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Topic 5: Electricity and magnetism
5.1 Charge and field
▪ charge comes in two forms + and –
▪ charge is a scalar quantity
▪ charge is quantized
▪ charge is conserved in a closed system
▪ charging neutral object
◊ Add electrons and make the object negatively charged.
◊ Remove electrons and make the object positively charged.
▪ charge is measured in Coulombs [C]
▪ Coulomb is defined as the charge transported by a current of one ampere in one second
▪ fundamental amount of charge is known as the electronic (or elementary) charge
▪ e =1.6 x 10-19 C
ratom ≈ 100000 x rnucleus
mnucleon ≈ 2000 x melectron
Coulomb’s law:
Electric force between TWO POINT charges q1 and q2 separated by distance r is:
Charges that are in a vacuum
𝑞1 𝑞2
𝐹=𝑘 2
𝑟
1 𝑞1 𝑞2
𝐹=
4𝜋𝜀0 𝑟 2
k = 8.99109 N m2 C−2
k is Coulomb’s constant.
0 = 8.8510-12 C2 N-1 m−2
0 is permittivity of free space
1/[40] = 1 / [48.8510-12] = 8.99109 = k.
Charges immersed in a different medium
1 𝑞1 𝑞2
𝐹=
4𝜋𝜀 𝑟 2
𝜀 is permittivity of the medium
Material
vacuum(0 )
air
paper
rubber
water
graphite
diamond
Permittivity 
/10-12C2 N-1m-2
8.85
~ 8.85
34
62
779
106
71
Coulomb’s law – extended distribution
 Coulomb’s law works not only for point charges, which have
no radii, but for any spherical distribution of charge at any radius.
 Be very clear that r is the distance between the centers of the charges.
Q
q
r
Electric field strength (at a point P) is the force per unit positive
point/test charge placed at that point. (it is a vector!)
𝐹
𝐸=
𝑞
𝐸 = 𝑁𝐶 −1
Direction of electric field is the direction of the force
on a positive test charge placed at that point.
𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑓𝑖𝑒𝑙𝑑 𝑟 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑝𝑜𝑖𝑛𝑡 𝑐ℎ𝑎𝑟𝑔𝑒 𝑄 𝑖𝑠:
𝑄
1 𝑄
𝐸=𝑘 2=
𝑟
4𝜋𝜀0 𝑟 2
𝑟𝑎𝑑𝑖𝑎𝑙𝑙𝑦 𝑜𝑢𝑡𝑤𝑎𝑟𝑑 𝑖𝑓 𝑄 𝑖𝑠 +
𝑟𝑎𝑑𝑖𝑎𝑙𝑙𝑦 𝑖𝑛𝑤𝑎𝑟𝑑 𝑖𝑓 𝑄 𝑖𝑠 −
Electric field due to charged conducting
sphere whether hollow or solid:
E=k
r
E=0
q
R
E=k
q
at the surface
R2
q
r2
Mapping fields – Electric field lines
Field lines are imaginary
◊ The lines starts on + charges and end on – charges
◊ An arrow is essential to show the direction in which a positive charge would move
◊ Where the field is strong the lines are close together.
◊ The lines never cross.
◊ The lines meet a conducting surface at 90°.
monopol
monopol
dipol
uniform
electric field
El. field at surface of a charged conductor is
perpendicular to the conductor’s surface.
Potential difference
 When electric charge is placed in electric field it will experience
electric force. Work W is done.
 Electric potential difference V (∆V) between two points A and B
is the amount of work W done per unit charge in moving a point
charge from A to B.
𝑊
𝑉=
𝑞
Q
q
A
𝑉 ≡ ∆𝑉 ≡ 𝑝. 𝑑. = 𝑣𝑜𝑙𝑡𝑎𝑔𝑒
 units of V are JC-1 which are volts V.
The potential difference between two points is one volt if one joule of energy is
transferred per coulomb of charge passing between the two points.
