Electricity and Magnetism Flashcards

1
Q

how did Faraday discover EM induction

A

a voltage was produced when magnetic field through circuit changed

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

modern version of Faraday’s experiment

A
  1. magnet at rest = no current in coil
  2. magnet moving relative to coil = current induced in coil
  3. second current-carrying coil moving relative to stationary coil = current induced in stationary coil
  4. second current-carrying coil at rest relative to outer coil = current induced in outer coil only when current in inner coil changes
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

magnetic flux

A

similar to electric flux

divide any surface into elements of area dA

magnetic flux through surface element is B.dA

total magnetic flux is sum of these contibutions i.e integral of B.dA

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

units of magnetic flux

A

Tm^2 = webber Wb

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

Faraday’s law

A

induced emf in a coil is proportional to the negative of the rate of change of magnetic flux through the coil

for coil with N turns, emf is N times bigger

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

emf

A

voltage generated by battery or by magnetic force

energy per unit charge

available by chemical energy for battery, magnetic for induction

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

B.A=BAcostheta where theta is

A

the angle between the direction of the area vector (normal to the plane of the coil) and the direction of the magnetic field

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

what can give rise to an induced emf

A

changing any of B, A or theta

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

right hand rule

A
  1. point thumb of right hand in +ve direction of area vector
  2. induced emf is +ve in direction of curled fingers
  3. positive charges in direction of emf
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

if emf is induced in a closed loop

A

it will cause a current

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

direction of induced conventional current

A

in same direction as induced emf

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

Lenz’s law

A

electrons flowing along wire represented as a current flow (+ to -) in opposite direction

current flow creates a magnetic field around wire

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

how to work out field direction

A

RH grip rule

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

RH grip rule

A

grip wire with right hand
point thumb in direction of (conventional) current flow
fingers wrap in direction of magnetic field

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

current induced by

A

a change in magnetic flux

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

current induced by a change in magnetic flux creates additional…

A

magnetic field, B induced

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

the induced current produces a magnetic field that tends to…

A

oppose the change in magnetic flux that gave rise to the induction

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

eddy current created when

A

a moving conductor experiences changes in a magnetic field generated by a stationary object

or when a stationary conductor encounters a varying magnetic field

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
19
Q

relative motion causes circulating eddies of current within the conductor which in turn…

A

create magnetic fields that oppose the effect of the applied magnetic field

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

eddy currents create losses due to

A

resistive heating (ohmic heating) as the heating power generated in an electrical conductor of resistance E through which a current I is flowing if P prop to I^2R

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
21
Q

eddy current use

A

brakes on trains

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
22
Q

how to reduce eddy currents

A

thin laminations (insulated layers) that minimise current flow

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
23
Q

induced emf - area of coil is A and magnetic field strength is B so magnetic flux through coil is

