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2. Energy In a Magnetic Field
AP Physics C
Montwood High School
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3. • The induced EMF set up by an inductor prevents
a battery from establishing an instantaneous
current.
• Therefore, a battery has to do work against an
inductor to create a current.
• Part of the energy supplied by the battery goes
into joule heat dissipated in the resistor, while the
remaining energy is stored in the inductor.
• Beginning with Kirchhoff’s equation for an RL
circuit in which the current is increasing, multiply
each term by the current I and rearrange:
0 0
dI dI
EMF I R L EMF I I I R L I
dt dt
− × − × = × − × × − × × =

4. • The rate at which energy is supplied by the battery,
EMF·I, is equal to the sum of the rate at which joule heat
is dissipated in the resistor, I2
·R, and the rate at which
energy is stored in the inductor, L·I·dI/dt.
• The equation is an expression of the law of conservation
of energy.
• If Um is the energy stored in the inductor at any time, then
the rate dUm/dt at which energy is stored in the inductor
is:
2 2
0
dI dI
EMF I I R L I EMF I I R L I
dt dt
× − × − × × = × = × + × ×
mdU dI
L I
dt dt
= × ×

5. • To find the total energy stored in the inductor, rewrite the
equation as:
• This equation represents the energy stored as magnetic
energy in the field of the inductor when the current is I.
• After the current has reached its final steady state value I,
dI/dt = 0 and no more energy is input into the inductor.
0 0
1 1 2
0
2
integrate
1 1 2
0.5
mU I
m m
I
m m
m
dU L I dI dU L I dI
I I
U L I dI U L L
U L I
+
= × × = × ×
= × × = × = ×
+
= × ×
∫ ∫
∫

6. • When there is no current, the stored energy Um is 0 J;
when the current is I, the stored energy Um is 0.5·L·I2
.
• When the current decreases from I to zero, the inductor
acts as a source that supplies a total amount of energy
0.5·L·I2
to the external circuit.
• If we interrupt the circuit suddenly by opening a switch
or yanking a plug from a wall socket, the current
decreases very rapidly, the induced EMF is very large,
and the energy may be dissipated in an arc across the
switch contacts.
• This large EMF is the electrical equivalent to the large
force exerted by a car running into a brick wall and
stopping very suddenly.

7. • It is important not to confuse the behavior of resistors and
inductors where energy is concerned.
• Energy flows into a resistor whenever a current passes through it,
whether the current is steady or varying; this energy is dissipated
in the form of heat.
• Energy flows into an ideal, zero-resistance inductor only when the
current in the inductor increases.
• The energy is not dissipated; it is stored in the inductor and
released when the current decreases.
• When a steady current flows through an inductor, there is no
energy flow in or out.

8. • The energy in an inductor is actually stored in the
magnetic field within the coil, just as the energy
of a capacitor is stored in the electric field
between the plates.
• You can also determine the energy per unit
volume, or energy density, stored in a magnetic
field.
• Consider a solenoid whose inductance is given by
L = μo·n2
·A·l.
• The magnetic field of the solenoid is given by
B = μo·n2
·I.

9. • Substituting:
2 2 2
2
2
2
2
0.5 0.5
0.5
0.5
m o
o
o
m o
o
m o
o
U L I n A l I
B
B n I rearrange I
n
B
U n A l
n
B
U n A l
n
µ
µ
µ
µ
µ
µ
µ
= × × = × × × × ×
= × × =
×
 
= × × × × × ÷
× 
 
= × × × × × ÷
× 

10. • Because A·l is the volume of the solenoid, the
energy stored per unit volume in a magnetic field
is:
2
2
2 2
2
0.5
0.5
m o
o
m
o
B
U n A l
n
B
U A l
µ
µ
µ
= × × × × ×
×
= × × ×
2
2
m
m
o
U B
u
A l µ
= =
× ×