B
▪ 1 electron-Volt (eV) is the amount of work done when an elementary
charge e is moved through a potential difference of 1volt.
1 eV = W = q ∆V = (1.6x10-19 C) (1V)
1 eV = 1.6x10-19 J
energy!!!
▪ electronvolts are almost exclusively used in atomic and nuclear physics.
Capacitor: uniform electric field (the one that has constant magnitude and
direction) is generated between two oppositely charged parallel plates.
Edge effect is minimized when the length is long compared to their separation.
▪ ① Work done by electric force 𝑭
If a charge, q, is moved on its own from A to B, through a potential
difference, ∆V, the work done on it by electric force is equal to the
decrease in its electric potential energy which is converted into kinetic
energy:
W = Fd = q Ed = q ∆V = ½ mv2
▪ ② Work done by external force 𝑭𝒆𝒙𝒕
on charge against an electric field, from B to A is stored in the charge
as the change in electrical potential energy, U.
(𝐹𝑒𝑥𝑡 has to be equal to electric force in magnitude, opposite in direction)
W = Fext d = qEd
W = q ∆V
qEd = q ∆V ⇒ 𝑬 =
∆𝑽
⇒ (E) = NC-1 = Vm-1
𝒅
▪ Positive charge accelerates from higher to lower potential (from positive to negative).
▪ Negative charge accelerates from lower to higher potential. (from negative to positive)
Drift speed
▪
▪
▪
▪
▪
Imagine a cylindrical conductor that is carrying an electric current I.
The cross-sectional area of the conductor is A
It contains charge carriers each with charge q.
n is charge carriers density
We assume that each carrier has a speed v
v t
A
Q
v
Through any time interval t, only the charges Q between the two
black cross-sections will provide the current I.
The volume containing the charge Q is V = Avt.
Thus Q = nVq = nAvtq.
Finally, I = Q / t = nAvq.
I = nAvq
current vs. drift velocity
▪ Current is the rate at which charge flows past a given cross-section
𝑄
𝐼 =
𝑡
1A =
1C
1s
▪ Electrical resistance , R is the ratio of the potential difference across the resistor/conductor
to the current that flows through it.
𝑅=
𝑉
𝐼
1 (ohm) = 1V/1A
▪ The resistance of a conducting wire depends on four main factors:
In conclusion, we could say that a short fat
cold wire makes the best conductor.
• length, L
• cross-sectional area, A
• material/resistivity, ρ
• temperature, T:
• if temperature is kept constant: 𝑅 = 𝜌
𝐿
𝐴
If you double the length of a wire, you will
double the resistance of the wire.
If you double the cross sectional area of a
wire you will cut its resistance in half.
If you double the radius of a wire you will
cut its resistance in quarter.
Electrical Resistance
▪The different types of resistors have different schematic symbols.
fixed-value resistor
2 leads
potentiometer
3 leads
variable resistor
2 leads
light-dependent
resistor (LDR)
2 leads
thermister
2 leads
As temperature increases
resistance decreases
As brightness increases
resistance decreases
▪ OHM’S LAW: Current through resistor/conductor is proportional to potential difference on the
resistor if the temperature/resistance of a resistor is constant.
𝑉
or:
𝑉
𝐼
= 𝑐𝑜𝑛𝑠𝑡
𝐼 =𝑅
I – current through resistor,
V – potential difference across R
𝑖𝑓 𝑅 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡.
Ohmic and Non-Ohmic conductors
How does the current varies with potential difference for some typical devices?
diode
current
filament lamp
current
current
metal at const. temp.
potential
difference
𝐼
1
=
𝑖𝑠 𝑐𝑜𝑛𝑠𝑡. → 𝑅 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡.
