5. ELECTRICITY AND MAGNETISM - Jnoodle · The IB Physics Compendium 2005: ... 5. ELECTRICITY AND MAGNETISM 5.1. Electric charge • Atoms consist of ... E = F/q 1 which in the IB

The IB Physics Compendium 2005: Electricity and magnetism

Thomas Illman and Vasa övningsskola

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5. ELECTRICITY AND MAGNETISM

5.1. Electric charge

• Atoms consist of heavy, positive protons and neutrons in the nucleus and light, negative

electrons around it

• the two types of negative and positive electric charge are a fundamental property of

materia, like mass

• the net charge is conserved, like mass (except that mass and energy can be converted to

each other (relativity))

• masses always attract each other, but charges of the same type repel; different types

attract

• the unit of charge is 1 coulomb = 1 C; the charge of one electron = e = - 1.6 x 10-19

C (we

can sometimes also use e = the elementary charge = 1.6 x 10-19

C and then the charge of the

proton is e, the charge of one electron is - e.)

• since the sign of the charge denotes its type ("positive" or "negative") but no direction,

charge is a scalar quantity.

Conductors, semiconductors and insulators

A material which electrons can move easily through is a conductor, one where this is more difficult

is an insulator. Metals are good conductors because metal atoms have a few electrons in the outer

shell which are not very strongly attached to any particular nucleus. Semiconductors are materials

where the possibility of conduction of charge depends strongly on some factor (direction,

temperature, light, other).

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In a piece of metal the "unwanted" outer shell electrons are not connected to any particular metal

nucleus and can easily be set in motion by any electric force acting on them. As a result of this,

electrons may then be moving through the metal conductor at some drift velocity which may not be

very high (compare to swithcing on the water in a garden hose - even if the water starts to move

almost immediately, a water molecule does not immediately travel from the tap to the end of the

hose).

When traveling through the metal the electrons will collide with the metal "cations" (positive ions)

formed by the nuclei and the inner shell electrons. In these collisions they lose some of the kinetic

energy they are given by the external battery or other causing the flow of electrons.



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Electrification by friction and contact

By rubbing materials against each other some electrons can be moved from one object to each

other, which means one will have a positive and the other a negative net charge. This works best

with insulators where the net charge on the surface of the material is not easily spread out through

the whole object.

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If a charged object is brought to contact with a conductor with no net charge, this conductor will

also be charged (but the net charge on the first object will decrease).



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Electrostatic induction

If an electrically charged object is place near another object where charges can move easily (a piece

of metal), charges in this object will be attracted or repelled. If an object is allowed to touch another

conductor or some charges are led to or from it from the earth, a conductor can be charged without

touching it.



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The electroscope

A simple instrument to show the presence of electric charge is based on light pieces of conductors

(metal) which all are in contact with each so that if the electroscope plate is touched by a charged

object, the net charge is distributed over all inner parts of the instrument (but the outer parts are kept

insulated).

Some of the inner metal parts are then easy to move by a repulsive force, which can be seen (gold

leaves moving apart, or a metal needle turning in other types of electroscopes).

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If the conductor is hollow, the charge will be distributed on the outside of it, and the inside left

uncharged (it will form a "Faraday's cage").

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This explains why it is relatively safe to sit in a car or an airplane in a thunderstorm, or why radios

and cel phones may not work inside metal cages or buildings.

5.2. Electric force and field

Coulomb's law for electric force

F = kq1q2/r2 where k = 1/4ππππεεεε0 [DB p.7]

where q1 and q2 are the charges, r the distance between them (or the distance between the centers of

them if they are not very small "point charges").

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The Coulomb constant k = 8.99 x 109 Nm

2C

-2 in vacuum and approximately the same in air. In

other materials a k-value can be calculated from the relevant ε-value (electric permittivity). The k-

value and the permittivity in vacuum (or air) are given in the data booklet. The ε-value for other

materials is given when necessary. In vacuum or air ε0 = 8.85 x 10-12

Fm-1

(F the unit 1 farad, not

explained here but a SI-unit). Some table list relative permittivity (εr) values, where the actual

permittivity ε = εrε0.

This can be compared to Newton's law of universal gravity F = Gm1m2/r2 but :

• we have charges instead of masses

• k is much greater than G, but mostly electrical forces are not noticed since ordinary

materia consists of both positive and negative charges, and the Coulomb forces usually

cancel out

• unlike the G-value, the k-value depends on the material (it is much different in water than

in air or in oil).

The Coulomb formula gives the magnitude of the force on either of the charges q1 and q2. The

directions of the forces are opposite (repelling or attracting) because of Newton's III law.

Note:

• if we have more than one charge present, we may have to split up the force(s) from some

of them into components parallel or perpendicular to suitably chosen directions

Electric field

Coulomb's law gives the force acting on a charge q1 caused by q2. If we want to describe what force

would act on an imagined small positive test charge q1 here called just q, we can define the

electric field strength as



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E = F/q1 which in the IB data booklet is given as:

E = F / q [DB p.7]

a vector quantity with the unit 1 NC-1

Using Coulomb's law for the field caused by a charge q2 we get

E = F/q1 = (kq1q2/r2) / q1 = kq2/r

2 which in the IB data booklet is given as:

E = kq / r2 [DB p.7]

Notice that like in Mechanics where m sometimes means the mass of a planet causing a

gravitational field and sometimes the mass of a spacecraft in that field, here q also sometimes

means a "big" charge causing a field, sometimes a small test charge in that field. If we further

compare this to the force of gravity and remembering than mass is replacing charge we get

• F/m = g = the gravitational field strength (near earth the usual gravity constant 9.81 ms-2

which is the same as 9.81 Nkg-1

; compare this to the unit 1 NC-1

!!

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Note that since the imagined small test charge is positive the field is directed away from a positive

charge, and towards a negative charge. The field of this type can be called a radial field.

The field lines drawn do not exist in reality (like the charge causing the field does), they are graphic

descriptions of what would happen (what force would act) if the small positive test charge was

placed in a certain place

• the closer the field lines are, the stronger is the field (nearer the charge; the further

away, the weaker)

Electric field patterns for other situations

If we have two or more charges, the field in a certain point is the sum of the fields caused by the

charges. Since the field E is a vector quantity, directions are relevant and it may be necessary to

split the field vector into suitably chosen components.



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• two point charges of different type: on a line through the charges, the field is from the

positive to the negative between them, away from the positive and into the negative on the

far side of them. In other regions, the field lines are bent curves since at any point it is the

resultant of a vector towards the negative and one away from the positive charge (remember

that the field is defined from a hypothetical small positive test charge - if a negative charge

is placed in the field, it will be affected by a force in the opposite direction to the field).

Since the distance r to the charge appears in the E = kq/r2 , the magnitudes of these vectors

vary. The bent lines do not follow any known mathematical function (they are not parabolas,

hypberbolas or other such curves) and have to be found by calculating the field in every

point in the plane separately (in practice by computer).

