Dr. Afaf Abdelhady
Physics B
September 10, 2012
Chapter 25
CURRENT, RESISTANCE and EMF
Electric Charge: (q or Q) An uncharged object becomes negatively charged if it gains
electrons and becomes positively charged if it loses electrons. All electrons carry the
same quantity of charge, so the more electrons an object gains or loses the greater is the
overall negative or positive charge upon it. The electric charge is measured in Coulomb
(C for abbreviation).
Cells and Batteries: All cells and batteries have two terminals, positive and negative.
Chemical reactions inside a cell create a small potential difference between these
terminals and this makes the electrons flow, from the negative terminal to the positive
terminal, along any conducting path that links them. The flow of electrons, or current,
may last for many hours. Potential energy lost by electrons as they flow from one
terminal to the other is changed into other forms, such as heat and light energies in the
light bulb for example.
Conventional Current: is defined as the opposite direction of the flow of the electrons.
Electric Current: When an electric charge, dq, flows from the positive pole of a battery
to the negative pole, through an external conductor, it produces an electric current, i . In
symbols,
dq = idt
⇒ i=
dq
dt
Where i is current in Amperes (A, commonly stated as amps for short), and t is time in
seconds. The current is also defined as “the charge flowing per unit time''. The current
flowing in a circuit is measured by an ammeter. The ammeter is connected in series in
the circuit.
(*) One Coulomb: is “the charge transported in one second by an electric current of one
ampere''.
Current, Drift Velocity, and Current Density
Suppose there are n moving charged particles
per unit volume. Assume that all particles move
with the same drift velocity v d . In a time
interval dt, each particle moves a distance v d dt .
The volume of the cylinder is Av d dt . If each
particle has a charge q, the charge dQ that
flows out of the end of the cylinder is
dQ = q (nAv d dt )
And the current is
I = dQ dt = nqv d A
The current per unit cross-sectional area is called the current density J:
J = I A = nqv d
1
Dr. Afaf Abdelhady
Physics B
September 10, 2012
The units of the current density are amperes per square meter ( A / m 2 ). The general
expressions for current and current density in case of q is positive or negative are:
I = dQ dt = n q Av d
J = I A = n q vd
We can also define a vector current density J that includes the direction of the drift
velocity:
J = nqv d
(vector current density)
There are no absolute value signs in the above equation. If q is positive, v d is in the same
direction as E ; if q is negative, v d is opposite to E . In either case, J is in the same
direction as E .
CAUTION: current density vs. current
1- Current density is a vector but current is not.
2- Current density describes how charges flow at a certain point, and the direction
tells you about the direction of the flow at that point. Its magnitude can vary
around the circuit.
3- Current describes how charges flow through an extended object such as a wire, it
has the same value at all points in the circuit
Resistivity
For some materials, especially metals, at a given
temperature, J is nearly directly proportional to
E and the ratio of their magnitudes is constant,
so this constant is called the resistivity
ρ= E J
(definition of resistivity)
Resistivity and Temperature
ρ (T ) = ρ0 [1 + α ∆T
]
⇒
∆T = T − T 0
Where ρ0 is the
resistivity at a
reference temperature
T 0 (usually taken to
be 20 o C ), and α is
called the temperature
coefficient of
resistivity.
2
Dr. Afaf Abdelhady
Physics B
September 10, 2012
Potential Difference: (P.D.) (or voltage), is '' a measure of the energy transfer per unit
charge''. When the charge flows through a lamp the electrical energy is transferred into
heat energy and light energy. The potential difference measures energy transfer in both
ways. The p.d. across a power supply measures the electrical energy transfer into the
circuit and the p.d. across a component measures the energy transfer from the circuit.The
potential difference between two points is “the work done in joules in moving one
coulomb of charge between them”. P.D. is measured by a voltmeter. In the circuit, the
voltmeter is connected in parallel with the element (lamp for example).
Example: a 12 V power supply (battery) transfer 12 joules of energy to each Coulomb of
charge.