𝑉
𝑅
potential
difference
potential
difference
devices are non-ohmic if
resistance changes
Devices for which current through them is directly proportional to the potential difference across device
are said to be ‘ohmic devices’ or ‘ohmic conductors’ or simply resistors. There are very few devices that
are trully ohmic. However, many useful devices obey the law at least over a reasonable range.
▪ When a current is flowing through a load such as a resistor, it dissipates energy in it. In collision with
lattice ions electrons’ kinetic energy is transferred to the ions, and as a result the amplitude of
vibrations of the ions increases and therefore the temperature of the device increases.
KE is transferred to thermal energy.
▪ Electric power is the rate at which energy is supplied to or used by a device.
▪ Electric power is the rate at which electric energy is converted into another form such as
mechanical energy, thermal energy, or light.
Power dissipated in a resistor/circuit:
W = qV → P = qV/t
P=IV
and I = q/t, so P = I V
Resistors in Series: connected in such a way that all components have
the same current through them.
Burning out of one of the lamp filaments in series or simply opening
the switch causes a break.
Equivalent resistance is greater that the greatest resistance in series.
𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + ⋯
Resistors in Parallel: are connected to the same two points of an
electric circuit, so all resistors in parallel have the same potential
difference across them.
The current flowing into the point of splitting is equal to the sum of
the currents flowing out at that point: I = I1 + I2 + …
A break in any one path does not interrupt the flow of charge in the
other paths. Each device operates independently of the other devices.
The greater resistance, the smaller current. Equivalent resistance is
smaller than the smallest resistance in series.
1
1
1
=
+
+⋯
𝑅𝑒𝑞 𝑅1 𝑅2
Internal resistance, r: some of the power/energy delivered by a cell is used/dissipated
in driving the current though the cell itself
Terminal voltage (the actual voltage delivered to the circuit): V = ε – Ir
𝜀
𝜀
𝑉
I=
=
=
𝑅𝑒𝑞 𝑅+𝑟
𝑅
To measure the current, we use an AMMETER.
• Ammeter is in SERIES with resistor R in order that whatever current passes through the resistor
also passes through the ammeter. Ram << R, so it doesn’t change the current being measured.
• In order to not alter the original properties of the circuit ammeters would ideally have no resistance,
to minimize the effect on the current that is being measured (Req = R+ Rammeter ≈ 𝑅). No energy would
be dissipated in ammeter.
To measure the potential difference across resistor, we use a VOLTMETER.
• Voltmeter is in PARALLEL with the resistor we are measuring. Rvoltmeter >> R so that it takes very little
current
from the device whose potential difference is being measured.
• In order to not alter the original properties of the circuit an ideal voltmeter would have infinite
resistance (1/ Rvoltmeter ) with no current passing through it and no energy would be dissipated in it.
Potential divider circuits
Consider a battery of  = 6 V.
Suppose we have a light bulb that can only use three volts.
How do we obtain 3 V from a 6 V battery?
A potential divider is a circuit
made of two (or more) series resistors
that allows us to tap off any voltage we
want that is less than the battery voltage.
The input voltage is the emf of the battery.
The output voltage is the voltage drop across R2.
 R = R1 + R2.
 I = VIN / R = VIN / (R1 + R2).
 VOUT = V2 = IR2
𝑉𝑜𝑢𝑡 =

𝑅2
𝑉
𝑅1 + 𝑅2 𝑖𝑛
𝑉1 =
𝑅1
𝑉
𝑅1 + 𝑅2 𝑖𝑛
R1
R2
potential
divider
Using a potential divider to give a variable pd
variable resistor circuit
a power supply, an ammeter, a variable resistor and a resistor.
When the variable resistor is set to its minimum value, 0 Ω, pd across the
resistor is 2 V and a current of 0.2 A in the circuit.
When the variable resistor is set to its maximum value, 10 Ω, pd across the
resistor is 1 V and a current of 0.1 A in the circuit.
Therefore the range of pd across the fixed resistor can only vary from 1 V to 2 V.