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Note: If we place a small positive test charge at rest in the field, it will initially be affected by a

force in the direction of the field in the point where it is placed, but its motion thereafter will not

generally follow a field line - the electric force is parallel to the field, and the acceleration is parallel

to the force (F = ma), but the new velocity v after a short time period t is v = u + at, where u and at

are vectors, and generally not parallel.

• two point charges of same type: if they both are positive, they will "bend away" from a

line where the distance to both is the same. If both are negative, the shape of the field lines

is the same but the direction opposite.

• a charged metal sphere: outside the sphere, the field is the same as if all the net charge

on the sphere was concentrated to its center; inside the sphere it is zero.

The field lines from a metal surface are always at a 90 degree angle to it (otherwise they

would have a component parallel to it, and this component would result in a force parallel to

the surface on any freely moving charges on it, and they would move until this is no longer

the case).

=> if the hollow metal object has another shape, the E-field lines still have to be

perpendicular to its surface. They will be closer together and the field stronger at sharp and

"pointy" places.

• two oppositely charged parallel plates: between the plates, the field is the resultant of

millions of field vectors each describing the effect of one small charge on either of the

plates. The "sideways" components cancel out and the field lines are parallel, going from the

positive to the negative plate. At the ends, outside the area between the plates, they are

slightly bent.

A homogenous or uniform field is one which in some area has the same direction and magnitude.

Can be produced by parallel metal plates.

5.3. Electric potential energy, potential and potential difference = "voltage"

Electric potential energy

The electric field between a positive and negative metal plate is homogenous and similar to the

gravitational field near the surface of a planet (so near that the facts that the planet surface is not flat

and the gravitational force and field get weaker far out in space can be disregarded).

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If a positive test charge q is "lifted up" from A to B or "falls down" from B to A, the change in its

potential energy caused by electrical forces can be calculated. (There may be a force of gravity and

gravitational potential energy involved also, but since the k-constant is much larger than the G-

constant it can usually be disregarded. Also, since we assume the situation to be independent of any

force of gravity, the plate pair can be turned any way we like; "up" just means towards the positive

plate and "down" towards the negative.)

The work done by or against the E-field is then

• W = Felectricx but since E = F/q we get F = qE and then

• W = qEx = the change in potential energy

(compare this qEx to mgh where charge q corresponds to mass m, the electric field strength E to

the gravitational field strength = the gravity constant g, and x or h symbolise how far "up" or

"down" the field we have moved.)

Electric potential V

For the force of gravity we had

• the gravitational potential V = Ep,gravitational / m

and this is here replaced by

• the electric potential V = Ep,electric / q

Remember that the gravitational potential V = Ep/m = mgh/m = gh is rarely used since most

applications of physics are placed near earth and the g-value always the same, so only the h-value is

interesting, for example as in the height difference between to places. We now get:

• the work = change in electric potential energy Ep = W = qEx

• but since the electric potential is defined as V = Ep/q = W/q = qEx / q we get

• V = Ex, which using "deltas" and a negative sign to show that if we move against the

field we gain potential energy and if we move with the field we lose potential energy:

E = - ∆∆∆∆V / ∆∆∆∆x [DB p.7]

Another way to write this is, now replacing ∆x with d for the distance between two charged plates:

E = V / d [DB p.7]

Comparing gravitational and electric quantities: A summary

Here we will for clarity let the big central mass or charge be represented by M or Q, the

hypothetical test- or other small mass or charge with m or q.



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GRAVIT. ELECTR.

Homogenous Point,planet UNIT Homogenous Point, sphere UNIT

F = mg F= GMm/r2 N F = qE F = kQq/r

2 N

g = F/m g = GM/r2 Nkg-1=ms-2 E = F/q E = kQr

2 NC-1=Vm-1

Ep = mgh Ep=-GMm/r J Ep = qEd Ep = kQq/r J

V = Ep/m= gh V = -GM/r Jkg-1 V = Ep/q =Ed V = kQ/r JC-1 = V

Quantities corresponding to each other (gravitation - electricity), in addition to this the universal

gravity constant G = 6.67 x 10-11

Nm2kg

-2 is replaced by the Coulomb constant k = 8.99 x 10

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Nm2C

-2 .

F - F g - E Ep - Ep V - V M,m - Q,q h - d

Potential difference = "voltage"

The potential difference V (if the potential in one point of comparison is zero) or ∆V between to

places in the uniform field or between the plates causing the field is

V = Ep / q

so its unit is 1 JC-1

which is called 1 volt = 1 V.

The potential difference between two points is what is commonly called the "voltage" between them.

It is extremely useful to remember this:

voltagevoltagevoltagevoltage ==== workworkworkwork orororor energyenergyenergyenergy perperperper chargechargechargecharge

for later applications.

Since we have E = V/d we can write the unit for electric field strength E as 1 Vm-1

in addition to

the earlier presented unit 1 NC-1

based on the definition E = F / q.

These units are the same : 1 Vm-1

= 1 JC-1

m-1

= 1 NmC-1

m-1

= 1 NC-1

The unit 1 electronvolt = 1 eV = an energy unit

If one electron with the charge q = e (or - e depending on which definition we follow) = 1.6 x 10-19

C is accelerated through a potential difference of 1 volt, it will get an energy = the work done = qV

= 1.6 x 10-19

C x 1 JC-1

= 1.6 x 10-19

J = 1 eV.

A situation confusing enough to make angels cry is the fact that V is used both as the symbol and

the unit for potential ( we can write V = 5.0 V ) and e both for the electron, the charge of an

electron, and in the unit eV for the energy of an electron.



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The unit 1 eV for energy is in atomic and nuclear physics also used for many other purposes than

just electrons. The energy a charge - electron or other - gets when accelerated by a potential

difference can be as kinetic energy, if air resistance and other forces are not considered:

qV = ½mv2

Electric potential from a point charge or charged sphere

For the gravitational force, a different formula for potential energy had to be used in situations

where an object was not staying near the surface of a planet but moving at significantly different

distances to it (or rather its center), meaning that the force of gravity on it was not constant. The

same can be found for electrical forces - and we can define electric potential V as:

V = kq/r where k = 1/4ππππεεεε0 [DB p.7]

The electric potential is a scalar, which is zero when r is infinitely large. If the potential difference

between two points is calculated, this potential difference ("voltage") can be related to the energy or

work W needed to transport a charge q against the field from one point to the other (or the energy

released in the opposite case) as before :

VA - VB = ∆∆∆∆V = qW

Electric potential from some charge systems

• point charge : the potential positive near a positive charge (which would repel a small

positive test charge - unlike gravity which is always attractive!) and negative near a negative

charge. The value follows a hyperbolic curve, approching positive or negative infinity near

the charge, and zero infinitely far from it.

• outside a hollow conducting sphere the potential follows a curve similar to that from a

point charge at the center of the sphere; inside the sphere the value of the potential is

constant at the value at its surface, since the field E inside it is zero, no resultant force would

act on a test charge and no work would be needed or released inside the sphere.

