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Transcript
E & B Fields
28 TH FEBRUARY – BG GROUP
What is a field?
A field is a physical quantity that
has a value for each point in
space and time.
For example, temperature at
the surface of the Earth is a
scalar field, it has a single value
at every point on the Earth.
Wind measurements give a
vector field, each point has both
a magnitude (wind speed) and
direction.
What is a field?
We represent vector fields using
field lines.
The lines themselves tell us about
the direction of the field.
The spacing of the lines tell us the
strength of the field; the closer
together the lines are, the stronger
the field is at that point.
Electric field strength
Electric field strength is defined as the force per unit charge.
𝐸=
𝐹
𝑄
𝐸 is the electric field (NC-1)
𝐹 is the force (N)
𝑄 is the charge
Radial electric fields
The field around a point charge is
radial, so the force the charge would
experience depends on π‘Ÿ 2 .
Coulombs law allows us to work out
the force of attraction or repulsion
between two point charges.
𝐹=
q
𝑄1 𝑄2
4πœ‹πœ–0 π‘Ÿ 2
𝐹 is the force (N)
𝑄 is the charge (C)
π‘Ÿ is the separation between the charges (m)
πœ–0 is the permittivity of free space
Lines of equipotential
Radial electric fields
The electric field strength at a
distance from a point charge is given
by:
𝐸=
𝑄
4πœ‹πœ–0 π‘Ÿ 2
𝐸 is the electric field strength (NC-1)
𝑄 is the charge (C)
π‘Ÿ is the separation between the charges (m)
πœ–0 is the permittivity of free space
q
Lines of equipotential
Radial electric fields
The distance between two point charges of +8.0nC and +2.0nC is 60mm.
At a point between them, the resultant electric field strength Is zero.
How far is this point from the +8.0nC charge?
+8.0nC
+2.0nC
q
q
60mm
Uniform electric fields
A uniform field can be produced
between two metal plates by
connecting them to a cell.
300V
The field strength is the same at all
points between the plates and is
given by:
𝐸=
𝑉
𝑑
𝐸 is the electric field strength (Vm-1)
𝑉 is the potential difference (V)
𝑑 is the separation of the plates (m).
0V
Lines of equipotential
Charges in electric fields
Charged particles move through electric fields like projectiles. The
particle will experience a force parallel to the field lines and accelerate
in this direction at a constant rate. The path is a parabola.
1000V
v = 2×106 ms-1
p
0.1m
s
0V
0.2m
Calculate the deflection of the proton, s.
Electric potential
Electrical potential, 𝑉, is the
electrical potential energy per unit
charge and is measured in Volts.
𝑉=
V
𝑄
4πœ‹πœ–0 π‘Ÿ
To move a charge from one
potential to another requires doing
work, Ξ”π‘Š.
Ξ”π‘Š = π‘žΞ”π‘‰
𝑉 is positive when the force is
repulsive and negative when the
force is attractive.
Gradient is the
field strength
Repulsive
r
V
Attractive
r
Gradient is the
field strength
Magnetic fields
A magnetic field is a region in which a force is exerted on a magnetic
material. A current flowing in a wire will produce a magnetic field. The
direction of this field can be found using the right-hand rule.
Force on a current
Motion
Field
A wire carrying a current in a
magnetic field will experience a
force. Fleming's left hand rule can
determine which direction this
force will be in and the magnitude
of this force can be determined by:
𝐹 = 𝐡𝐼𝑙
Current
First finger – Field
Second finger – Current
Thumb – Motion (force)
𝐹 is the force (N)
𝐡 is the magnetic flux density (T)
𝐼 is the current (A)
𝑙 is the length of wire in the field
Force on a current
In the situation shown, work
out the direction in which
the wire will move, and
calculate the force exerted
on the wire.
𝐡 = 0.25T
0.1m
𝐼 = 3A
Force on a current
The Force is the greatest when the wire and the field are perpendicular.
πœƒ
If πœƒ = 90°, 𝐹 = 𝐡𝐼𝑙
If πœƒ = 30°, 𝐹 = 𝐡𝐼𝑙 × 0.5
If πœƒ = 0°, 𝐹 = 0
For a wire at angle πœƒ to the field, the force acting on it is:
𝐹 = 𝐡𝐼𝑙 sin πœƒ.