The limited range is a significant limitation in the use of the variable resistor.
potential divider
The same variable resistor can be used but the set up is different and involves the
use of the three terminals on the variable resistor (sometimes called a rheostat.)
Terminals of rheostat resistor are connected to the terminals of the cell.
The potential at any point along the resistance winding depends on the position of
the slider (or wiper) that can be swept across the windings from one end to the other.
Typical values for the potentials at various points on the windings are shown for the
three blue slider positions.
The component that is under test (again, a resistor in this case) is connected in a secondary
circuit between one terminal of the resistance winding and the slider on the rheostat. When the
slider is positioned at one end, the full 2 V from the cell is available to the resistor under test.
When at the other end, the pd between the ends of the resistor is 0 V (the two leads to the
resistor are effectively connected directly to each other at the variable resistor). You should
know how to set this arrangement up and also how to draw the circuit and explain its use.
• Potentiometer – rheostat – variable resistor
Kirchhoff’s first law
\ˈkir-ˌkȯf\
current into a junction has "+" sign
current out of a junction has “ − " sign
for any junction ∑I = 0
OR:
the sum of the currents into a junction equals the
sum of the currents away from a junction
𝐼1 + 𝐼2 + 𝐼3 = 𝐼4 + 𝐼5
It is equivalent to a statement of conservation of charge.
Kirchhoff’s second law
−
in a complete circuit loop, the sum of the emfs in the loop is
equal to the sum of the potential differences in the loop or
the sum of all variations of potential in a closed loop equals
zero
𝑓𝑜𝑟 𝑎𝑛𝑦 𝑐𝑙𝑜𝑠𝑒𝑑 𝑙𝑜𝑜𝑝 𝑖𝑛 𝑎 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 ∑𝜀 = ∑𝐼𝑅
Equivalent to conservation of energy.
If a circuit has ONLY one cell you are lucky.
You use Kirchhoff's laws in simple form (like juniors).
Next four slides are the simplest possible way to do it -
If a circuit has MORE than ONE cell you are less lucky.
But you are seniors now.
You have to use Kirchhoff's laws in purest form. Later.
Find power of the source, current in each resistor, terminal potential, potential drop across each resistor
and power dissipated in each resistor.
1. step: find total/equivalent resistance
Req = 120 Ω
2. step: find current in main circuit: 𝐼1 =
𝜀
𝑅
I1 = ε ∕ Req = 0.3 A
3. step: to find currents in parallel branches use the Kirchhoff's laws:
voltage is the same across all resistors in parallel:
𝐼2 𝑅2 = 𝐼3 𝑅3 , and 𝐼1 = 𝐼2 +𝐼3
one could use 𝜀 = 𝐼1 r + 𝑉𝐴𝐵
𝑉𝐴𝐵 = 𝐼2 𝑅2 = 𝐼3 𝑅3
100𝐼2 = 50𝐼3 → 𝐼3 = 2𝐼2
0.3 = 𝐼2 +𝐼3 → 0.3 = 3𝐼2 → 𝐼2 = 0.1 A
80 Ω
100 Ω
50 Ω
6.7 Ω
potential drops
V = IR
0.3x80 = 24 V
0.1x100 = 10 V
0.2x50 = 10 V
0.3x6.7 = 2 V
power dissipated
P = IV
0.3x24 = 7.2 W
0.1x10 = 1 W
0.2x10 = 2 W
0.3x2 = 0.6 W
𝐼3 = 0.2 A
ε = Σ all potential drops: (Kirchhoff's law)
36 V = 2 V + 24 V + 10 V
power dissipated in the circuit =
power of the source
0.6 + 2 + 1 + 7.2 = 0.3x36
Resistors in series
Three resistors of 330  each are connected to a 6.0 V battery in series
R1
R2
R3

What is the voltage and current on each resistor?