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Equipotential lines or surfaces

A graphic way to illustrate electric potential are equipotential lines (or in a 3-dimensionsal

situation surfaces) for which we have :



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• they describe points where the potential has the same value

• they are always perpendicular to electric field lines

• the same work is needed/released when a charge is moved between two equipotential

lines or surfaces

• no work is needed/released when a charge is moved along one

• they can be compared to altitude curves on a map for gravity (strictly, gravitational

potential = altitude multiplied by the gravity constant)

Certain situations are commonlt studied:

• isolated point charge: the equipotential lines are concentric circles or in 3 dimensions

spherical surfaces

• charged conducting sphere: outside the sphere, the equipotential lines/surfaces are the same

as for the point charge

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• two point charges: near each of them they are approximately circles/spheres, between them

is a straight line (in 3 dimensions, a planar surface). Note that they are always perpendicular to

the field lines.

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• parallel oppositely charged plates: they are straigth lines parallel to the plates, or in 3-d

parallel planar surfaces

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5.4. Electric circuits: current, resistance, power

Electric current

So far we have mainly studied electrostatics, the physics of electric charges at rest. Since the

charges can be affected by forces, they may also move. We can then define electric current I as :

I = ∆∆∆∆ q / ∆∆∆∆t [DB p.7]

or simpler I = q / t = the amount of electric charge transported per time.

Unit: 1 coulomb/second = 1 Cs-1

= 1 ampere = 1 amp = 1 A. Since currents are easier to measure

than charges, it is the ampere which is used as a fundamental unit in the SI-system, and 1 C = 1As

Electric circuit, conventional current and electron flow

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An electric circuit consists of

• a source of "voltage" = potential difference, for example a battery



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• a resistor (or more complicated arrangements of components) where the energy/charge

supplied by the battery is used

• connecting wires between the positive terminal of the battery and the resistor or other

apparatus, and that and the negative terminal (for alternating currents the positive and negative

terminal may switch many times per second). Two wires (or something else doing their job) are

always needed to complete the circuit (unless the current is flowing to or from an enormous

body like the earth)

The "conventional" current is from the positive to the negative terminal (the way a positive

test charge would go), while the actual electron flow is in the opposite direction.

Electric resistance

For any circuit or component where the current I is caused by the potential difference V we define

the electric resistance R as :

R = V / I [DB p.7]

Unit: 1 VA-1

= 1 ohm = 1 Ω

• The resistance describes how "hard" it is to move charges through the resistor - the higher

R, the more "voltage" is needed to keep up a certain current. Good conductors have a low R,

good isolators a very high R

[We could have defined the inverse quantity to describe how well a component conducts electricity:

the electric conductivity κ (kappa) or G = I / V with the unit 1 AV-1

= 1 Ω-1 = 1 siemens = 1 S,

sometimes called 1 mho ("ohm" backwards!). In chemistry, the conductivity is related to the

amount of ions in a solution; a solution of an ionic-bonded or polar compound has a high κ, while a

solution of very clean water or a covalent, non-polar compound has a low κ.]

Ohm's law with V- and A-meters

The resistance R can be defined or measured for any component; but for metallic conductors at a

constant temperature R is constant. This is Ohm's law.

• if the conductor is ohmic, a graph of I as a function of V will give a straight line with the

gradient 1/R [ = κ ], while a graph of V as a function of I will give one with the gradient R.

• if the conductor is non-ohmic, the graphs will be other curves

To experimentally produce this curve we need a circuit with

• an amperemeter = A-meter connected so the current flows through it ("in series" with the

resistor). A good A-meter has a very low resistance which can be neglected.

• a voltmeter = V - meter connected "beside" the electron flow ("in parallel"). For a good

V-meter, very little of the current flows through since it has a high resistance. (The A- and

V-meters are based on magnetic phenomena studied later).

• a resistor



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• a "voltage" source. Either the resistor or the the voltage source is variable.

• connecting wires, with a negligibly small resistance

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The filament lamp

This device which is an ordinary "light bulb", has a spiral metal wire which is heated by the flowing

electrons colliding with the metal electrons until it glows brightly. The metal is tungsten with a high

melting point. Since the temperature changes radically when a filament lamp is turned on, the R is

not constant but increases with temperature. The result is that the slope of an I-V-curve decreases

with higher V. There are other light sources and components with different characteristics.

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Circuit diagram symbols: see data booklet

Electric power

We recall that power P = energy or work per time. Now:

potential difference V = W/q and current I = q / t so we obtain VI = (W/q)(q/t) = W/t = P :

P = VI = I2R = V

2/R [DB p.7]

where the unit 1 watt = 1 W = 1 VA. Notice that since P = W / t we get W = Pt so 1 Ws = 1 VAs =

1 J is a unit of work or energy.

More common is the unit 1 kWh ("kilowatthour") = 1000 W x 3600s = 3.6 MWs = 3.6 MJ. The

other relations are found by combining formulas:



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• R = V/I so V = RI giving P = VI = RI2

• R = V/I so I = V/R giving P = VI = V2/R

The power is said to be "dissipated" meaning that this amount of energy per time is lost to heat.

Equivalent (effective) resistance for series and parallel circuits

[Kirchoff's laws, not required in the IB :

KI. The sum of currents flowing into a point = the sum of currents flowing out of it.

KII. The sum of all potential differences around any closed loop in an electric circuit is zero.

"Voltage" sources usually counted positive and the "voltage drop" RI in resistors negative

]

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In a series circuit we have that

• the current only has one possible way, and is the same through both resistors R1 and R2,

that is I1 = I2 = I (KI applied to arbitrary points in the circuit)

• KII gives V - R1I -R2I = 0 or

• V = R1I + R2I which using R1 = V1/I => R1I = V1 and R2 = V2/I => R2I = V2 gives:

• the sum of the voltage drops is the total voltage drop in the resistors or V = V1 + V2

• if we would like to replace the resistors R1 and R2 with only one with the same resistance

R as they both have together, we get when dividing both sides in V = V1 + V2 with I:

• R = V/I = V1/I + V2/I = R1 + R2 so:

R = R1 + R2 [DB p.7]

For 3 or more resistors we have R = R1 + R2 + R3 + ...



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In a parallel circuit the current can take different ways and splits up as I = I1 + I2 according to KI.

