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Electromagnetic Induction!
March 11, 2014
Chapter 29
1
Notes
!  Exam 4 next Tuesday
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Covers Chapters 27, 28, 29 in the book
Magnetism, Magnetic Fields, Electromagnetic Induction
Material from the week before spring break and this week
Homework sets 7 and 8
!  I will be out of town next week
•  Prof. Joey Huston will teach lectures
March 11, 2014
Chapter 29
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Magnetic field
!  The force that a magnetic field exerts on a charge moving with velocity
v is given by

 
FB = qv × B
!  The direction is sideways (RH rule).
!  The magnitude of the force
FB = qvBsin θ
!  If the charge moves perpendicular to the magnetic field then
F = qvB
Physics for Scientists & Engineers 2
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Review: Magnetic field
!  The unit of magnetic field strength the tesla (T)
Ns
N
1T =1
=1
Cm
Am
!  Another unit of magnetic field strength that is often used but is not an
SI unit is the gauss (G)
1 G = 10-4 T
10 kG = 1 T
!  Typically the Earth’s magnetic field is about 0.5 G at the surface.
!  A charged particle with charge q and mass m traveling
with speed v perpendicular to a constant magnetic field B will travel in
a circle with radius r given by
qB
,
so
that
and
angular
frequency
ω=
p
mv
Br =
m
r=
q
qB
Physics for Scientists & Engineers 2
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Hall effect
!  If we run a current i through a
conductor of width h in a constant
magnetic field B, we induce a
voltage VH across the conductor
that is given by
d
VH VH dhne VH hne
B=
=
=
dv
di
i
h
!  where n is the number of electrons per unit volume and e is the
charge of an electron
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magnetic dipole moment
!  We define the magnitude of the magnetic dipole moment of a coil to
be
µ = NiA
!  We can express the torque on a coil
in a magnetic field as
  
τ =µ×B
!  The magnetic potential energy of a magnetic dipole in a magnetic field
is given by
 
U = − µ ⋅ B = − µ B cosθ
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Magnetic field of a wire
!  Magnetic field of a long, straight wire
with current i:
µ 0i
B(r⊥ ) =
2π r⊥
!  Magnetic field in the center of a current-carrying loop of radius R
!  The force between two currentcarrying wires is given by
µ0i1i2 L
F12 =
2π d
Physics for Scientists & Engineers 2
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Solenoid and Toroid
!  The magnetic field inside an ideal solenoid is
given by
B = µ0in
•  The density of the coil loops is given by n
!  The magnetic field inside an ideal toroidal magnet is given
by
µ0 Ni
B=
2π r
•  The number of loops
is given by N
Physics for Scientists & Engineers 2
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Induction - Faraday’s Experiments
!  A magnetic field is generated when electrical charges are in
motion
!  An electric field is generated when a magnetic field is in
motion (relative to a conductor) or otherwise changes as a
function of time
!  Consider a wire loop connected
to an ammeter and a bar magnet
!  While the magnet is stationary,
no current flows in the loop
!  The experiments relevant to this
chapter were performed in the
1830s by the British chemist and physicist Michael Faraday
and independently by the American physicist Joseph Henry
March 11, 2014
Chapter 29
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Faraday’s Experiments
!  If the magnet is moved toward the
loop a counterclockwise current
flows in the loop as indicated by
the positive current in the
ammeter
!  If the magnet is reversed so the
south pole points toward the loop
and moved toward the loop,
current flows in the loop in the
opposite direction as indicated by
the negative current in the
ammeter
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Chapter 29
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Faraday’s Experiments
!  If the north pole of the magnet
points toward the loop, and the
magnet is then moved away from
the loop, a negative clockwise
current, as indicated on the
ammeter, is induced in the loop
!  If the south pole of the magnet
points toward the loop, and the
magnet is moved away from the
loop a positive current is induced
as indicated by the ammeter
March 11, 2014
Chapter 29
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Faraday’s Experiments
If a constant current is
flowing through loop 1, no
current is induced in loop 2.
If the current in loop 1 is
increased, a current is
induced in loop 2 in the
opposite direction
If current is flowing in loop
1 in the same direction as
before and is then
decreased, the current
induced in loop 2 flows in
the same direction as the
current in loop 1
March 11, 2014
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Law of Induction
!  From these observations we see that a changing magnetic
field induces a current in a loop
!  We can visualize the change in magnetic field as a change in
the number of magnetic field lines passing through the loop
!  Faraday’s Law of Induction states that:
A potential difference is induced in a loop when the number
of magnetic field lines passing through the loop changes with
time
!  The rate of change of magnetic field lines determines the
induced potential difference
March 11, 2014
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Magnetic Flux
!  When we introduced Gauss’s Law for the electric field, we
defined the electric flux as
 
ΦE = 
∫∫ E • dA
!  For the magnetic field, we can define magnetic flux in
analogy as
 
ΦB = 
∫∫ B• dA
!  Integration of the magnetic flux over a closed surface yields
zero:
 