▪ R = R1 + R2 + R3
R = 330 + 330 + 330 = 990 
▪ I = V / R = 6 / 990 = 0.0061 A
▪ The current I = 0.0061 A is the same in each resistor.
▪ voltage/potential difference across each resistor:
V = I R1 = I R2 = I R3 = (0.0061)(330)
= 2.0 V
▪ In series the V’s are different if the R’s are different.
Resistors in series and parallel
Three resistors of 330  each are connected to a 6.0 V cell in parallel.
What is the voltage and current on each resistor?
▪ 1/R = 1/R1 + 1/R2 + 1/R3
R = 110 
1/R = 1/330 + 1/330 + 1/330 = 0.00909
▪ The voltage on each resistor is 6.0 V, since the resistors are in
parallel. (Each resistor is clearly directly connected to the battery).
▪ I1 = V1 / R1 = 0.018 A
▪ I2 = V2 / R2 = 6 / 330 = 0.018 A
▪ I3 = V3 / R3 = 6 / 330 = 0.018 A
▪ In parallel the I’s are different if the R’s are different.
Kirchhoff’s rules – solving the circuit (less lucky)
EXCELLENT EXPLANATION of KIRCHOFF’s LAWS – thank you Kiran
Kirchhoff’s rules – solving the circuit
EXAMPLE: Suppose each of the resistors is R = 2.0 , and the emfs are
1 = 12 V and 2 = 6.0 V. Find the voltages and the currents of the circuit.
▪ I1 – I2 + I3 = 0 (1)
+ –V2 + –V4 + 1 + – 2 = 0,
– – V – V = 0
2
3
4
▪ –V
▪
–V
1
1
–
2
+ –V2 + –V4 + 1 + – 2 = 0 & V=IR  –2I1 + –2I1 + –2I2 + 12 + –6 = 0 (2)
– V3 – V4 = 0 & V=IR 
–– 6
We now have three equations in I:
(1)  I3 = I2 – I1.
I1 = 1.8 A
(2)  3 = 2I1 + I2.
I2 = -0.6 A
I3 = -2.4 A
(3)  3 = -I2 + -I3
– 2I3 – 2I2 = 0 (3)
Finally, we can redraw our currents:
resistor voltages: V = IR
V1 = 1.8(2) = 3.6 V.
V2 = 1.8(2) = 3.6 V.
V3 = 2.4(2) = 4.8 V.
V4 = 0.6(2) = 1.2 V
PRACTICE: A filament lamp is rated at “4.0 V, 0.80 W”
on its package. The potentiometer has a resistance
from X to Z of 24  and has linear variation.
R1
R2
(e) The potentiometer is adjusted so that the meter shows 4.0 V.
What is the current in the Y-Z portion of the potentiometer?
SOLUTION:
▪V2 = 4.0 V because it is in parallel with the lamp.
I2
= V2 / R2
= 4 / 13.71 = 0.29 A
(f) The potentiometer is adjusted so that the meter shows 4.0 V. What is the current
in the ammeter?
SOLUTION: The battery supplies two currents.
▪The red current is 0.29 A because it is the I2 we just calculated in (e).
▪The green current is 0.20 A found in (c).
▪The ammeter has both so I = 0.29 + 0.20 = 0.49 A.
Primary and secondary cells
primary cells – non rechargeable cells
The cells are used until they are exhausted and then thrown away. The original chemicals
have completely reacted and been used up, and they cannot be recharged. Examples include AA
cells (properly called dry cells) and button mercury cells as used in clocks and other small low
current devices.
secondary cells – rechargeable cells
When the chemical reactions have finished, the cells can be connected to a charger. Then the
chemical reaction is reversed and the original chemicals form again. When as much of the reconversion as is possible has been achieved, the cell is again available as a chemical energy
store.