Charges in magnetic fields
𝑣 = 2×106 ms-1
Forces act on charged particles in a
magnetic field. This force is given
by:
𝐹 = π΅π‘žπ‘£
According to Fleming’s left hand
rule, the force is always
perpendicular to the direction of
travel – this results in circular
motion. The force is equal to the
centripetal force, given by:
π‘šπ‘£ 2
𝐹=
π‘Ÿ
p
𝐡 = 0.20 T
×
×
×
×
×
×
×
×
×
×
×
×
Calculate the radius of curvature, r.
Flux, flux density & flux linkage
Magnetic flux
Magnetic flux density
Magnetic flux linkage
Symbol: πœ™
Symbol: 𝐡
Symbol: Ξ¦
Unit: Wb (Weber)
Unit: T (Tesla)
Unit: Wb (Weber)
Think of it as the total
number of field lines
The number of field
lines per square meter
Total number of field
lines within a coil
multiplied by the
number of turns
The descriptions above lead us to the following relationships:
πœ™ = 𝐡𝐴
Ξ¦ = π‘πœ™ = 𝐡𝐴𝑁
Coil not perpendicular to B
If the coil is not perpendicular to
the magnetic flux, you can use the
following equations to calculate πœ™
or Ξ¦.
Coil
πœ™ = 𝐡𝐴 cos πœƒ
Ξ¦ = 𝐡𝐴𝑁 cos πœƒ
πœƒ
In each case, πœƒ is the acute angle
between the normal of the coil
and the magnetic flux.
Normal to the coil
Electromagnetic induction
When a conductor moves through a magnetic field, its electrons will
experience a force. This means they will accumulate at one end of the
conductor inducing an emf just like a battery. An emf is produced when
lines of flux are β€˜cut’, but a current will only be induced if the circuit is
complete.
×
×
×
×
×
×
×
×
×
×
×
×
Positive charge
Negative charge
Faraday’s law
The induced emf (πœ–) is directly proportional to the rate of change of flux
linkage.
flux change ΔΦ
ΔΦ
πœ–=
=
=𝑁
time taken
Δ𝑑
Δ𝑑
Ξ¦
πœ–
Gradient = πœ–
Area = Ξ¦
time
time
Faraday’s law
An aeroplane with a wingspan of 45m is travelling at 950km/h
perpendicular to the magnetic field of the Earth. The magnetic field
strength is 5×10-5 T. Calculate the emf induced across the wingtips.
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
45m
264m
Finally, use πœ™ = 𝐡𝐴 and πœ– = 𝑁
Ξ”πœ™
Δ𝑑
to calculate the emf
Lenz’s law
The direction of the
induced emf is given by
Lenz’s law. It states that
the induced e.m.f is
always in such as a
direction as to oppose the
change that caused it.
You can work out the
direction of the emf by
using Fleming’s left hand
rule – with your right
hand.
𝑣 = 0.2ms-1
0.1m
0.3m
𝐡 = 0.25T
Calculate the flux cut by the wire, and the emf
generated.
Flux linkage, emf and phase
Ξ¦ / Wb
For a rotating coil, flux linkage and induced voltage are 90° out of phase.
Flux linkage is given by Ξ¦ = 𝐡𝐴𝑁 cos πœ”π‘‘ and induced emf can be found
with πœ– = π΅π΄π‘πœ” sin πœ”π‘‘ where πœ” = 2πœ‹π‘“.
t/s
N
πœ–/V
S
t/s
Transformers
An alternating voltage in the
primary coil produces an
alternating magnetic flux. The flux
travels through the iron core and
induces an alternating voltage in
the secondary coil. For an ideal
transformer:
𝑉𝑝 𝑁𝑝 Is
=
=
𝑉𝑠 𝑁𝑠 Ip
For a real transformer:
efficiency =
𝑉𝑠 𝐼𝑠
𝑉𝑝 𝐼𝑝
Primary
1200 turns
0.25A
240V
Secondary
500 turns
Efficiency
83%
Calculate the current in the
secondary coil
Velocity selector
A proton is accelerated through a 16kV potential before entering a
velocity selector. The velocity selector consists of two parallel plates
with a potential difference of 62kV and a separation of 0.2m. Calculate
the magnetic field such that the proton is not deflected.
16kV
62kV
0V
×
×
×
×
×
×
p
0.2m
×
×
×
×
×
×
0V