A

ΦB = B.A = BAcos theta

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
24
Q

induced emf - as the loop is rotating uniformly Φ(t)=

A

Φ0 + wt

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
25
induced emf - according to faraday's law ε=
- dΦB/dt = BAwsinΦ this is the basis of AC generators
26
AC generator
mechanical energy input to a generator turns the coil in the magnetic field voltage proportional to the rate of change of the area facing the mag field is generate in the coil sinusoidal voltage output
27
commercial AC generators
single power station sized generator B field provided by electromagnet to get large B reverse stationary and rotating bit so rotating B field and stationary coils
28
slidewire generator - flux through the loop is changing because
area of the loop in the uniform field is changing as the rod is slid to the right
29
induced emf in slidewire generator
A and B point into page so ΦB=BA area and width increasing uniformly A(t) = Lw(t) = A0 + Lvt faraday's law: ε= - dΦB/dt = -d/dt[B(A0+LVT)] = -BLv
30
magnitude of induced emf in slidewire generator
-BvL
31
the -BvL emf was caused only by
moving conductor
32
isolated conducting rod emf
same emf developed between its ends
33
force on a charge Q in magnetic field B
F=Qv x B
34
emf between ends of rod
ε= integral of E.dL where path integral is along the rod
35
motional emf - using F=QE
sub in E=F/Q to integral of E.dL and then sub in F= Qv x B and then cancel Qs so ε=vBL
36
changing flux causes an induced emf and hence current, how?
changing magnetic flux causes an induced electric field in the conductor, must be non-conservative since it does net work on a charge as it is driven around loop i.e. integral E.dL = - dΦB/dt
37
changing current will give rise to
changing magnetic field produced by the first circuit and hence a changing magnetic flux through the second Faraday's law then tells us there will be an induced current in the second circuit
38
induction between two adjacent coils
current 1 in coil 1 creates a magnetic field in region of coil 2 magnetic flux through coil 2 is N2ΦB2 (N is no of turns in coil) which is prop. to B and therefore I1
39
mutual inductance
proportionality constant M such that N2ΦB2=MI1 unit Henry (H)
40
what does value of mutual inductance depend on
properties (geometric etc) of the two coils
41
reciprocity theorem
other way round same thing ie N1ΦB1=MI2
42
solenoid
used interchangeably with coil
43
induction happens when
current in one of the coupled coils changes
44
mutual inductance and emf
start with N2ΦB2=MI1 take time derivative of both sides use faraday's law to get emf induced in coil 2 (same applies if started with other way round)
45
transformers - if current flowing through coil 1 is I1=I0coswt, emf induced in coil 2 is
ε2= -MdI1/dt = MI0wsinwt
46
induction applications
transformers used: powering low-voltage devices from mains, generation of high voltage eg ignition sparks in car engines also used: metal detectors, wireless chargers and security tags
47
self inductance L
Proportionality factor depends on number of turns, geometry and material inside coil, unit is henry NΦB=LI
48
flux through a single turn
ΦB=BA =μ0NI/l piR^2
49
if inductor has a material in its core, we need to
replace vacuum magnetic permeability μ0 with μrμ0
50
if the current in the inductor varies, then so does
the flux through its N turns NΦB=LI (take time derivs and use faraday's law)
51
potential change or voltage across an inductor
ε=-N dΦB/dt = -L dI/dt
52
electric field lines
originate at positive charges and direction is direction of the net field at that point never cross each other density of field lines at a location indicates magnitude of field there
53
outward electric flux
positive charge inside the box
54
electric flux
analogous to magnetic flux ΦE= integral E.dA
55
Gauss' law
electric flux over the surface of a volume V is prop. to electric charge contained in V ΦE= integral over surface of V E.dA = Qenclosed/ε0
56
an inductor opposes...
(by generating the induced emf) any change in the current flowing through it So work must be done by an external source to change the current
57
resistors vs inductors
resistor with current I: energy is dissipated inductor with current I: energy is stored
58
* Going round a loop, if an inductor is traversed in the same direction as the direction of flow of positive (conventional) current
the “potential change” (the potential at the output node of the inductor minus the potential at the input node) is -L dI/dt
59
* Gauss’s law provides an easy way of finding
the electric field for charge distributions that exhibit a high degree of symmetry
60
symmetry in some problems allows an intelligent choice of
integration (Gaussian) surface
61
General recipe for using Gauss’s law to calculate E
1.From the symmetries of the problem, deduce the direction of the electric field you want to calculate 2. Set up a surface (“Gaussian surface”), chosen such that the calculation of the electric flux through the surface is easy 3. Compute the electric flux through the Gaussian surface, which will involve the unknown electric field E as a variable 4. Equate the flux to (enclosed charge)/ε0 5. Calculate E
62
in equilibrium, the electric field inside a conductor is
zero, and any excess charge on the conductor resides solely on its surface
63
the work done by the E field when moving a charge from A to B is
path independent
64
the component of the E field that is parallel to the surface is
continuous, the general boundary condition on the E field
65
electrostatic force is a
conservative force
66
work done by the force round a closed path is
0
67
potential for a conservative force
potential energy per unit charge, volts
68
work done by the electric force when a unit charge moves from a to b
Vab= Va - Vb
69
equipotentials
always normal to electric field lines (No work to move charges along equipotentials so no electric field component along equipotentials)
70
capacitor
A capacitor is a device for storing electric charge * It has two conductors which are electrically separated * One is positively charged, the other negatively
71
various geometric possibilities for a capacitor including:
