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Transcript
Induced EMFs and
Electric Fields
AP Physics C
Montwood High School
R. Casao
• A changing magnetic flux induced an EMF and a
current in a conducting loop.
• An electric field is created in the conductor as a
result of the changing magnetic flux.
• The law of electromagnetic induction shows that
an electric field is always generated by a changing
magnetic flux, even in free space where no charges
are present.
• The induced electric field has properties that are
very different from those of an electrostatic field
produced by stationary charges.
• Consider a conducting loop of radius r in a uniform
magnetic field that is perpendicular to the plane of
the loop.
• If the magnetic field
changes with time,
Faraday’s law tells us
that an EMF given by
-dΦm
EMF =
dt
is induced in the loop.
• The induced current produced implies the presence
of an induced electric field E, which must be
tangent to the loop since all points on the loop are
equivalent.
• The work done in moving a
test charge q once around the
loop is equal to W = q·EMF.
• The electric force on the charge is F = q·E, the
work done by this force in moving the charge
around the loop is W = q·E·2·π·r, where 2·π·r is
the circumference of the loop.
• The two equations for work
are equal to each other:
q·EMF = q·E·2·π·r, so
EMF
E=
2πr
• Combining this equation for the electric field,
Faraday’s law, and the fact that magnetic flux
Φm = B·A = B·π·r2 for a circular loop shows that
the induced electric field is:
-1 dΦm
-1
E=
=

2    r dt
2  r

d B    r2

dt
-  r2 dB -r dB
E=

= 
2    r dt 2 dt
• The negative sign indicates that the induced
electric field E opposes the change in the magnetic
field.
• An induced electric field is produced by a
changing magnetic field even if there is no
conductor present.
• A free charge placed in a changing magnetic field
will experience an electric field of magnitude:
-r dB
E= 
2 dt
• The EMF for any closed path can be expressed as
the line integral of E•ds over the path.
• The electric field E may not be constant, and the
path may not be a circle, therefore, Faraday’s law
of induction can be written as:
-dΦm
E•ds=
dt
• The induced electric field E is a non-conservative,
time-varying field that is generated by a changing
magnetic field.
• The induced electric field E can’t be an electrostatic field because if the field were electrostatic,
hence conservative, the line integral of E•ds
over a closed loop would be zero (dΦm/dt = 0).

Electric Field Due to a Solenoid
• A long solenoid of radius R has n turns per unit
length and carries a time-varying current that
varies sinusoidally as I=Io  cos  ω  t  , where Io is
the maximum current and ω is the angular
frequency of the current
source.
• A. Determine the
electric field outside
the solenoid, a
distance r from the
axis.
• Take the path for the line
integral to be a circle
centered on the solenoid.
• By symmetry, the magnitude of the electric field E is
constant and tangent to the
loop on every point of
radius r.
• The magnetic flux through
the solenoid of radius R is:
Φm =B  A =B    R2
• Applying Faraday’s law:

dΦm d  B  A 
E•ds =
=
dt
dt

d B    R
2

E•ds =

dB
E•ds =    R 
dt
dt
2

• The electric field E is constant at all points on
the loop:
2 dB
E•ds =    R 
dt
2 dB
E  ds     R 
ds  2    r
dt
2 dB
E 2 r   R 
dt


  R 2 dB
E

2    r dt
R 2 dB
E

2  r dt

• The magnetic field inside the solenoid is:
B = μo·n·I
2
2 d μ nI

• Substituting: E  R  dB  R   o
2  r dt
2r
dt
2
 μo  n  R d  I 
E

2r
dt
μo  n  R d  Io  cos  ω  t  
E

2r
dt
μo  n  Io  R 2 d  cos  ω  t  
E

2r
dt
2
d ω  t
μo  n  Io  R
E
 sin  ω  t 
2r
dt
2
μo  n  Io  R 2
dt
E
 sin  ω  t   ω 
2r
dt
2
μo  n  Io  R
E
 ω  sin  ω  t 
2r
• The electric field varies sinusoidally with time, and
its amplitude fall off as 1/r outside the solenoid.
• B. What is the electric field inside the solenoid, a
distance r from its axis?
• Inside the solenoid, r < R, the magnetic flux
through the integration loop is Φm = B·π·r2.


dΦm d  B  A 
E•ds =
=
dt
dt
E•ds =

d B    r
2

dt
dB
E•ds =    r 
dt
2 d  μo  n  I 
E  ds    r 
dt

2

E
 ds    r
2
E
 ds    r
2
E

E
 ds    r
 μo  n 
dt
d  Io  cos  ω  t  
ds    r2  μo  Io  n 
2
dt
d  cos  ω  t  
dt
 μo  Io  n  sin  ω  t  
d ω  t
dt
dt
2
ds    r  μo  Io  n  sin  ω  t   ω 
dt

E   ds    r
E
 μo  n 
d  I
2
 μo  Io  n  ω  sin  ω  t 

ds  2    r
E  2    r    r  μo  Io  n  ω  sin  ω  t 
2
  r  μo  Io  n  ω  sin  ω  t 
E
2  r
2
E
r  μo  Io  n  ω  sin  ω  t 
2
• The amplitude of the electric field inside the
solenoid increases linearly with r and varies
sinusoidally with time.