(*) One Volt: is “the potential difference between two points when one joule of energy is
needed to move one Coulomb of charge between the two points''. In symbols,
V =
U
q
Where U is the energy in Joules, and q is the charge in Coulombs.
Resistance
Consider a conductor of uniform cross section A and length L. The current density is
uniform, and the electric field is constant along the length. So, the total current I = J A ,
and the potential difference V = E L . Since
E =ρ J
V ρI
ρL
=
or V =
I
L A
A
This shows that when ρ is constant, the total current I is proportional to the potential
difference V.
Ohm's Law: Experiments show that: ''At constant temperature, the current passing
through a conductor is directly proportional to the potential difference across its ends''.
In symbols:
V ∝I
⇒
V = IR
Where R is a constant representing the resistance of the circuit components (may be a
lamp or any equipment). It represents the opposition of a material to an electric current
flow, which is measured in ''Ohm'', and its abbreviation is the Greek letter Ω (Omega).
Comments:
* There are two main uses of resistors in electrical and electronic circuits:
(a) They convert electrical energy into heat energy. For example, the element in an
electric kettle or iron, heat bulbs, etc.
(b) They ''help'' other components in a circuit, by controlling the current level in the
circuit in the various parts of the circuit. The other components, such as diodes
and transistors, can then do their job. The resistors in an electronic circuit may
become slightly warm, but they are not heaters.
3
Dr. Afaf Abdelhady
Physics B
September 10, 2012
*If a conductor obeys Ohm's law, it is called Ohmic conductor. A graph of potential
difference against current (V:I) is a straight line passing through the origin. The gradient
of the graph is the resistance of the conductor.
* If a conductor does not obey Ohm's law, it is called non-Ohmic conductor. An example
of non-Ohmic conductor is the flash light bulb, in which the resistance increases with
increasing the temperature, but not in a linear relation.
*Resistance of a Metal Wire: The resistance, R, of a resistor (e.g. metallic wire) is given
by:
R =ρ
Where ρ is the resistivity ( ≡
1
σ
L
,
A
[ ρ ] = Ω.m
) of the conductor, which depends on the type of
material. (L ) is he length of the resistor in meters and (A ) is the cross-sectional area of
the wire in m2.
From the formula above, it is noted that:
(a) The longer the wire, the higher will be the resistance. Also,
(b) Te thinner the wire, the higher will be the resistance.
Another factor which affects the resistance of a conductor is its temperature. Generally,
for most metals, their resistances increase with the rise in temperature. However, for
semi-conductors, and thermistors, their resistances decrease with a rise in temperature.
Resistance Varies with Temperature: If a conductor has a resistance R 0 at a
temperatureT 0 , then its resistance R at a temperature T is:
R = R [1 + α ∆T
]
Where ∆T = T −T 0 and α is the temperature coefficient of resistivity. Usually α varies
with temperature and so this relation is applicable only over a small temperature range.
α has the units of K-1 or Co -1 .
Power dissipated in the circuit: Rate of electrical energy transfer
du
dq
P=
=V
= IV = i 2 R =V 2 / R
dt
dt
Also, the power dissipated due to a resistance is equal to: P = i 2 R =
The unit of the power is watt (1 J s = 1 W ) .
4
V2
R
Dr. Afaf Abdelhady
Physics B
September 10, 2012
Electromotive Force and Circuits:
There can be no steady motion of charge in an
incomplete circuit.
How to maintain a steady current in a complete circuit? If a charge q goes around a
complete circuit and returns to its starting point, the potential energy must be the same at
the end of the round trip as the beginning.
There is always a decrease in the potential
energy when charges move through an
ordinary conducting material with
resistance. So, there must be some part of
the circuit in which the potential energy
increases (a device that makes the current
moves from lower potential to higher
potential). The influence that makes this
action is called electromotive force (emf)
and the device is called a source of emf. So,
emf is the “energy needed to move one
coulomb of a charge around the whole
circuit". It is measured in volts and its
symbol is ε .
Batteries, electric generators, solar cells, and
fuel cells are all examples of sources of emf.