* Parallel plates * Coaxial cylinders * Concentric spheres
72
* If charges +Q and –Q are present on the two plates there will be a
potential difference V between the plates * The plate with +ve charge is at higher potential
73
The electric field between the plates of a capacitor is proportional to
the magnitude of Q so potential difference between the plates is also proportional to Q
74
capacitance is defined by the ratio
C = Q/V
75
SI unit of capacitance
Farad 1F=1C/V 1 farad is a huge capacitance
76
the potential difference between the plates is given by
V=Ed
77
instead of vacuum, most practical capacitors have
an insulating material between the electrodes such material is called a dielectric
78
why use a dielectric
1. makes it easy to get conductors close together (larger capacitance) without danger of electrical contact 2. gives bigger capacitance because of what dielectrics do to the electric field inside the capacitor 3. can withstand higher voltages than air allowing higher stored charge and energy
79
what does a dielectric do?
adding a dielectric to a charged capacitor reduced the potential difference between plates removing it restores potential difference
80
if charge stays the same with a dielectric and voltage reduces then...
capacitance must be increased without dielectric: C0=Q/V0 with: C=Q/V
81
dielectric constant (or relative permittivity) of material
εr=C/C0
82
for fixed charge, presence of a dielectric reduces the
potential difference by a factor of εr V=V0/εr
83
for fixed free charge on the plates, if the voltage across the plates decreases then so
must the electric field in the gap E=E0/εr
84
net surface charges
all +ve charges slightly shifted one way, -ve other way surface charges not free to move ie bound charges **dielectric is polarised **
85
if a dielectric with dielectric constant εr is inserted between the plates, electric field becomes
E=E0/εr= Q/A / εrε0 ε=εrε0 is permittivity of the material
86
capacitance for parallel capacitors
voltage for both capacitors is same so Ceq=Q/V = Q1+Q2/V = Q1/V +Q2/V = C1 + C2
87
capacitance for capacitors in series
Ceq= Q/V = Q/V1+V2 1/Ceq = V1+V2/Q = V1?Q +V2/Q = 1/C1 +1/C2
88
combining capacitors
think of like combining resistors with rules: parallel - Ceq=C1+C2 series - Ceq= 1/C1 + 1/C2
89
rate of charge transfer to capacitor's left plate
I=dQ/dt
90
RC circuit charging - voltages (loop rule)
ε = VR+VC sub in VR=IR VC=Q/C I=dQ/dt end up with diff.eqn.
91
energy stored in capacitor is energy stored in
the electric field In a plate capacitor, this E-field fills a volume Ad resulting in an enery density
92
energy density in magnetic field
uB = 1/2B^2/μ **μ is permeability**
93
energy density in electric field
uE=1/2 εE^2 **ε is permittivity**
94
in AC circuits, voltage and current vary
sinusoidally
95
AC circuit sinusoidal form
V(t)=V0cos(wt+phi0)
96
AC allows more efficient
power transmission (because of transformers)
97
phasors
graphical representations of AC quantities vector rotating anticlockwise instantaneous value is projection onto horizontal axis
98
phasor - resistor across AC source
V0/R cos(wt+phi0) voltage and current in phase parallel on phasor diagram
99
phasor - inductor across an AC source
I(t)=I0cos(wt+phi0) V(t)=L dI/dt = -LI0w sin(wt+phi0) voltage leads current by 90 degrees
100
phasor - capacitor across an AC source
V(t)=Vocos(wt+phi0) I(t)=dQ/dt = d/dt CV(t) = -CV0wsin(wt+phi0) voltage lags current by 90 degrees
101
Lenz's law tells us that induced emf resists...
the buildup of current when a voltage is applied to an inductor takes a finite time for current to rise to max so voltage across inductor leads current
102
for a capacitor, the current that flows to charge up capacitor, causes voltage
across the capacitor to rise so voltage lags the current
103
rms voltage
Vo/root 2
104
rms values are important for calculating
dissipated power when AC signals are used
105
impedance, Z
ratio of complex voltage and current Z=V/I
106
impedance for a resistor
Z=R simply resistance and it is real
107
impedance for inductors and capacitors
Z is complex describing both the relative magnitude and phase of voltage and current
108
impedance of an inductor
start with sinusoidal current I=I0exp(iwt) voltage is V=L DI/dt = iwLI Z=V/I = iwL
109
impedance of a capacitor
start with sinusoidal voltage V=V0exp(iwt) Q(t)=CV(t) I=dQ/dt = iwCV Z=V/I = 1/iwC
110
Z for resistor
Z=R
111
Z for inductor
Z=iwL
112
Z for a capacitor
Z= 1/iwC
113
phasor - impedance for resistor
V and I parallel
114
phasor - impedance for inductance
V leads I by 90 degrees
115
phasor - impedance for capacitor
V lags I by 90 degrees
116
impedances in series
exactly like resistors
117
impedance in parallel
exactly like resistors
118
series LCR circuit
loop rule differentiate wrt t divide by L use I(t)=dQ(t)/dt
119
series LCR circuit is same as
damp, driven harmonic oscillator
120
LC circuit energy oscillates between being
stored in capacitor (in E field) and inductor (B field)
121
impedance analysis
impedance of all the components adds up (series) Z=R+L+C
122
greatest current
resasonace I=V/Z
123
maximum |I| when
|Z| is minimal
124
what w when |Z|minimal?
w=w0=1/sqrt(LC)
125
expensive way to have voltage source whose voltage is independent of current
build huge redundancy into voltage source
126
alternative way to have voltage source whose voltage is independent of current
use feedback to keep voltage constant standard feedback component: op-amp
127
key parts of an op-amp
non-inverting input V+ inverting input V- voltage source,G=open-loop gain Vout
128
output of opamp is
input multiplied by a factor of G
129
ideal open loop gain
infinity
130
ideal input resistance
infinity
131
ideal output resistance
0
132
negative feedback example
if bicycle is heading too far right, you steer left
133
non-inverting amplifier
voltage source attached to non-inverting part
134
infinite input impedance means
no current flowing through inputs
135
infinite G means
voltage at inverting and non-inverting inputs are the same
136
exact parameters of op-amp don't matter as long as
G>>1 Rin>>1 Rout<<1
137
wide tolerances of op-amp mean
very cheap to manufacture
138