11. • Although the magnetic energy density equation
was derived for a solenoid, it is valid for any
region of space in which a magnetic field exists.
What Happens to the Energy in the Inductor?
• Consider the RL circuit shown. Recall that the current in
the right-hand loop decays exponentially with time
according to the equation:
where Io = EMF/R is the
initial current in the circuit
and τ = L/R is the time
constant.
t
oI I e τ
−
= ×

12. • Show that all the energy initially stored in the
magnetic field of the inductor appears as internal
energy in the resistor as the current decays to
zero.
• The rate at which energy is dissipated in the
resistor, dU/dt (or the power), is equal to I2
·R,
where I is the instantaneous current.
• To find the total energy dissipated in the resistor,
integrate the equation from t = 0 s to t = ∞.
2
2
2 2
R t R t
L L
o o
dU
I R I e R I R e
dt
− × − × × 
 ÷= × = × × = × ×
 ÷
 

13. 2
2
2
2
0 0
2
2
0
2 2
00
2 2 2
2
2 2
R t
L
o
R t
L
o
R t
L
o
u u
o o
dU I R e dt
dU I R e dt
U I R e dt
R t R t R
let u then du d dt
L L L
L
dt du
R
L L
U I R e du I R e
R R
− × ×
− × ×
∞ ∞
− × ×
∞
∞ ∞
= × × ×
= × × ×
= × × ×
− × × − × × − × 
= = = × ÷
 
−
= ×
×
− −
= × × × × = × × ×
× ×
∫ ∫
∫
∫

14. • Note that this is equal to the initial energy stored in the
magnetic field of the inductor.
The Coaxial Cable
• A long coaxial cable consists of two concentric
cylindrical conductors of radii a and b and length l.
• The inner conductor is assumed to be a thin cylindrical
shell.
( )
2
2 0
0
2 2
0 2
2 2
0.5
2 2
u o
o
o o
o
I LL
U I R e e e
R
I L I L
U e e L I
∞
∞
∞
− ×−
= × × × = × −
×
− × − ×
= × − × ≅ × ×

15. • Each conductor carries a
current I (the outer one being a
return path).
• Calculate the self-inductance L
of the cable.
– To obtain L, we must know the
magnetic flux through any cross-
section between the two
conductors.
– From Ampere’s law, it is easy to
see that the magnetic field
between the conductors is given
by
2
o I
B
r
µ
π
×
=
× ×

16. – The magnetic field is zero outside
the conductors and zero inside the
inner hollow cylinder.
– The magnetic field is zero outside
the conductors because the net
current through a circular path
surrounding both wires is zero.
– The magnetic field inside the
inner conductor because there is
no current within the hollow
inner cylinder.
• The magnetic field is
perpendicular to the shaded
rectangular strip of length l and
width b – a.

17. • Dividing the rectangle into
strips of width dr, the area of
each strip is l·dr and the
magnetic flux through each
strip is B·dA = B·l·dr.
• The total magnetic flux through
any cross-section is:
2
1
2
m
b
o
m
a
b
o
m
a
B dA
I
l dr
r
I l
dr
r
µ
π
µ
π
Φ = •
×
Φ = × ×
× ×
× ×
Φ = × ×
×
∫
∫
∫

18. • The self-inductance of the
coaxial cable is:
( )
ln
2
ln ln
2
ln
2
bo
m a
o
m
o
m
I l
r
I l
b a
I l b
a
µ
π
µ
π
µ
π
× ×
Φ = ×
×
× ×
Φ = × −
×
× ×
Φ = ×
×
ln
2
ln
2
m o
o
I l b
L
I I a
l b
L
a
µ
π
µ
π
Φ × ×
= = ×
× ×
×
= ×
×

19. • Calculate the total energy
stored in the magnetic field of
the cable.
2
2
2
0.5
0.5 ln
2
ln
4
m
o
m
o
m
U L I
l b
U I
a
I l b
U
a
µ
π
µ
π
= × ×
×
= × × ×
×
× ×
= ×
×

20. • For the homework, remember that the current I
as a function of time for an RC circuit that is
increasing is:
( ) 1
R
t
L
EMF
I t e
R
−
× 
 ÷= × −
 ÷
 