All such devices convert energy of some
form (mechanical, chemical, and so on) into
electric potential energy and transfer it into
the circuit. An ideal source of emf maintains a constant potential difference between its
terminals, independent of the current through it.
The increase in potential energy ( qV ab ) is just equal to the non-electrostatic work W ( qε )
or
5
Dr. Afaf Abdelhady
Physics B
September 10, 2012
V ab = ε
(ideal source of emf)
The potential difference between the ends of the
wire is V ab = IR , then we have
ε =V ab = IR
(ideal source of emf)
Note:
ELECTROMOTIVE FORCE OF A
BATTERY: It is the "total potential difference
across its terminals when it is not supplying a
current". It is equal to the sum of all the
potential differences across the components of
the circuit.
Internal resistance
The potential difference across a real source in a circuit is not equal to the emf. The
reason is that any real source has internal resistance, denoted by r. So, the potential
difference is
(terminal voltage)
V ab = ε − Ir
The potential V ab , called the terminal voltage, is less than emf because of the term Ir
representing the potential drop across the terminal resistance r.
The current in the external circuit is still determined by V ab = IR . So,
ε − Ir = IR
or
I =
ε
R +r
That is, the current equals the source emf divided by the total circuit resistance (R + r ) .
Note:
1- Power output of a source is P =V ab I = ε I − I 2 r .
2- Power input to a source is P =V ab I = ε I + I 2 r .
3- Power delivered to a resistor is P =V ab I = I 2 R =
6
V2
R
Dr. Afaf Abdelhady
Physics B
September 10, 2012
The term I 2 r is the rate at which electrical energy is dissipated in the internal resistance
of the source.
7
Dr. Afaf Abdelhady
Physics B
September 10, 2012
Chapter 26
DIRECT- CURRENT CIRCUIT
Laws for Series Circuits:
1. The same current (I) pass through each part of the
circuit.
2. The applied P.D is equal to the sum of the P.D. across
the separate resistors:
V = V1 + V2 + V3 +
= IR1 + IR2 + IR3 +
= IRtotal
3. The total (equivalent) resistance is equal to the sum of the separate resistance:
Rtotal = R1 + R2 + R3 +
.
(*) The effective resistance increases ⇒ decreases in the total current in the
circuit.
Laws for Parallel Circuits:
1. The P.D. across each resistor is the same.
2. The total current is equal to the sum of the current in
the separate resistors:
I total = I1 + I 2 + I 3 +
V
Rtotal
=
V
R1
+
V
R2
+
V
R3
+
3. The total resistance of a number of resistors in parallel is less than the value of
any of the separate resistors and is given by:
1
Rt
=
1
R1
+
1
R2
+
1
R3
+
The reciprocal of the equivalent resistance equals the sum of the reciprocals of
their individual resistances.
(*) The effective resistance decreases ⇒ increases in the total current in the
circuit.
8
Dr. Afaf Abdelhady
Physics B
September 10, 2012
(**) When two resistors only are connected in parallel, then the combined (total or
equivalent) resistance (R) is given by
R + R2
1
1
1
=
+
= 1
Rt
R1
R2
R1 R 2
i .e .
Rt =
r e s i s t a n c e s m u lt i p li e d
R1 R 2
=
re s is ta n c e s a d d e d
R1 + R 2
The combined resistance is less than that of either resistor by itself.
Ammeters and Voltammeters:
Ammeter (A)
Voltammeter (V)
1- measures the current in Amperes. measures potential difference in Volts.
2- connected in series in the circuit. connected in parallel with the component.
3- has a very low resistance.
has a very high resistance.
Potential Changes Around a Circuit
The net change in potential
around the circuit must be zero;
in other words, the algebraic
sum of the potential differences
and emf's around the loop is
zero. So,
ε − Ir − IR = 0
If we take the potential to be
zero at the negative terminal
of the battery, then we have
a rise ε and a drop Ir in the
battery and an additional drop
IR in the external resistor, and
as we finish our trip around the
loop, the potential is back
where it started.