To reverse the chemical processes we need to return energy to the cell using electrons as the
agents, so that the chemical action can be reversed. When charging, the electrons need to travel in
the reverse direction to that of the discharge current and you can imagine that the charger has to
force the electrons the “wrong” way through the cell.
capacity
of a cell is the quantity used to measure the ability of a cell to release charge: if a cell can supply a
constant current of 2 A for 20 hours then it said to have a capacity of 40 amp-hours (40 A h).
The implication is that this cell could supply 1 A for 40 hours, or 0.1 A for 400 hours, or 10 A for 4 hours.
However, practical cells do not necessarily discharge in such a linear way and this cell may be
able to provide a small discharge current of a few milliamps for much
Internal resistance, emf and terminal voltage of a cell
The materials from which the cells are made have electrical resistance in just the same way as the metals
in the external circuit. This internal resistance has an important effect on the total resistance and current
in the circuit.
model for a real cell
Assumptions:
▪ internal resistance is constant
(for a practical cell it varies with the state of discharge)
▪ emf is constant
(which also varies with discharge current).
Kirchhoff’s second law
the emf of the cell supplying energy to the circuit = the sum of the pds
ε = IR + Ir
𝐼
If the pd across the external resistor is V, then ε = V + Ir or
V = ε – Ir
V, which is the pd across the external resistance, is equal to the terminal pd across real cell
(in other words between A and B).
The emf is the open circuit pd across the terminals of a power source –
in other words, the terminal pd when no current is supplied.
Thus the TERMINAL VOLTAGE (the actual voltage delivered to the circuit) is: V = ε - Ir
In the true sense, electromotive force (emf) is the work (energy) per unit charge made available
by an electrical source.
▪ Electromotive force (emf) is the work (energy) per unit charge made available by
an electrical source.
▪ Electromotive force, (ε) is the power supplied to the circuit per unit current
P = I V ⟹ V = P/ I
▪ Energy supplied by the source = ε Q
electric potential =
power
current
(coming from W = Q ∆V)
Total energy supplied by the source = energy used in the resistors:
ε Q = V 1 Q + V2 Q + …..
divide it by time ⟹ ε I = V1 I + V2 I + …..
ex: the cell supplies 8.0 kJ of energy when 4 kC of charge moves completely across the circuit with
constant current. Find ε:
ε = energy/charge = 2 V
▪ Power supplied by the source will be dissipated in the circuit:
Power of the source = sum of the powers across the resistors:
Pout = Σ Pi
⟹ ε I = V1 I + V2 I + …..
Emf and internal resistance of the battery (made up of two cells)
When voltage sources are connected in series in the same polarity,
their emfs and internal resistances are add.
Two voltage sources with identical emfs connected in parallel have a net emf equivalent to one
emf source, however, the net internal resistance is less, and therefore produces a higher current.
COULOMB’S/ NEWTON’S LAW
Electric (Coulomb’s law)/gravitational force (Newton’s law of
gravitation) between two POINT charges/masses is proportional
to the product of two charges/masses and
inversely proportional to the distance between them squared.
Both laws can be applied to the objects that are spherically
symmetrical (do not look like potatoes).
In that case we consider that the charge/mass is concentrated
at the center.
Electric /gravitational force is a vector. If two or more point
charges/masses act on some charge/mass, the net force on that
charge/mass is a vector sum of all individual forces acting on
that charge/mass.
The electric/gravitational field at point P is defined as the force per unit charge/mass placed at that point.
F
=E
q
F
= g
m
𝑀
g = G 𝑅2
M – mass of the planet
R – radius of the planet
𝑀
gEarth = G 2 = 9.80 m s-2
𝑅
Direction of electric field due to charge Q at
some point P is equal to the direction of the
force on a positive charge placed at that point.
Electric force acting on a charge q placed there is:
𝐹=𝑞𝐸
Direction of gravitational field due to mass M
at some point P is toward mass M.
Gravitational force acting on a mass m placed there is:
𝐹 =𝑚𝑔
commonly called “weight of mass m”
▪ Uniform electric field, E
▪ Uniform gravitational field, g
If a positive charge q is released at point A,
force F = qE will accelerate it in the direction of
the field toward point B.
