductance- II

Inductance- II
Plan:
 Energy Storage in a Magnetic
field
Energy density and the
magnetic field
Recap
di
L 
dt
di
L  L
dt
• Units : volt-second/ampere
(Henry)
N B
L
i
N B  0 N h b
L

ln
i
2
a
2
Toroid
• Solenoid
L  0 n 2 A
l
Filling the magnetic materials
can increase the inductance
L   m L0
Inductors in series
Leq  L1  L2
• Inductors in parallel
1
1 1
 
Leq L1 L2
Energy storage in magnetic
field
• Energy is stored in the electric
field due to the charges.
• Similar manner, there is energy
stored around a current carrying
wire, where magnetic field
exists.
Example
• Work done in separating the
current carrying wires is stored
in the energy of the magnetic
field of the wire.
• This energy can be recovered
by allowing the wires to move.
Energy in Magnetic fields
• When the current is turned on, it has to
work against the back emf.
• This energy is recovered if the current
is turned off.
• This energy can be regarded as the
energy stored in the magnetic field.
• Work done on a charge dq against
the back emf in one trip around
the circuit
W  dq
•Work done per unit time
dW
 I
dt
dW
dI
  LI
dt
dt
The energy stored in the
magnetic field.
• Work done, from zero to build a
current I
W   LI dI
1 2
W  LI
2
Energy density and magnetic
field
UB
uB 
volume
UB
uB 
AL
1 2
LI
2
uB 
AL
1 2
LI
2
uB 
AL
• For a solenoid
L  l 0n A
2
1 2
2
I l 0 An
2
uB 
AL
uB 
1
2 0
B
2
• A solenoid plays a role for magnetic field
similar to that of the parallel plate capacitor
for the electric fields.
uB 
1
2 0
Solenoid
B
2
1
2
uB   0 E
2
Capacitor
A length of cupper wire carries a current of 10
A, uniformly distributed. Calculate (a)
magnetic energy density just outside the
surface of the wire. The wire diameter is 2.5
mm.
0 I
B
2r
1  0 I 
uB 


20  2r 
2
uB =
3
1J/m
Find the magnetic energy density of a
circulating
electron
in
the
hydrogen
atom.Electron circulates about the nucleus in a
circular path of radius 5.29 x 10-11m at a
frequency f of 6.60 x 1015 Hz (rev/s).
uB 
B
1
2 0
B
2
q
I   qf
T
0 I
B= 12.6 T
2R
uB = 6.32 x 107 J/m3
A long coaxial cable carries a current I
(the current flows down the surface of
the inner cylinder, radius a, and back
along the outer cylinder, radius b. Find
the magnetic energy stored in a section
of length l.
I
a
b
I
 0 I
ˆ
B

2r
uB 
1
2 0
1  0 I 
uB 


2 0  2r 
0 I
uB  2 2
8 r
2
B
2
2
• Energy in a cylindrical shell of
length l, radius r and thickness
ds is
2
0 I
dU  2 2 dr rd dz
8 r
0 I
U    2 2 2lr dzdr
8

r
0 a
l b
2
0 I
U    2 2 2lr dzdr
8 r
0 a
l b
2
0 I l  b 
U
ln  
4
a
2
1 2
U  LI
2
 0l  b 
L
ln  
2  a 
0 I
  
dzd r
2r
0 a
l b
b
Area element
is dr dz
r
a
Calculate the energy stored in a
section of length l of the
solenoid.
1 2
W  LI
1
2
2 2
2
W   0 n R lI
L   0 n R l
2
2
2
Calculate the energy stored in the
toroidal coil.
1 2
uB 
B
2 0
 0 NI
B
2r
UB 
1
2 0
B
d


2
UB 
1
2 0
2 b
B
d


2
2
  0 NI 
UB 

 dr rdh


2 0 0 a  2r 
1
1
b
2 2
UB 
 0 n I h ln
4
a
A long wire carries a current I
uniformly distributed over a cross
section of the wire.
(a) Find the magnetic energy of a
length l stored within the wire.
0 I r
B
2
2 R
1  0 Ir 
uB 

2 
20  2R 
2
0 I
2
uB  2 4 r
8 R
2
R l 2
0 I 2
U B     2 4 r dr rd dz
8

R
0 0 0
2
0 I l
UB 
16
2
Find the inductance of the length
l of the wire associated with the
flux inside the wire
1 2
U B  LI
2
0 I l
UB 
16
2
 0l
L
8
Two long parallel wires, each radius
a, whose centers are at a distance d
apart carry equal current in opposite
directions. Neglecting the flux within
the wire themselves, find the
inductance of a length l of such a
pair of wires.
I
a
d-y
d
y
I
a

d a

a
 0 I   0 I 

  
 ldy
 2y   2 d  y  
 0 Il
d a
d a
ln y a  ln d  y a

2
 0l d  a
L
ln

a
A uniform magnetic field B fills a
cylindrical volume of radius R. A metal
rod of length L is placed as shown. If B
is changing at the rate dB/dt, find the
emf that is produced by the changing
magnetic field and that acts between the
ends of the rod.
Facts
 
• For electrostatics  E  dl  0
Time varying
magnetic fields
 
E

d
l

A
B
 
E

d
l

0

Path
dependent
Electric field at a distance r from
center
d
 E 2r   BA
dt
r dB
E
2 dt
Along the triangle via path
AOBA
  O  B  A 
 E  dl   E  dl   E  dl   E  dl
A
O
B
A 

 
E

d
l

0

0

E

d
l


A
B
A 



d
  B  A  0  0   E  dl
dt
B
O
B
A 



d
  B  A  0  0   E  dl
dt
B
 
dB
A
  E  dl
dt B
A
• A is the area of the triangle AOB
If you choose other path
Via
Path
ACBA
B
A
C
 
 E  dl 
B

AC
  A 
E  dl   E  dl
B
B 
A 




d
  B  da   E  dl   E  dl
dt
AC
B
A 



d
B  da   RE   E  dl

dt
B
Important laws
 
divE 
0

divB  0

dB
CurlE  
dt


CurlB  0 J