Kirchhoff's Rules
There are two terms that we will use. A junction in
a circuit is a point where three or more conductors
9
Dr. Afaf Abdelhady
Physics B
September 10, 2012
meet. Junctions are also called nodes or branch points. A loop is any closed
conducting path.
Kirchhof's rules are the following two statements:
Kirchhof's junction rule: The algebraic sum of currents into any junction is zero.
That is,
∑I
=0
(valid at any junction)
This rule is based on conservation of electric charge.
Kirchhof's loop rule: The algebraic sum of the potential differences in any loop,
including those associated with emf's and those of resistive elements, must equal
zero. That is,
∑V
=0
(valid for any closed loop)
This rule is based on the electrostatic force is conservative.
R-C Circuits
A circuit that has a resistor and a capacitor in series is
called an R-C circuit. We consider a simple circuit for
charging a capacitor.
Charging a Capacitor
We begin with the capacitor initially uncharged, the potential
difference across it is zero, but across R is equal to the battery
emf. The current across R is I 0 = ε R . Then, we close the
switch, completing the circuit and permitting current around
the loop to begin charging the capacitor. As the capacitor
charges, its voltage increases and the potential difference
across R decreases, corresponding to a decrease in current.
The instantaneous potential difference v ab and v bc are
v ab = iR
v bc = q C
10
Dr. Afaf Abdelhady
Physics B
September 10, 2012
Using Kirchhoff's loop rule, we find
ε − iR − q C = 0
Solving for i, we find
ε
i =
R
−
q
RC
After a long time the capacitor becomes fully charged, the current decreases to zero, and
the
potential difference across R becomes zero. Then
ε
R
=
q
RC
and Qf = C ε
The entire battery emf appears across the capacitor
and v bc = ε .
Note that the final charge Qf does not depend on R.
We can derive general expression for the charge and
the current as a function of time. Since i = dq dt ,
then
dq ε
q
1
= −
=−
(q − C ε )
dt
R
RC
RC
dq
dt
=−
q −C ε
RC
And then we integrate both sides
dq '
dt '
=
−
∫0 q ' − C ε ∫0 RC
q
ln(
t
q −C ε
t
)=−
−C ε
RC
then
t
−
q −C ε
= e RC
−C ε
Solving for q, we find
q = C ε (1 − e
−
t
RC
)=Qf (1 − e
−
t
RC
)
The instantaneous current i is just the time derivative
of q
t
−
dq
ε − R tC
RC
i =
dt
=
R
= I 0e
e
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Dr. Afaf Abdelhady
Physics B
September 10, 2012
The charge and current are both exponential functions of time.
Discharge the capacitor
Using Kirchhoff's loop rule with ε = 0
i = dq dt = − q RC
The current is now negative; this is because positive charge q is leaving the left
hand capacitor plate (fig.26.23b), so the current is in the direction opposite to that
shown in the figure. Rearrange the equation and integrate
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Dr. Afaf Abdelhady
Physics B
September 10, 2012
dq '
1
'
∫Q q ' = − RC ∫0 dt
0
q
t
ln
q
t
=−
Q0
RC
q=Q0e t
RC
The instantaneous current i is the derivative of this with respect to time
i = dq dt = −
Q 0 −t
e
RC
RC
= − I 0e −t
RC
Time constant
After a time equal to RC, the current in the R-C circuit has decreased to 1/e of its
initial value. The product RC is therefore a measure of how quickly the capacitor
charges. We call RC the time constant, or the relaxation time, of the circuit,
denoted by τ :
τ = RC
Energy for R-C circuit
While the capacitor is charging, the instantaneous rate at which the battery
delivered energy to the circuit is
P = εi
Since
ε = iR +
q
C
Multiply this equation by i
ε i = i 2R +
iq
C
This means that of the power (ε i ) supplied by the battery, part (i 2 R ) is dissipated
in the resistor and part (iq C ) is stored in the capacitor.
13
Dr. Afaf Abdelhady
Physics B
September 10, 2012
Kirchhoff’s Laws
Junction rule: “The sum of the currents entering any junction must be equal to the sum
of the current leaving that junction.” In symbols ∑ I in = ∑ I out . It is a statement of
conservation of charge.