We can apply KII in 3 different closed loops:

Loop 1 : through the battery V and R1 but not R2:

• V - R1I1 = 0 so V = R1I1 = V1, the voltage over R1


• V - R2I2 = 0 so V = R2I2 = V2, voltage over R2

Loop 3 : through R1 and R2 e.g. clockwise but not through the battery V:

• now the hypothetical loop passes through R1 against the direction of the current; we are

actually following the "current" -I1 through it

• so R1(-I1) + R2I2 = 0 giving R1I1 = R2I2 or V1 = V2

All these lines of reasoning can then be followed by this:

• the potential drop is the same no matter which resistor we follow the current through: V =

V1 = V2

• for the equivalent resistance R we then get R = V/I and then I = V/R so

• V/R = I = I1 + I2 and

• V/R = V1/R1 + V2/R2 but since V1 = V2 =V we get

• V/R = V/R1 + V/R2 and dividing both sides with V

• 1/R = 1/R1 + 1/R2 , (with more similar terms for 3 or more resistors in parallel)

1/R = 1/R1 + 1/R2 [DB p.7]

Note : If we have both serial and parallel connections combined, we can stepwise replace them with

effective resistances until the effective resistance of them all is arrived at.

5.5. Electromotive force (emf) and internal resistance

The source of electric potential difference is some device which gives a certain amount of

energy/charge to the moving charges. This energy may come from chemical reactions in a battery.

Since the electrons are moving around the circuit out from one terminal and eventually back into the



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other, they must also in some way move through the battery (possibly attached to ions moving in

solutions in the battery or otherwise). Even if the resistance in the circuit outside the battery - the

"outer circuit", with its "outer" or external resistance R (which may consist of a complicated set of

serial and/or parallel connections which give this total effective resistance) there will always be

some internal resistance r in the battery. This causes a potential drop (= loss of some energy per

charge):

electromotive force = emf = ε = rI + RI

where:

• emf is not at all a force, but a "voltage" = energy/charge; the one supplied by the original

source of energy (for example chemical reactions). In Finnish "lähdejännite", in Swedish

"källspänning".

• rI = the potential drop in the battery. Note that it depends on I - the higher current is

drawn from the battery, the more loss inside it. (This is why in older cars the headlights

would be dimmed when a lot of current was drawn to the starter engine).

• RI = the potential difference = "voltage" available in the external circuit connected to the

terminals of the battery. Also called terminal voltage. (In Finland symbolised U =

napajännite = polspänning). This is what earlier was symbolised V in R = V/I, the voltage

over the external circuit, so we can write:

emf = rI + V

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5.6.* Capacitors

Capacitors (Sw. kondensator, Fi. kondensaattori) are devices where electric charge can be stored on

plates or sheets of metal (often wrapped to a small cylinder) with a thin layer or air or insulating

material in between. The amount of stored charge is much smaller than in a battery, but the

advantage is that it can be taken out of there very quickly, in a small fraction of a second, which

makes them useful in alternating current (AC) circuits where the direction of the current may

change many times per second.

For the capacitor, the capacitance (kapacitans, kapasitanssi) is



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C = Q / U

in the unit 1 farad = 1 F = 1CV-1

, if we use the symbol U instead of V for "voltage" = potential

difference and as usual Q for the amount of electric charge stored in the capacitor (the charge is of

opposite signs on opposing plates, but there is equally much positive and negative).

If the area of the plate (only one is counted, not both the positively and the negatively charged one)

is A and the distance between the foils d then

C = εεεεrεεεε0A/d

where ε0 = the electric permittivity in vacuum (same as in F = 1/4πε0)(q1q2/r2). εr = the relative

permittivity = a number without unit which gives the correction for the vacuum value. Found in

MAOL's tables for various substances.

The energy stored in a capacitor (in the unit J) is

E = ½QU

(which using C =Q/U can be written E = Q2/2C or E = ½CU

2 ) where the voltage U is in volts and

the charge Q in coulombs.

Capacitors in series and parallel

Capacitors can be connected in series or in parallel like resistors. Then we have:

series => same charge, parallel = > same voltage

Compare to resistors:

series => same current, parallel => same voltage

For the total capacitance we have the formulas

series 1/Ctot = 1/C1 + 1/C2 + .... parallel Ctot = C1 + C2 + ...

(Note that is opposite to the formulas for resistors! There it was

series Rtot = R1 + R2 + ..., parallel 1/Rtot = 1/R1 + 1/R2 + ...

5.7. Magnets and magnetic fields

Magnetic poles

Magnetism is the long-known phenomenon that pieces of certain materials (like an iron-rich ore,

magnetite) turn towards north? The ends of magnetic materials can be called North and South poles,

and like electric charges,



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like magnetic poles repel, opposite poles attract

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Magnetic field lines

Electric fields do not materially exist, but describe what would happen to (in what direction a force

would act on) a small positive test charge placed in a certain point.

Magnetic field lines similarly do not exist but

the magnetic field B describes in what direction the north end of a small test compass would

point if placed in a certain point

• the unit of the magnetic field B (a vector quantity) is 1 tesla = 1 T, which will be

explained later, as will why the quantity also can be called flux density.

A problem here is that we do not have isolated N or S poles - a magnetic piece of material always

has both poles, and if it is sawed in two pieces these will also have both N and S poles.

We can also note that the test compass is not accelerated by a force along a field line, only turned

into the correct direction (as if by a torque rather than a force).

Magnetic field lines from permanent magnets

• a bar magnet : the field lines go out from the N pole and into the S end and are otherwise

shaped like the electric field lines around a + and - charge.

• Earth : the field lines are shaped as if a bar magnet was placed inside the Earth with a

"magnetic north pole" near the geographic N pole, although physically this is a S pole, since

it attracts the N poles of compass needles.

Magnetic fields caused by currents (all!)

In the 1800s it was discovered that compass needles also react to wires carrying electric current.

Later it has been revealed that all magnetic fields are caused by currents.



21

• bar magnets : in the atom, electrons orbit the nucleus and this is like a current around it.

In most materials the magnetic fields from the atoms are cancelled out since they are in

random directions; some materials can be magnetised = have the small fields in more or less

the same direction.

• Earth : flows of molten rock under high pressure in a plasma state act as convection

currents inside the earth, powered by the heat from nuclear reactions inside Earth. These

cause the "permanent" magnetic field, which every few million years or so changes

direction, possible as a result of chaotic processes.

e06b

More common examples of current-caused magnetic fields are:

• straight wires : when looking in the conventional direction of the current, the magnetic

field lines are in concentric circles clockwise around the wire. We may also use

For a long thin straight wire carrying the current I the magnetic field B at a distance r from the wire

is

B = µµµµ0I / 2ππππr [DB p. 7]

where the direction of the field is given by the first right hand rule and µ0 = the magnetic

permeability in vacuum = 4π x 10-7

TmA-1

. In air the same value can approximately be used, for

other materials the value is different.

Right hand rule 1 : grip the wire with your right hand, the thumb in the direction of the current: the

bent fingers will indicate the magnetic field

In this context it may be convenient to introduce a way to show in a graph

the direction of any vector quantity (current, field, ...) as a CROSS if perpendicularly into the page,

a DOT if perpendicularly out of the page.



22

e06c

• flat circular coil : if the straight wire is bent to a circle, the field will be in one direction

inside it and in the opposite outside it.