If a mass q is released at point A, force F = mg will
accelerate it in the direction of the field toward point B.
Work done on charge by force F along
displacement d is converted into kinetic energy.
W = Fd = qEd = ½ mv2
(remember: const E, const. F so W = Fd)
Work done on mass by force F along displacement d is
converted into kinetic energy.
W = Fd = mgd = ½ mv2
(remember: const E, const. F so W = Fd)
Magnetic field
▪ We call the lines along which the magnets align themselves the magnetic field lines.
▪ magnetic field 𝑩 is measured in Tesla (T).
|B| =
Fmag
qv
1 T(Tesla) =
N∙s
C∙m
Direction at any location is the direction in which the north
pole of the compass needle points at that location.
N
S
N
▪ Strength of the B-field is proportional
to the density of the field lines.
N
S
▪ At either pole of the earth the B-field is thus the greatest.
▪ Magnetic field lines don’t start or stop.
▪ There are no magnetic charges (monopoles)
Outside magnet: N → S
Inside magnet: S → N
(always closed loops)
S
N
S
Sketching conventions for drawing direction of 3-D vector.
How do you draw a vector that is
directed toward you or away from you
View from head of a vector
OR
B
OUT of Page
toward you
View from tail of a vector
OR
INTO Page
away from you
Magnetic field caused by a straight line current
RHR 2: The direction of the magnetic field produced by electric current is given
by the right-hand rule 2:
If a wire is grasped in the right hand with the thumb in the direction of current flow,
the fingers will curl in the direction of the magnetic field.
Magnetic field B around a wire with current I
B=
𝜇0 𝐼
2𝜋 𝑟
Magnetic Field B Inside of a Solenoid
B = 𝜇0 n I
𝜇 0 = the permeability of free space 4p×10-7 T·m/A The magnetic field is concentrated into a
I = current [A]
nearly uniform field in the centre of a long
r = distance from the center of the conductor
solenoid. The field outside is weak and
n = N/L number of turns of wire per unit length diverging
Magnetic field caused by a straight line current
RHR 2: The direction of the magnetic field produced by electric
current is given by the right-hand rule 2:
If a wire is grasped in the right hand with the thumb in the direction
of current flow, the fingers will curl in the direction of the magnetic field.
Magnetic field is decreasing
away from wire
Most of the time you’ll find this
much easier to draw
Just keep that in mind!!!!
B
r
Current I OUT
Lines of B
Determining magnetic field direction – wire loop
Solenoids
A solenoid consists of several current loops stacked together.
In the limit of a very long solenoid, the magnetic field inside is
very uniform, and outside is almost zero (B  0 )
N
S
N
S
I
▪ If we place an iron core inside the solenoid we have an electromagnet.
▪The ferrous core enhances the strength of the B-field
▪ Another way: RHR for solenoids
Grasp the solenoid with your right hand in such a way that
your fingers curl in the direction of the current.
▪ Your extended thumb points in the direction of north pole.
I
I
Force on a charge moving in a B-field
Lorentz force (Hendrick Antoon Lorentz,
Dutch physicist of the late 19th and early 20th century
▪ Moving charge produces a magnetic field.
▪ Moving charge placed in an external magnetic field will feel
a magnetic force
Surprise, or isn’t?
Interaction of fields !!!
a stationary charge in a magnetic field will
feel no magnetic force because the charge
will not have its own magnetic field.