Loop rule: “The algebraic sum of the changes in potential encountered in a complete
traversal of any loop of a circuit must be zero.” In symbols ∑ Vi = 0 . It is a statement of
conservation of energy.
I
a
I
+
+
a
b
R
b
R
∆V =V b −V a = +iR
∆V =V b −V a = −iR
⇒ V b − iR =V a
⇒ V b + iR =V a
a
a
b
b
έ
1-
Find Va − Vb .
έ
∆V =V b −V a = −ε
∆V =V b −V a = +ε
⇒ V b + ε =V a
⇒ V b − ε =V a
10 Ω
2A
a
b
20 V
10 V
Start with point a and move to the right, one can have
V a − 20 + 2 × 10 − 10 = Vb
2-
⇒
V a − Vb = 10 volts
A- Find the values of the currents in figure (3).
Ans: At the junction b
I1 + I 2 + I 3 = 0
Loop I implies
+7I 1 − 10 − 4 = 0 ⇒ I 1 = 2A
Loop II implies
4 + 6 − 5I 2 = 0 ⇒ I 2 = 2 A
from (2) and (3) in (1), one can find
I 3 = − I 1 − I 2 = −4 A
7Ω
5Ω
a
(1)
(2)
4V
I1
I
I3
(3)
10 V
14
II
b
6V
I2
Dr. Afaf Abdelhady
Physics B
September 10, 2012
So, the current I3 will be in the opposite direction.
B- With three different methods, prove that V a −V b = +4V .
Through loop I, one finds V a − I 1 (7) + 10 =V b ⇒V a −V b = 14 − 10 = 4 Volts
Through loop II, one finds V a − I 2 (5) + 6 =V b ⇒V a −V b = 10 − 6 = 4 Volts
Through the middle connection, one finds V a − 4 =V b ⇒V a −V b = 4 Volts
2Ω
I1
c
I2
7Ω
I3
1Ω
6Ω
I
II
a
b
E1
3-
1Ω
E2
1
1
The two branch currents in the circuit shown in figure are I 1 = A and I 2 = A .
3
2
A- Determine the emfs, E1 and E2.
B- With three different methods, prove that
V a −V b = 3 volts.
Ans: At the junction c
5
(1)
I1 + I 2 = I 3 ⇒ I 3 = A
6
Loop I implies
E 1 − I 1 (1 + 2) − 6I 3 = 0 ⇒ E 1 = 6 volts (2)
Loop II implies
− E 2 + I 2 (1 + 7) + 6I 3 = 0 ⇒ E 2 = 9 volts (3)
15
Dr. Afaf Abdelhady
Physics B
September 10, 2012
Example: If the current I in figure (5) is equal to 4.0 A, then the potential difference
between point 1 and 2, i.e. (V2 - V1), is:
(a)
- 24 Volts.
(b)
24 Volts.
(c)
40 Volts.
(d)@ - 40 Volts.
(e)
Zero.
V 1 − 3I − 12 − 5I + 4 =V 2
⇒V 2 −V 1 = −40 Volts
Example: For the circuit shown in figure (4), if V a −V c = 20 V = 20 Volts, what is the
potential difference V b −V d ?
(a)
(b)
(c)
(d)
(e)
-55 Volts.
25 Volts.
-25 Volts.
35 Volts.
55 Volts.
In Loop I, 65 − 15I 1 − 20 = 0 ⇒ I 1 = 3A
∴
V b −V d =V b −V c +V c −V d = 40 + 3 × 5 = 55 V
Another
For large loop
V b − 40 − 3 × 15 =V d ⇒V b −V d = 55 Volts
Example: A capacitor of capacitance C is discharging through a resistor of resistance R.
In terms of RC , when will the energy stored in the capacitor reduces to one fifth of its
initial value?