• solenoid = long coil with several loops bunched together so that the fields inside the

solenoid point in one direction, and outside it are like those around a bar magnet

For a solenoid of length l with the number of turns of wire N the magnetic field inside the solenoid

is given by

B = µµµµ0NI / l = µµµµ0nI [DB p.7]

where it can be noted that the B-value can be increased by inserting some materials like an iron core

into the solenoid, replacing µ0 with the larger µiron.

e06d

Right hand rule nr 2 : grip the solenoid with your right hand, use the four bent fingers to follow the

current in the solenoid: the thumb will indicate the North pole of this electromagnet.

5.8. Magnetic forces

Magnetic force on moving charges

If a straight wire with the length l carrying the current I is placed in a homogenous magnetic field B,

the force F acting on it will be

F = I l B sin θθθθ [DB p. 7]



23

where θ = the angle between the direction of the current (opposite to the direction where negative

electrons move!) and the direction of the magnetic field. If the angle = 90 degrees, then F = I l B.

e07a

The direction of the magnetic force is given by

Right hand rule nr 3 : thumb in the direction of I, four fingers in the direction of B gives the

direction of the force F in a third dimension out of the palm of the hand = in the direction where

the four fingers can easily be bent.

• This makes it possible to express the unit of the magnetic field B in other units. Solving

for B in a case with the angle = 90 degrees gives B = F/Il and therefore 1 tesla = 1 T = 1

N/Am = 1 NA-1

m-1

.

• Compare this to the electric field E = F / q giving the unit (which has no separate name) 1

N/C = 1 NC-1

.

• In both cases, the unit of the field is the unit of force divided by the unit of what will be

affected by a force if placed in the field - in the electric case a charge, in the magnetic case

moving charges as a piece of current-carrying wire.

In the formula for the magnetic force the factors Il can be replaced as:

• I l = (q / t) l but since the distance l that a charge moves through the wire in the

chosen time t = its velocity v we get

• I l = q v so we can write

F = q v B sin θθθθ [DB p. 7]

or F = q v B for the 90 degree case. This would describe the unit

1 T = 1 N/(Cms-1

)-1

indicating that magnetic forces act on moving charges

Magnetic force on two parallel wires

If two parallel wires, assumedly very long and and thin, are near each other then they will act on

each other with a forces that following Newtons's III law are of the same but opposite directions.



24

e07b

• We study a length l of the parallel wires with currents in the same direction

• The wire carrying I1 causes the field B1 at the location of I2 a distance r away.

• This field B1 = µµµµ0I1 / 2ππππr is directed "downwards" (right hand rule 1)

• The force acting on the wire carrying I2 will then be F = I2lB1

• It is directed towards the wire carrying I1 (right hand rule 3)

• combining gives F = I2lB1µ0I1 / 2πr or :

F / l = µµµµ0I1I2 / 2ππππr [DB p.7]

and in the corresponding way, the same force but in the opposite direction will be acting on the I1-

wire (as Newton's III law tells us).

Note : We may use the rule that currents in the same direction attract, in opposite directions

repel - if we are not confused by this being contrary to the usual "opposite attract, same

repel" rule which is valid for positive (+) and negative (-) charges as well as for North and

South magnetic poles.

Defining the unit 1 amp

The unit for current, 1 ampere = 1 amp is defined from a situation like this: two infinitely long and

thin parallel wires 1 m apart in vacuum carrying the current I which by definition is 1 amp if the

force between them is 2 x 10-7

N per meter of the wire pair.

The simple DC-motor (and A- and V-meters)



25

e07c

For a rectangular loop of wire (or several loops) in a magnetic field carrying a current a force will

be acting on the opposite sides, in opposite directions but producing a torque in the same

clockwise/anticlockwise direction. The magnetic field can be made homogenous almost for all

angles by having magnet poles shaped like "half-pipes" towards each other. The direction of the

current needs to change every half turn of the loop, which is achieved by the commutator and brush

contact arrangement in the picture. (An alternating current, one which periodically changes

directions could have been used directly - see the chapter about alternating currents).

If instead of this a needle is attached to the loop(s) and a spiral spring which counteracts the torque

by the magnetic force with one directly proportional to the angle turned is attached to this, then the

loop with the needle will not rotate but turn an angle proportional to the current through it. This can

be used as an ammeter (amperemeter). The ammeter is connected in series with the studied

component so that the same current passes them; ammeters have a low resistance so as to decrease

this current as little as possible.

A simple ammeter without a spiral spring opposing the magnetic torque is very sensitive and this

instrument, called a galvanometer, can be used to detect the presence of very small currents.

If an ammeter is connected in series to a large known resistor then the needle reading will be

directly proportional to the voltage over this instrument, a voltmeter.

Rvoltmeter = Vvoltmeter/Ivoltmeter => Vvoltmeter = RvoltmeterIvoltmeter

The scale of the instrument can be marked to directly show the voltage value, and different resistors

give different maximum readings. The voltmeter is connected in parallel with the studied

component so that the voltage over them is the same. Since the resistance of the voltmeter is high



26

only a small part of the current through the studied component will divert through the voltmeter.

Since

Runknown = Vunknown/Iunknown => Vunknown = RunknownIunknown

any change in Iunknown will affect the value we wish to measure.

The magnetic force as centripetal force

If a moving charge enters a homogenous magnetic field at a 90o angle then it will be affected by a

magnetic force perpendicular to the velocity and this force can act as a centripetal force:

e07d

We will then have

• qvB = mv2/r and cancelling a factor v

• qB = mv/r which can be used to solve for r:

• r = mv/qB = p/qB with p = mv = momentum

• or p = qBr

• or if the factor v was not cancelled qvB = mv2/r giving

• mv2 = qvBr and then ½mv

2 = Ek = ½qvBr



27

5.9. Induction

Previously we have studied

• magnetic fields causing forces on moving charges (=> motor)

Now we will focus on the opposite:

• moving magnets causing moving charges = currents (=> generator)

Induced emf in straight wire

If we move a straight piece of wire quickly between the poles of a U-shaped magnet a small current

will be shown on a digital microammeter (or a galvanometer = sensitive ammeter).

e08a

To analyse this we focus on a straight wire of length l moving with the velocity v in a homogenous

magnetic field B directed into the plane of the page.

e08b [switch signs, minus to left and plus to right!]

In a metal wire electrons with the charge q = e (or -e) can move. We will get:



28

• the force acting on an electron is Fmagnetic = qvB

• this force will make electrons drift towards one end of the wire

• they will then cause an electric field parallel to the wire

• this field E = Felectric/q so the electrons are affected by Felectric in the opposite direction to

Fmagnetic

• the more electrons gather in one end of the wire, the stronger E

• eventually there will be an equilibrium where Fmagnetic = Felectric so

• qvB = qE giving E = Bv

• the potential difference or emf (or "voltage") V = ε between the ends of the wire will then

follow the formula E = V / d where now V = ε and d = l so

• E = ε/l which gives εεεε/l = Bv and therefore

εεεε = Blv [DB p.7]

If the wire is moving in a homogenous field it will in the time t sweep a rectangular area A with

• one side = the length of the wire = l

• another side = the distance traveled by the wire = vt

• from this we get A = lvt giving v = A / lt

• then ε = Blv can be written ε = BlA / lt = BA / t

The quantity BA is called magnetic flux ΦΦΦΦ with the unit 1 weber = 1 Wb = 1 Tm2

• it follows from this that B = Φ / A wherefore the magnetic field intensity B also can be

called the magnetic flux density ; normally "density" would mean mass/volume but

"density" can be used in a more general sense as something per length, area or volume.

