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FLUX THROUGH A SINGLE TURN COIL
B
The flux through the coil is given by
Ф=AB
Where A is the area of the coil
But what happens to the flux if the coil turns (perpendicular to the field) ?
FLUX THROUGH A SINGLE TURN COIL
B
When the coil begins to turn.
We can see that the flux through the coil becomes
less and when it is horizontal it is zero.
B
FLUX THROUGH A SINGLE TURN COIL
θ
Generalizing the result, taking a line normal to the plane of
the coil (the dotted line)
Where θ is the angle between the field and this normal line
The flux through the coil is given by
Ф=ABcosθ
B
Flux Linkage
When the coil has more than one turn the flux is
equal to the sum of the fluxes through the
individual turns.
We call this the flux linkage though the whole
coil.
FLUX LINKAGE
B
The flux linkage through the whole coil is given by
NФ=NABcosθ
Where N is the number of turns in the
coil
This formula is handy later when we consider the emf induced in a rotating coil.
Faraday’s Law
• Accurate experiments show that the e.m.f.
induced in a coil increases with the rate of
change of magnetic flux through it.
e.m.f.
rate of change of flux
The rate of change of flux is measured in Webers per
second (Wm-1)
ε


t
Faraday’s law of electromagnetic
induction

ε 
t
ε N
So combining these relationships

ε N
t
The units of the SI system combine in such a way that
the constant of proportionality is 1

ε N
t
The expression to the right of the = sign is just the
rate of change of flux linkage
Faraday’s Law
ε

N
t
ε is the e.m.f. Induced (V)
N is the number of turns on the coil
∆Φ is the change in flux though each turn of the coil. (Wb)
∆t is the time taken for the flux change.(s)
Note that in this equation the total change in flux linkage in the coil is N∆Φ.
Sometimes you may see this written as ∆NΦ.
It follows that 1 weber is the flux linkage in a coil if an emf of 1V falls
evenly to zero in 1 second
Using Faraday’s Law
A coil of 200 turns and 3cm in diameter lies perpendicular to a uniform
magnetic field with a flux density of 2 x10-2 T. The field falls evenly to 0T in
1s. Calculate the emf generated:.
1. Calculate the flux
through 1 turn of the coil
  BA
  .02  .0152   1.414 105Wb
2 Now apply Faraday’s law
V

εN
t
(1.414105 )
Ε  200
 2.83 103V
1
Examples
• A coil of 450 turns and 10cm in diameter lies
perpendicular between the poles of an electromagnet.
The field rises evenly from 0 to 4 x10-3T in 0.5s.
Calculate the e.m.f generated:
• A Helmholtz coil with 2000 turns of diameter 12cm is set
perpendicular to a uniform magnetic field of 0.5T. The
field is reduced to 0 in 0.2 seconds. Calculate the e.m.f
generated.
The e.m.f. rotating coil
• Now we can apply Faraday’s law to the
rotating coil
The Flux linkage is:
NФ=NABcosθ
The e.m.f. rotating coil
NФ=NABcosθ
We need to find the rate of change of flux through the coil
The angular speed of the coil is
t
θ


t
NФ=BANcosθ
Becomes
NФ=BANcosωt
The e.m.f. is the rate of change of flux
linkage
N
εE 
t
And as δt →0
dN
εE 
dt
d ( BAN cos t )
εE 
dt
εE  BAN ( sin t )
Differentiation gives
The minus sign is a
consequence of Lenses Law
and gives the direction of the
emf. This can be ignored
εE  BAN sin t
The Rotating Coil
The e.m.f. generated by a rotating coil is given by:
ε
 BAN sin t
B
Note that this A SINUSOIDAL RELATIONSHIP. The e.m.f generated
is varying:
The peak output of
the generator is
BANω when the
sinωt = 1
εE  BAN sin t
ε
T
t
This curve is
characteristic of an a.c.
generator
Questions
• A circular coil with 300 turns and a diameter of 5cm is rotated
at a frequency of 5Hz perpendicular to a magnetic field of
3.0 x 10-3 T. Calculate the e.m.f. generated at a quarter cycle.
• A Helmholtz coil with 1000 turns of diameter 15 cm is set
vertically in the Earth magnetic field and rapidly rotated with a
frequency of 20Hz. Calculate the peak e.m.f. you would
expect to generate. (Take the horizontal component of the
Earth’s magnetic field to be 3.2 x 10-5T)