∵U = U o e −2t /τ
1
5
∴ U o = U o e −2t /τ ⇒ t = (
16
ln 5
)τ = 0.80 RC
2
Dr. Afaf Abdelhady
Physics B
September 7, 2012
CHAPTER 27
MAGNETIC FIELDS AND MAGNETIC FORCES
Properties of Magnet
1- A magnet attracts iron pieces or iron fillings. The filings cling near the ends, at the (poles}) of
the magnet.
2- The magnetization is zero at the middle of the magnet and maximum at the poles.
3- When a magnet is freely suspended so that it can swing in a horizontal plane, it comes to rest in
N-S direction. N-pole points towards the north, and S-pole points towards the south.
4- First law of magnetism: '' like poles repel, unlike poles attract”.
5- When a magnet is broken into pieces, each piece is found to be a magnet with two poles, i.e. no
monopole magnet.
What causes magnetism? It is the spinning of the electrons in the atoms of materials, and the behavior
of these electrons, which give a material its magnetic properties.
Magnetic field: ''It is the region around a magnet in which a magnetic force is exerted''. We can
describe magnetic interactions in two steps:
1- A moving charge or a current creates a magnetic field in the surrounding space (in addition to
its electric field)
JG
2- The magnetic field exerts a force F on any other moving charge or current that is present in the
field.
1
Dr. Afaf Abdelhady
Physics B
September 7, 2012
JJG
JJG
Magnetic field is a vector quantity and denoted by B . At any position the direction of B is defined
as the direction in which the north pole of a compass needle tends to point.
The Magnetic Force on a Moving Charge: The magnetic force, FB , on a test charge, q o , moving
with velocity v in a magnetic field, is
JJG
G JJG
F = qv × B
The magnitude of the magnetic force is F = qvB sin φ , where φ is the angle measured from the
direction of v to the direction of B . The force is perpendicular both to the velocity and to the
magnetic field. Since the force is proportional to the velocity, charges at rest do not experience
N
magnetic forces. [B] = T (Tesla) =
. Also, 1T =104 Gauss.
A.m
EXTRA NOTICES ABOUT THE FORCE
1- The charge should be in motion (i.e. current).
2- Positive and negative charges experience forces in opposite directions.
3- The force is greatest when the charge moves perpendicular to the magnetic field and zero when the
charges move parallel to the field.
4- The size of the force also depends on the magnitudes of the magnetic field and the electric charge
(current) and on the speed of the moving charge.
5- The magnetic force does not accelerate the charge, but deflect it.
Magnetic Field Lines and Magnetic Flux
The magnetic flux lines or lines of force represent the magnitude and direction of the magnetic field,
i.e. it is vector quantities.
2
Dr. Afaf Abdelhady
Physics B
September 7, 2012
Note that:
1- The density of lines is proportional to the strength of
the field (i.e. the lines are close together where the field
is strong and vice versa).
2- The lines go from north pole to the south pole.
3- The lines never cross each other.
3
Dr. Afaf Abdelhady
Physics B
September 7, 2012
Magnetic Flux and Gauss's Law for Magnetism
WE can divide any surface into elements of area dA. For each
JJG
element we determine B ⊥ , the component of B normal to the
surface at the position of that element. We define the magnetic
flux d φB through this area as
JJG JG
d φB = B ⊥dA = B cos ϕdA = B .d A
JJG
Where ϕ is the angle between the direction of B and the line
perpendicular to the surface.
The total magnetic flux through the surface is the sum of the contributions from the individual area elements: JJG JG
φB = ∫ B ⊥dA = ∫ B cos ϕdA = ∫ B .d A
JJG
Magnetic flux is a scalar quantity. In the special case in which B is uniform over a plane surface with
total area A, B ⊥ and ϕ are the same at all points on the surface, and
φB = B ⊥ A = BA cos ϕ
The SI unit of magnetic flux is called weber (Wb).
1 Wb = 1 T .m 2 ,1 T = 1 N A
Gauss's law for magnetism
The total magnetic flux through a closed surface is always zero.
