Download Chap 21. Electromagnetic Induction Sec. 1

Chap 21. Electromagnetic Induction
Sec. 1 - Magnetic field
Magnetic fields are produced by electric currents:
• They can be macroscopic currents in wires.
• They can be microscopic currents ex: with electrons in atomic orbits.
• Magnetic field sources are essentially dipolar in nature, having a north
and south magnetic pole
• Magnetic field lines are continuous and originate from the north pole
and terminate at the south pole.
• The interaction of magnetic field with charge leads to many practical
applications.
~
F~ = q~v × B
Bar Magnet
• Magnetic field sources are essentially dipolar in nature, having a north
and south magnetic pole
• Magnetic field lines are closed lines and permanent magnets can be
made from ferromagnetic materials.
• Magnetic monopoles do not exist. Compare with a electric charge above
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Lorentz Force (F) Both electric field and magnetic field can be defined
in terms of force on charge in the Lorentz force law.
~
F~E = q E
~
F~B = q~v × B
|F | = qvBsinθ
(1)
• The SI unit for magnetic field (B)is the Tesla and it is a vector quantity
A smaller magnetic field unit is the Gauss (1 Tesla = 10,000 Gauss).
• The SI unit for Electric field (E) is the Newton/Coulomb and it is a
vector quantity
~ if the charge q is
• The direction of the electric force is in direction of E
~
positive. If q is negative then direction of F~E is reverse of E.
• The direction of the magnetic part of the force is given by the right
hand rule.
• The magnetic force on a free moving charge is perpendicular to both
the velocity of the charge and the magnetic field.
~
• The force is maximum when ~v is perpendicular to B
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4
Force on a current carrying wire
5
Ampere’s Law
The magnetic field in space around an electric current is proportional
to the electric current which serves as its source, just as the electric field
in space is proportional to the charge which serves as its source. Ampere’s
Law states that for any closed loop path, the sum of the length elements
times the magnetic field in the direction of the length element is equal to
the permeability times the electric current enclosed in the loop.
Application We can use Ampere’s Law to obtain the following
• The magnetic field inside a solenoid.
• The magnetic field around a current carring wire.
• The magnetic field inside a toriodal coil.
• The magnetic field inside a conductor.
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The magnetic field around a current carring wire
~ tangent to the circle of
Use Ampere’s Law: Find the component of B
radius r and sum its value around a closed path.
[B|| is a constant at a circle of radius r (symmetry of the problem)].
Z
µo I
B|| dl = µoI
B||2πr = µoI
B|| =
2πr
RIGHT HAND RULE 2: If the thumb gives the direction of the current
then the curled fingers give the direction of the magnetic field.
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Solenoid: The magnetic field inside a loop of N turns
~ tangent
Use Ampere’s Law: In the path shown, Find the component of B
to the path of the rectangle and sum its value around a closed path.If I is
the current in each loop then the total current enclosed in the area is
Ienc = In = I
Z
B|| dl = µoIenc
B||l = µo
N
∗l
L
IN ∗ l
L
B|| =
µoIN
L
RIGHT HAND RULE 2: If the curled fingers gives the direction of the
current then the thumb give the direction of the magnetic field.
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Faradays Law: Any change in the magnetic environment of a coil of
wire will cause a voltage (emf) to be ”induced” in the coil. No matter how
the change is produced, the voltage will be generated.
• The change could be produced by changing the magnetic field (B)
strength, moving a magnet toward or away from the coil, moving the
coil into or out of the magnetic field, rotating the coil relative to the
magnet etc:
• Changing the area (A) of the coil so that the flux varies.
• Faraday’s law is a fundamental relationship which comes from
Maxwell’s equations. The induced emf is equal to the negative of the
rate of change of magnetic flux times the number of turns in the coil.
The emf induced in the coil is given as
dφ
d(B⊥A)
=−
dt
dt
Remember to take the perpendicular component of B to the surface area
you consider. The parallel component of B does not induce emf.