If the B is not perpendicular to the area A, then we can use

ΦΦΦΦ = BA cos θθθθ [DB p.7]

where the Θ = the angle between B and the normal to A (not to the "surface" A).

• If B is perpendicular to A then we have Θ = 0o giving cos Θ = 1 and Φ = BA

• if B is parallel to the surface A then Θ = 90o and cos Θ = 0 so Φ = 0

Graphically, the flux ΦΦΦΦ is represented by the number of field lines (crosses or dots if into or

out of page) and B by the number of crosses or dots per given area

One way to use this emf to cause a current in an electric circuit would be to let the moving wire be

in (assumedly frictionless) contact with rails which are connected by a stationary wire parallel to the

moving one.



29

e08c

The ε = BA / t would then become ε = Φ / t or the induced emf would be the change in flux per

time. (can be written εεεε = ∆∆∆∆ΦΦΦΦ / ∆∆∆∆t)

It is more practical to have a closed circuit - which may be rectangular, circular or other and change

the flux in it to cause an induced emf.

Changing the flux in a circuit: Faraday's and Lenz' laws

The change in flux can be achieved by changin any of the variables in ΦΦΦΦ = BA cos θθθθ. The effects

are easier to detect if a solenoid with N turns of wire is used instead of a single loop, and stronger

the faster the change is done.

1. Changing B : the magnetic field near the pole of a bar magnet is stronger the closer to the

magnet we are. By moving the magnet and the circuit relative to each other (moving either the

magnet or the loop/ solenoid or both) we can affect the B-value.

2. Changing A : for a single loop of wire placed between the parts of a strong U-shaped magnet a

small current pulse may be detected on a digital microammeter if the loop is very quickly made

smaller or larger.

3. Changing ΘΘΘΘ : this can be achieved by rotating the loop or solenoid in a magnetic field and is

most common in technical applications (generators).

For this (sometimes called magnetic flux-linkage) we have Faraday's law

εεεε = - N ∆∆∆∆ΦΦΦΦ / ∆∆∆∆t [DB p. 7]

The induced emf will cause an current in the wire which will cause a magnetic field around the

wire, and affect the flux through the area A. We have two possibilities:

• either the induced current causes an additional change in flux which causes more emf

to cause more current .... and so on making it possible to get an infinitely high current

(or lots of free energy) by starting the process with a small input work. This is not

happening in our universe, and would be against the law of conservation of energy



30

• or - which is the case - the induced current causes a change of flux opposed to the

initial change which caused it. Therefore the minus in the formula = LENZ' LAW

5.10. Alternating currents

AC generators

A rotating loop of wire (rectangular or other) in a homogenous magnetic field works as an AC

generator.

e09a

[Not required in the IB: Since the induced emf depends on the change of flux Φ and this flux in turn

is Φ = BAcosΘ rotating the loop or coil in a magnetic field will produce an emf which follows a

sine function, since the rate of change in (the derivative of ) the cosine function is a negative sine

function, where the angle rotated Θ = 2πft. If f = the frequency of rotation then:

• Φ(t) = BAcos(2πft) taking the time derivative to get

• Φ'(t) = -BAsin(2πft)2πf

where

• the quantity -2πfBA has the unit HzTm2

• using ε = Blv => B = ε/lv we get the unit 1 T = 1 V/(m2s

-1) so

• the unit HzTm2 = s

-1V/(m

2s

-1) = V = 1 volt

This is the voltage V0 referred to below. For a constant resistance R this leads to a similar function

for the current, with the constant V0/R = I0 in front of the sine function.

Peak and rms values

The alternating voltage and current as a function of time both follow a sine function:

I(t) = I0sin(2ππππft) and V(t) = V0sin(2ππππft)

where the I0 and V0 are the "peak" values of the quantities and f = the frequency of the alternating

current; in Europe 50 Hz, in the US 60 Hz.. The effective value of them - sometimes called the rms

value referring to a statistical "root mean square" concept, but we can think of it as an average - can

be found using:



31

• the power P =VI as a function of time P(t) = V(t)I(t) giving

• P(t) = V0sin(2πft) I0sin(2πft) = V0I0sin2(2πft)

• if we plot the function y = sin2x we find geometrically that an average for y = one which

gives the same area under the curve as under a horizontal line at the average is ½y (cut off

peaks and use them to fill the 'troughs' - all on the positive side above the x-axis)

• analogous to that the average or effective power Prms = ½V0I0

• we can use a formula Prms = VrmsIrms if we define :

e09b

Irms = I0 /√√√√2 Vrms = V0/√√√√2 [DB p.7]

resulting in Prms = VrmsIrms = (I0 /√2)(V0/√2) = ½V0I0 = Prms

Note : It is the rms value of the voltage, not the peak value, which is indicated for the ordinary

household electricity, e.g. 230 V in Europe or 110 V in the US.

5.11. Transformers

We have earlier noticed that it is possible to induce a current in a loop of wire or a solenoid = bunch

of loops by one of these:

• changing the magnetic field B

• changing the area A

• changing the angle Φ

If the field B is caused by a permanent bar magnet, changing it means bringing it closer to or further

away from the loop or solenoid. But if the field is caused by another solenoid nearby, this field can

be varied by varying the current in the other solenoid - which is exactly what happens in a solenoid

connected to an AC source.

Without supplying a strict proof we, can notice that the number of loops N affects the induced emf

= voltage. By varying the number of loops in the primary coil = Np (to which the AC source is

connected) and that in the secondary coil Ns we can from a given input voltage in Vp (which

usually would be an rms value, not a peak value) get different output voltages Vs in the secondary

coil :



32

Vp / Vs = Np / Ns [DB p.7]

or : where there are many turns of wire, there is a high voltage

An ideal transformer would convey all the input power Pp to output power Ps in which case

VpIp = VsIs or

Vp/Vs = Is/Ip [not in DB]

that is, when the voltage is increased, the current is decreased and vice versa. High currents can be

achieved by connect an AC source to a primary coil with high Np in a transformer with a very low

Ns, giving a high current which dissipates a lot of power P = RI2, sometimes demonstrated by

melting a nail.

In practice the efficiency of any transformer is less than 100% but can be rather high if it is

equipped with an iron core:

e10a

• a step-up transformer is one where Vs > Vp

• a step-down transformer is one where Vs > Vp

Electricity from power plants is transformed up to high voltages where the current is lower and the

power loss P = RI2 minimized; then transformed down to the voltage delivered to the consumer

(which may be further transformed down to e.g. 12 V and possibly converted to DC for some

devices).