Symbolically
JJG JG
B
v∫ .d A = 0
If the element of area dA is at right angles to the field lines, then B ⊥ = B ; calling the area dA ⊥ , we
have
B=
d φB
dA ⊥
That is, the magnitude of magnetic field is equal to flux per unit area across an area at right angles to
JJG
the magnetic field. For this reason, magnetic field B is sometimes called magnetic flux density.
4
Dr. Afaf Abdelhady
Physics B
September 7, 2012
A Beam of Charged Particles in a Magnetic Field: Consider a beam of positively charged particles
.If this beam enters a magnetic field at right angle to v with velocity
its direction of motion, it will experience a force perpendicular to
both velocity and magnetic field, i.e. it will be deflected.
’ Circular motion: The beam will move in circular path if the
velocity is small and the magnetic field is strong. In that case:
Magnetic force = Centripetal force
qvB = m
v2
R
⇒ B =
mv
qR
where R is the radius of the circle.
Also, the angular frequency ( ω ) and the periodic time (T ) of a rotating
charged particle are:
v qB
=
,
R
m
2π 2π m
=
T =
ω qvB
ω=
) The beam will move in helical (spiral) path if the velocity vector has
an angel with the magnetic field.
) The beam of negative charges will be deflected in the reverse
direction.
Application of motion of charged particles
’ Velocity selector: In a region of crossed or perpendicular magnetic
field B and electric field E
Perpendicular to v, the forces cancel when
qvB = qE ⇒ v =
E
B
Only particles with speeds equal to E/B can pass through without being
deflected by the fields. By adjusting E and B appropriately, we can
select particles having a particular speed for use in other experiments.
Because v does not depend on the charge, a velocity selector works also
for electrons or other negatively charged particles.
Thomson's e/m experiment
5
Dr. Afaf Abdelhady
Physics B
September 7, 2012
’ When a particle is accelerated through
a potential difference, V , energy
conservation requires that the kinetic
energy equals the loss of electric
potential energy
qV =
q
1
v2
mv2 ⇒
=
m 2V
2
In this experiment, a beam of electron is
used, so
eV =
1
e
v2
mv2 ⇒
=
⇒v =
2
m 2V
2eV
m
The electron passes between the plates and strike the screen at the end of the tube, which is coated with
a material that fluoresces at the point of impact. The electrons pass straight through the plates when
E
=
B
2eV
m
so
e
E2
=
m 2V B 2
All the quantities on the right side can be measured, so the ratio e/m of charge to mass can be
determined.
The Magnetic Force on a Current Carrying Wire: The
magnetic force FB on a segment of wire, L , carrying current, I,
in a magnetic field, B , is given by
FB = q o v × B = q o
q
L
× B = o L × B = I (L × B )
t
t
The total force on the wire is the vector sum of the forces on the
segments.
6
Dr. Afaf Abdelhady
Physics B
September 7, 2012
Assignments (text Book, University Physics) (12 edition)
Chapter 21:
5,13,24,25,31,44,46,47,60,64,70.
Chapter 22:
1, 3,8,9,13,16,19,23.
Chapter 23:
1,5,6,19,21,28,31,32,38,41,47,48.
Chapter 24:
1,2,3,6,10,12,14,16,22,24,25,35,38,41,47.
Chapter 25:
12,3,10,11,16,22,25,28,31,32,37,44,47,48,53.
Chapter 26:
4,8,11,21,40,42,45,50,51,52.
Chapter 27:
1,4,5,6,10,11,21,25,26,28,32,34,37.
7
Dr. Afaf Abdelhady
Physics B
September 7, 2012
Assignments (text Book, University Physics) (13 edition)
Chapter 21:
7,15,64,25,31,75,47,56,62.
Chapter 22:
1, 3,8,9,11,14,17,21.
Chapter 23:
1,5,19,23,24,28,34,39,45.
Chapter 24:
2,3,6,10,14,16,18,22,24,25,33,35,36,39.
Chapter 25:
2,3,10,18,23,26,28,29,33,40,43,46,51.
Chapter 26:
4,8,14,25,42,44,47,52,53.
Chapter 27:
1,4,5,6, 11,21,25,26,28,38,39.
8