∫∫ B• dA = 0
!  This result is often termed Gauss’s Law for Magnetic Fields
!  There are no free magnetic charges, no magnetic
monopoles, no separate north poles or separate south poles
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Magnetic Flux
!  Consider the special case of a flat loop
of area A in a constant magnetic field B
!  We can re-write the magnetic flux as
Φ B = BAcosθ
!  If the magnetic field is
perpendicular to the plane of the loop
Φ B = BAcos0° = BA
!  If the magnetic field is parallel to the plane of the loop
Φ B = BAcos90° = 0
!  The unit of magnetic flux is the weber (Wb) given by
1 Wb = 1 T m 2
March 11, 2014
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Faraday’s Law of Induction
!  We can then recast Faraday’s Law of Induction in terms
of the magnetic flux as
The magnitude of the potential difference, ΔVind, induced in a
conducting loop is equal to the time rate of change of the
magnetic flux through the loop.
!  Faraday’s Law of Induction is thus contained in
dΦ B
ΔVind = −
dt
!  We can change the magnetic flux in several ways including
changing the magnitude of the magnetic field, changing the
area of the loop, or by changing the angle the loop with
respect to the magnetic field
March 11, 2014
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Induction in a Flat Loop
!  Let’s explore Faraday’s Law for the case of a flat loop and
a magnetic field that is constant in space, but varies in time
!  The magnetic flux for this case is
Φ B = BAcosθ
!  The induced emf is then
dΦ B
d
ΔVind = −
= − (BAcosθ )
dt
dt
!  Carrying out the derivative we get
ΔVind
dB
dA
dθ
= − Acosθ
− Bcosθ
+ ABsin θ
dt
dt
dt
!  Taking dθ/dt = ω we get
dB
dA
ΔVind = −Acosθ
− Bcosθ
+ ω ABsinθ
dt
dt
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Induction in a Flat Loop - Special
Cases
!  If we leave two of the three variables (A,B,θ) constant, then
we have the following three special cases
•  We leave the area of the loop and its orientation relative to the
magnetic field constant, but vary the magnetic field in time
dB
A,θ constant: ΔVind = −Acosθ
dt
•  We leave the magnetic field as well as the orientation of the loop
relative to the magnetic field constant, but change the area of the
loop that is exposed to the magnetic field
dA
B,θ constant: ΔVind = −Bcosθ
dt
•  We leave the magnetic field constant and keep the area of the loop
fixed as well, but allow the angle between the two to change as a
function of time
A, B constant: ΔVind = ω ABsinθ
March 11, 2014
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Changing Magnetic Field
!  A direct current of 600 mA is delivered to an ideal solenoid,
resulting in a magnetic field of 0.025 T
!  Then the current is increased according to
i(t) = i0 (1+ 2.4s −2t 2 )
PROBLEM
!  If a circular loop of radius 3.4 cm with 200 windings is
located inside the solenoid and perpendicular to the
magnetic field, what is the induced voltage at t = 2.0 s
in this loop?
SOLUTION
!  First, we compute
the area of the coil
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Changing Magnetic Field
!  Since it is circular, its area is πR2
!  However, there are N windings in the coil
!  The effective area of the coil is then
A = N π R 2 = 200π (0.034 m)2 = 0.73 m 2
!  The magnetic field inside an ideal solenoid is
B = µ0in
!  Because the magnetic field is linearly proportional to the
current, we obtain the time dependence of the magnetic
field in this case
B(t) = B0 ⎡⎣1+ ( 2.4s −2 )t 2 ⎤⎦ , B0 = 0.025 T
!  The area of the loop and the angle are kept constant so
dB
A,θ constant: ΔVind = −Acosθ
dt
March 11, 2014
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Changing Magnetic Field
!  For the induced voltage we then find
A,θ constant: ΔVind = −Acosθ
dB
dt
dB
ΔVind = −Acosθ
dt
d⎡
= −Acosθ ⎣ B0 ⎡⎣1+ ( 2.4s −2 )t 2 ⎤⎦ ⎤⎦
dt
= −Acosθ ⎡⎣ B0 ( 2.4s −2 ) 2t ⎤⎦
= −AB0 cosθ ⎡⎣ 2 ( 2.4s −2 )t ⎤⎦
= −(0.73 m 2 )(0.025 T)(cos0°)(4.8s −2 )t
at t = 2.0 s, ΔVind = −0.17 V
March 11, 2014
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Potential Difference Induced by a Moving
Loop
!  A rectangular loop of width 3.1 cm and
depth 4.8 cm is pulled out of the gap
between two permanent magnets, with a
field of 0.073 T throughout the gap
PROBLEM
!  If the loop is removed with a constant
velocity of 1.6 cm/s, what is the induced
voltage in the loop as a function of time?
SOLUTION
!  The magnetic field as well as the orientation of the loop
relative to the field remains constant
!  What changes is the area of the loop that is exposed to the
magnetic field
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Potential Difference Induced by a Moving
Loop
!  With the narrow gap, there will be very
little field outside the gap
!  The area of the loop exposed to the field is
A (t ) = ( w ) ( d (t ))
!  d(t) is the depth of the part of the loop in
the magnetic field at time t
d (t ) = d0 − vt
!  While the entire loop is in the magnetic field
no potential difference is produced
!  Letting the time of arrival of the right edge of the loop at the
right end of the gap be t = 0, we have
A (t ) = ( w )( d (t )) = w ( d0 − vt )
!  This holds until the left edge of the loop reaches the right
end of the gap, after which the exposed area is zero
March 11, 2014
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Potential Difference Induced by a Moving
Loop
!  The left edge arrives at time
d 4.8 cm
tf = =
= 3.0 s
v 1.6 cm/s
!  The induced potential difference is
dA
ΔVind = −Bcosθ
dt
d
= −Bcosθ ⎡⎣w ( d0 − vt ) ⎤⎦
dt
= wvBcosθ
= (0.031 m)(0.016 m/s)(0.073 T)cos ( 0° )
=3.6⋅10-5 V = 36 µV
!  During the time interval between 0 and 3 s, a constant
potential difference of 36 μV is induced, and no potential
difference is induced outside this time interval
March 11, 2014
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