EM F (V ) = −
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10
Voltage generated in a moving coil
The force acting on the charges due to the change in the magnetic flux if
the wire sweeps an area of A = lx is
EM F (V ) = −
d(B⊥A)
d(x)
= −Bl
sinθ = −Blvsinθ
dt
dt
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AC Generator
The turning of a coil in a magnetic field produces motional emfs in both
sides of the coil which add. Since the component of the velocity perpendicular to the magnetic field changes sinusoidally with the rotation, the
generated voltage is sinusoidal or AC. This process can be described in
terms of Faraday’s law when you see that the rotation of the coil continually changes the magnetic flux through the coil and therefore generates a
voltage. The emf induced for N number of turns is
EM F (V ) = −2N Blvsinθ
From rotational motion we can write v = ωr = ωh/2 and θ = ωt.
EM F (V ) = −N Blhω sin(ωt) = −N BAω sin(ωt)
The maximum induced on peak voltage is when sin(ωt) = 1
(Vpeak ) = −N Blhωsin(ωt) = −N BAω
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Inductance
Inductance is typified by the behavior of a coil of wire in resisting any
change of electric current through the coil. Arising from Faraday’s law,
the inductance L may be defined in terms of the emf generated to oppose
a given change in current:
EM F (V ) = −L
dI
dt
The unit of inductance (L) is Henry.
The inductance of a solenoid of N turns can be given as
dI
d(B⊥A)
= −N
dt
dt
dI
d(µoN IA/l)
−L
= −N
dt
dt
EM F (V ) = −L
This gives
µo N 2 A
L=
l
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Transformers
A transformer makes use of Faraday’s law and the ferromagnetic properties
of an iron core to efficiently raise or lower AC voltages. Remember very
important ... it of course cannot increase power so that if the
voltage is raised, the current is proportionally lowered and
vice versa.
If Vs and Is is the voltage and current of the secondary with Ns turns
and Vp and Ip is the voltage and current of the primary with Np turns,
then
N s V s Is
=
=
N p V p Ip
The power in the primary and secondary is equal
Pp = P s = I s V s = I p V p
In Step up transformer Vs > Vp and in a step down transformer Vp > Vs
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Energy stored in a inductor and a magnetic field
The energy stored in an inductor is given as
Z
Z
Z
Z
dI
1
U = P.dt = IV dt = I(L )dt = LIdI = LI 2
dt
2
Use the inductance and current derived for a soleniod and it will lead us
to the energy stored in a magnetic field.
1 µoN 2A Bl 2 1 B 2
(
) =
U=
Al
2 l
µo N
2 µo
The energy density is the energy per unit volume V = Al
1 B2
U
u= =
V
2 µo
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AC circuits - AC response of Resistors
For ordinary currents and frequencies the behavior of a resistor is that
of a dissipative element which converts electrical energy into heat. It
is independent of the direction of current flow and independent of the
frequency. So we say that the AC impedance of a resistor is the same as
its DC resistance. That assumes, however, that you are using the rms or
effective values for the current and voltage in the AC case.
The phasor diagram denotes the phase difference of the current and the
voltage in the ciruit. The phasor diagram for the same is given below
Resistance, Impedance for a resistance is Xr = R
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AC circuits - AC response of Capacitors
You know that the voltage across a capacitor lags the current because the
current must flow to build up the charge, and the voltage is proportional
to that charge which is built up on the capacitor plates.
1
Impedance for a capacitor is Xc = ωC
For DC currents ω = 0 and hence the Impedance of the capacitor for DC
currents is
1
=∞
Xc =
ωC
There is no current in the circuit after the capacitor is charged. Therefore
if there in an RC circuit hooked to a DC source, all the voltage is dropped
across the capacitor. Voltage across the resistor is zero since the current is
zero VR = IR = 0R = 0
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AC circuits - AC response of Inductors
You know that the voltage across an inductor leads the current because
the Lenz’ law behavior resists the buildup of the current, and it takes a
finite time for an imposed voltage to force the buildup of current to its
maximum.
Impedance for a inductor is XL = ωL
For DC currents ω = 0 and hence the Impedance of the inductor for DC
currents is
XL = ωL = 0
Therefore if there in an RL circuit hooked to a DC source, all the voltage
is dropped across the resistor. Voltage across the resistor is E = VR = IR.
Voltage across the inductor is VL = IXL = I0 = 0.
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AC circuits - AC response of LRC Circuits