Force F felt by a charge q traveling at velocity v through a B-field
of strength B is given by
𝐹 = 𝑞𝑣𝐵 sin 𝜃
𝜃 is the angle
between 𝑣 and 𝐵
Force on q due to
presence of B
Force on a charge moving in a B-field
Magnetic force acting on a charge q
in a magnetic Field B: 𝐹 = 𝑞𝑣𝐵 𝑠𝑖𝑛 𝜃
Magnetic force on a wire carrying current I
in a magnetic field B: 𝐹 = 𝐵𝐼𝐿𝑠𝑖𝑛 𝜃
q = charge [C]
v = velocity [m/s]
B = magnetic field [T]
𝜃 is the angle between 𝑣 and 𝐵
I = current [A]
L = length [m]
B = magnetic field [T]
𝜃 = 𝜃 is the angle between 𝐼 and 𝐵
RHR 1: The direction of the magnetic force on a charge/current is given by the right-hand rule 1:
Outstretch fingers in the direction of v (or current I).
Curl fingers as if rotating vector v (I ) into vector B.
Magnetic force on a positive charge (or I) is in
the direction of the thumb.
Magnetic force on a negative charge
points in opposite direction.
▪ 𝐹 is perpendicular to 𝑣 and 𝐵 and thus to the plane of 𝑣 and 𝐵
▪ 𝐹 is in the opposite direction for a (–) charge.
Charge q in elec. field E and mag. field B
The electric force: Felec = Eq
● is always parallel to the direction of the
electric field.
● acts on a charged particle independent of the
particle’s velocity (even at rest).
● does the work when moving charge.
The work, is converted into kinetic energy which
is, in the case of conductors, transferred to
thermal energy through collisions with the lattice
ions causing increased amplitude of vibrations
seen as rise in temperature.
v
F
F
● Force is perpendicular to the direction
of the motion, so the work done by
magnetic force is zero.
W = ΔKE = 0
Hence change in kinetic energy of the charge is 0,
and that means that mag. force cannot change
the speed of the charge. Magnetic force can only
change direction of the velocity – therefore it acts
as centripetal force.
The electric field accelerates
charged particles.
v
The magnetic force: Fmag = qvB sin
● is always perpendicular to the direction
of the magnetic field
● acts on a charged particle only when the particle
is in motion and only if v and B do not point in
the same or opposite direction
(sin 00 = sin 1800 = 0).
B
In the presence of magnetic field, the moving
charged particle is deflected (dotted lines)
The trajectory of a charge q in a uniform magnetic field B
A charge q traveling at velocity v perpendicular to magnetic field B.
Show that r = mv / qB.


Force is perpendicular to v
sin θ = 1
 Magnetic force (F=qvB) does no work!
 Speed is constant (W = Δ KE )
 Circular motion
In the case the charge q is subject to the uniform field B, centripetal force Fc
is magnetic force forcing the charge to move in a circle:
𝐹𝑐 = 𝐹𝑚𝑎𝑔
Positive charge q in magnetic field B
𝑣2
𝑚
= 𝑞𝑣𝐵
𝑅
𝑅=
𝑚𝑣
𝑞𝐵
● massive or fast charges – large circles
● large charges and/or large B – small circles
B = magnetic field [T]
R =is the radius of the path
Fma is magnetic force on the charge directed toward
the centre of the circular path
m = mass [kg]
v = velocity [m/s]
q = charge [C]
Force between wires carrying current
Current-carrying wires create magnetic fields
Magnetic fields exert a force on current-carrying wires
Current carrying wires exert forces on each other!
Two current-carrying wires exert magnetic forces on one another
We already saw that if we put a current carrying wire into a magnetic field it will
feel a force....so what will happen when we put two current carrying wires
together!?!?! One will create magnetic field that the other will feel a force from,
and vice versa! Let us see what is going on.
Force between wires carrying current
Use RHR #2 to find the direction of the
magnetic field at point P
I up
I up
F
B
F xP
I down
I up
use RHR #1 to find the force on second wire
Conclusion: Currents in same direction
attract!
Use RHR #2 to find the direction of the
magnetic field at point P
use RHR #1 to find the force on second wire
B
F
xF
P
Conclusion: Currents in opposite direction
repel!