[Note : A simple transformer can be converted to a primitive metal detector by removing the iron

core and turning one coil 90o. It will then not work as a transformer since the field lines from the

primary are mainly parallel to the loop area in the secondary, so no emf is induced there. But is a

(not too small) metal object is place nearby, eddy currents will be induced in it, and these will

induce some small emf in the secondary coil.]



33

5.12.* AC circuits

Self-inductance

Change in flux through a loop or a solenoid gives induced emf. If we connect or disconnect a

solenoid to a voltage source, the current I through it will increase or decrease quickly, causing a

"self-inductance" which opposes the change. For this:

εself = emfself = - L∆∆∆∆I/∆∆∆∆t

where L = inductance (induktans, induktanssi), unit 1 henry = 1 H = 1 VsA-1

= 1 Ωs. For a solenoid

with the cross-section area A, the length l and the number of turns we have when µ0 = the magnetic

permeability that

L = µµµµ0N2A/l

The energy stored in the magnetic field around a coil or solenoid when the current I runs through it

is:

E = ½LI2

AC circuits and impedance = "AC resistance"

When a circuit is connected to an AC power source the total resistance is affected not only by

resistors but also by capacitors and solenoids. The total

resistance for AC is called

• impedance Z = U/I (compare R = U/I), unit 1 ohm

Capacitors cause

• capacitive reactance XC = 1/ωωωωC where ωωωω = 2ππππf and f = the frequency in Hz.

This "reactance" has the same unit as resistance and impedance and only means the resistance

caused by the capacitors in the circuit. (Note: Direct current can not go through a capacitor since

the metal foils in it are separated by a layer of insulating material. High frequency AC can, since the

charges that cannot go through the capacitor are stored on one of the plates or foils, but soon the

current changes direction and they move back through the circuit and are stored on the opposite foil

etc.)

• Solenoids cause inductive reactance XL = ωωωωL

The total impedance is (can be shown with complex number mathematics)

Z = √√√√(R2 + (XL - XC)

2)

If there is no capacitor in the circuit Z = √(R2 + XL

2); inserting C = 0 would in principle give

infinitely high XC and therefore also Z but that would represent a circuit which is broken (there is a

place with insulating material that the current would have to go through) but where this break in the

circuit is an infinitely bad capacitor, with so small area or high d that C = εrε0A/d ≈ zero). For an

unbroken circuit with no capacitor in it, the term for capacitive reactance is just left out.



34

If there is a resistor and a capacitor then L = 0 gives XL = 0 and Z = √(R2 + (-XC)

2) = √(R

2 + XC

2)

and if there is only a resistor then Z = √(R2) = R. (But L is never quite = 0 since any closed circuit

contains at least one "loop" of wire).

Phase shift

When there is L and C in the circuit the current I and voltage U as a function of time will not be in

phase; the phase shift ϕϕϕϕ in radians (2π means one whole "wavelength" of the sine curve ) is given

by

tan ϕϕϕϕ = (XL - XC) / R

The AC circuit is in resonance when (XL - XC) = 0 so that Z = R which happens when XL = XC or

ωL = 1/ωC which gives ω = 1/√LC or

f = 1/2ππππ√√√√(LC)

This can be used radio tuning where the C-value of a capacitor is changed either by turning plates

so that the area opposing a plate with opposite charge is changed, or by changing the distance

between plates); thereby the resonance frequency is changed and difference stations can be tuned in.



1




electrons around it


materia, like mass




attract


C (we












e01a






hose).






2

e01b





the whole object.

e01c





3

e01d





touching it.

e01e



4

The electroscope




insulated).



e01f

e01g



e01h





5






e02a


2C





Fm-1








cancel out


in air or in oil).



Note:



Electric field






6


E = F / q [DB p.7]



E = F/q1 = (kq1q2/r2) / q1 = kq2/r


E = kq / r2 [DB p.7]








!!

e02b







away, the weaker)







7

e02c











e02d



8














the case).



"pointy" places.





slightly bent.








e03a



9

























E = - ∆∆∆∆V / ∆∆∆∆x [DB p.7]


E = V / d [DB p.7]






10

GRAVIT. ELECTR.



2 N


2 NC-1=Vm-1





Nm2kg


9

Nm2C

-2 .





V = Ep / q








in addition to




= 1 JC-1

m-1

= 1 NmC-1

m-1

= 1 NC-1




= 1.6 x 10-19

C x 1 JC-1

= 1.6 x 10-19

J = 1 eV.






11




qV = ½mv2











VA - VB = ∆∆∆∆V = qW










e03c






12




lines or surfaces






spherical surfaces



e03d



the field lines.

e03e



13



e03f


Electric current



I = ∆∆∆∆ q / ∆∆∆∆t [DB p.7]






e04a





14










Electric resistance



R = V / I [DB p.7]

Unit: 1 VA-1

= 1 ohm = 1 Ω






= 1 Ω-1 = 1 siemens = 1 S,
















• a resistor



15



e04b

The filament lamp






e04c


Electric power



P = VI = I2R = V

2/R [DB p.7]







16









]

e04d









• R = V/I = V1/I + V2/I = R1 + R2 so:

R = R1 + R2 [DB p.7]




17

e04e













V1 = V2


• V/R = I = I1 + I2 and




1/R = 1/R1 + 1/R2 [DB p.7]









18






charge):


where:



"källspänning".








emf = rI + V

e05a

5.6.* Capacitors










19

C = Q / U












E = ½QU














Magnetic poles






20


e06a
























21




the same direction.





e06b





is




TmA-1










22

e06c






is given by




e06d










23



e07a







N/Am = 1 NA-1

m-1

.


N/C = 1 NC-1

.










1 T = 1 N/(Cms-1

)-1







24

e07b






















25

e07c
























26


Since






e07d

We will then have




• or p = qBr



2 = Ek = ½qvBr



27

5.9. Induction








e08a







28





Fmagnetic


























moving one.



29

e08c














smaller or larger.




εεεε = - N ∆∆∆∆ΦΦΦΦ / ∆∆∆∆t [DB p. 7]









30




AC generators


generator.

e09a







where



-1) so


-1V/(m

2s

-1) = V = 1 volt



Peak and rms values






be found using:



31








e09b

Irms = I0 /√√√√2 Vrms = V0/√√√√2 [DB p.7]




5.11. Transformers














coil :



32




VpIp = VsIs or





melting a nail.



e10a






devices).








33

5.12.* AC circuits

Self-inductance








permeability that

L = µµµµ0N2A/l


is:

E = ½LI2






Capacitors cause







etc.)



Z = √√√√(R2 + (XL - XC)

2)









34


2) = √(R

2 + XC

2)



Phase shift



by




f = 1/2ππππ√√√√(LC)






1




electrons around it


materia, like mass




attract


C (we












e01a






hose).






2

e01b





the whole object.

e01c





3

e01d





touching it.

e01e



4

The electroscope




insulated).



e01f

e01g



e01h





5






e02a


2C





Fm-1








cancel out


in air or in oil).



Note:



Electric field






6


E = F / q [DB p.7]



E = F/q1 = (kq1q2/r2) / q1 = kq2/r


E = kq / r2 [DB p.7]








!!

e02b







away, the weaker)







7

e02c











e02d



8














the case).



"pointy" places.





slightly bent.








e03a



9

























E = - ∆∆∆∆V / ∆∆∆∆x [DB p.7]


E = V / d [DB p.7]






10

GRAVIT. ELECTR.



2 N


2 NC-1=Vm-1





Nm2kg


9

Nm2C

-2 .





V = Ep / q








in addition to




= 1 JC-1

m-1

= 1 NmC-1

m-1

= 1 NC-1




= 1.6 x 10-19

C x 1 JC-1

= 1.6 x 10-19

J = 1 eV.






11




qV = ½mv2











VA - VB = ∆∆∆∆V = qW










e03c






12




lines or surfaces






spherical surfaces



e03d



the field lines.

e03e



13



e03f


Electric current



I = ∆∆∆∆ q / ∆∆∆∆t [DB p.7]






e04a





14










Electric resistance



R = V / I [DB p.7]

Unit: 1 VA-1

= 1 ohm = 1 Ω






= 1 Ω-1 = 1 siemens = 1 S,
















• a resistor



15



e04b

The filament lamp






e04c


Electric power



P = VI = I2R = V

2/R [DB p.7]







16









]

e04d









• R = V/I = V1/I + V2/I = R1 + R2 so:

R = R1 + R2 [DB p.7]




17

e04e













V1 = V2


• V/R = I = I1 + I2 and




1/R = 1/R1 + 1/R2 [DB p.7]









18






charge):


where:



"källspänning".








emf = rI + V

e05a

5.6.* Capacitors










19

C = Q / U












E = ½QU














Magnetic poles






20


e06a
























21




the same direction.





e06b





is




TmA-1










22

e06c






is given by




e06d










23



e07a







N/Am = 1 NA-1

m-1

.


N/C = 1 NC-1

.










1 T = 1 N/(Cms-1

)-1







24

e07b






















25

e07c
























26


Since






e07d

We will then have




• or p = qBr



2 = Ek = ½qvBr



27

5.9. Induction








e08a







28





Fmagnetic


























moving one.



29

e08c














smaller or larger.




εεεε = - N ∆∆∆∆ΦΦΦΦ / ∆∆∆∆t [DB p. 7]









30




AC generators


generator.

e09a







where



-1) so


-1V/(m

2s

-1) = V = 1 volt



Peak and rms values






be found using:



31








e09b

Irms = I0 /√√√√2 Vrms = V0/√√√√2 [DB p.7]




5.11. Transformers














coil :



32




VpIp = VsIs or





melting a nail.



e10a






devices).








33

5.12.* AC circuits

Self-inductance








permeability that

L = µµµµ0N2A/l


is:

E = ½LI2






Capacitors cause







etc.)



Z = √√√√(R2 + (XL - XC)

2)









34


2) = √(R

2 + XC

2)



Phase shift



by




f = 1/2ππππ√√√√(LC)






1




electrons around it


materia, like mass




attract


C (we












e01a






hose).






2

e01b





the whole object.

e01c





3

e01d





touching it.

e01e



4

The electroscope




insulated).



e01f

e01g



e01h





5






e02a


2C





Fm-1








cancel out


in air or in oil).



Note:



Electric field







6

E = F / q [DB p.7]



E = F/q1 = (kq1q2/r2) / q1 = kq2/r


E = kq / r2 [DB p.7]








!!

e02b







away, the weaker)







7

e02c











e02d



8














the case).



"pointy" places.





slightly bent.








e03a



9

























E = - ∆∆∆∆V / ∆∆∆∆x [DB p.7]


E = V / d [DB p.7]






10

GRAVIT. ELECTR.



2 N


2 NC-1=Vm-1





Nm2kg


9

Nm2C

-2 .





V = Ep / q








in addition to




= 1 JC-1

m-1

= 1 NmC-1

m-1

= 1 NC-1




= 1.6 x 10-19

C x 1 JC-1

= 1.6 x 10-19

J = 1 eV.






11




qV = ½mv2











VA - VB = ∆∆∆∆V = qW










e03c






12




lines or surfaces






spherical surfaces



e03d



the field lines.

e03e



13



e03f


Electric current



I = ∆∆∆∆ q / ∆∆∆∆t [DB p.7]






e04a





14










Electric resistance



R = V / I [DB p.7]

Unit: 1 VA-1

= 1 ohm = 1 Ω






= 1 Ω-1 = 1 siemens = 1 S,
















• a resistor



15



e04b

The filament lamp






e04c


Electric power



P = VI = I2R = V

2/R [DB p.7]







16









]

e04d









• R = V/I = V1/I + V2/I = R1 + R2 so:

R = R1 + R2 [DB p.7]




17

e04e













V1 = V2


• V/R = I = I1 + I2 and




1/R = 1/R1 + 1/R2 [DB p.7]









18






charge):


where:



"källspänning".








emf = rI + V

e05a

5.6.* Capacitors










19

C = Q / U












E = ½QU














Magnetic poles






20


e06a
























21




the same direction.





e06b





is




TmA-1










22

e06c






is given by




e06d










23



e07a







N/Am = 1 NA-1

m-1

.


N/C = 1 NC-1

.










1 T = 1 N/(Cms-1

)-1







24

e07b






















25

e07c
























26


Since






e07d

We will then have




• or p = qBr



2 = Ek = ½qvBr



27

5.9. Induction








e08a







28





Fmagnetic


























moving one.



29

e08c














smaller or larger.




εεεε = - N ∆∆∆∆ΦΦΦΦ / ∆∆∆∆t [DB p. 7]









30




AC generators


generator.

e09a







where



-1) so


-1V/(m

2s

-1) = V = 1 volt



Peak and rms values






be found using:



31








e09b

Irms = I0 /√√√√2 Vrms = V0/√√√√2 [DB p.7]




5.11. Transformers














coil :



32




VpIp = VsIs or





melting a nail.



e10a






devices).








33

5.12.* AC circuits

Self-inductance








permeability that

L = µµµµ0N2A/l


is:

E = ½LI2






Capacitors cause







etc.)



Z = √√√√(R2 + (XL - XC)

2)









34


2) = √(R

2 + XC

2)



Phase shift



by




f = 1/2ππππ√√√√(LC)




Documents

5. ELECTRICITY AND MAGNETISM - Jnoodle · The IB Physics Compendium 2005: ... 5. ELECTRICITY AND MAGNETISM 5.1. Electric charge • Atoms consist of ... E = F/q 1 which in the IB