SCHEME OF WORK
WEEK:
1. Revision/ Electromagnetic waves. Type of radiation in electromagnetic in spectrum description and uses.
2. Gravitational field and law gravitational potential, escape velocity, potential energy in Gravitational field.
3. Electric field- Coulombs law electric field intensity, electric potential, capacitor and capacitance, Definition arrangement and application.
4. Current Electricity- Primary and secondary cells, Defects of simple cells. Daniel cells, Leclanche, Lead acid accumulator, Alkaline- cadmium cell.
5. Electrolysis- Electrolysis, Electrodes, ions, Faradays laws, electro chemical equivalent.
6. Electric measurement- Resistivity, conductivity, conversion of galvanometer to Ammeter and Voltmeter, method, shunt and electro chemical equivalent.
7. Magnetic field- Magnetic and magnetic material, temporary and permanent magnets, magnetization, demagnetization, theory of magnetization, magnetic- feux, earth magnetic field.
8. Magnetic field around current carrying conductors- straight conductor, circular conductor or solenoid Application- electromagnet, uses of electromagnet – electric bells, telephone earpiece.
9. Electromagnetic field- fleming left hand rule, Application-D.C motors, moving coil Galvanometer.
10. Electromagnetic induction- induced current, laws of electromagnetic induction, induction coil, A/C and D.C, Generator transformers and power transmission and distribution, Eddy current.
11. Revision
1ST TERM
WEEK 1
TOPIC: GRAVITATIONAL FIELD/LAW
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. State the Newton's law of universal gravitation.
ii. Define force of gravity
iii. Explain weightlessness and solve simple problems on gravitational potential.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: CONCEPT OF GRAVITATIONAL FIELD
Is a region of space surrounding the earth where object of masses experience force of gravity.
Newton's Law: states that every particle in the universe attracts every other particle with a force that is proportional to the product of their masses and inversely proportional to the square of their separation.
F α m1m2 = F = Gm1m2
r2 r2
note that G = 6.67 x 10-11 Nm2/kg2
example: what is the magnitude of force attraction between two 5kg ball, 50mm apart?
Relationship between "G" and "g" with earth
Fg = mg; and F = Gmm
R2
Mg = GMm
R2
G = GM
R2
At an altitude i.e. g (acc. At a point) is = g 1-2h
R
Which implies that the acceleration due to gravity decreases with increased rise in altitude.
Weightlessness occurs when
h = R
2
Also when a plane descends, or a body in a lift with i.e. a = g
Where R is reaction
W - R = ma
R = W- ma = mg - ma
M (g-a)
When R = 0, mg - ma =0
Mg = ma
And a=g
EVALUATION:
State Newton's law of universal gravitation.
ASSIGNMENT:
Two spheres of masses 100kg and 900kg respectively have their centres separated by a distance of 0.61m. Calculate the magnitude of the force of attraction between them [G = 6.67 x 10-11 Nm2kg-2]
SUB-TOPIC:
BEHAVIOURAL OBJECTIVES: At the end of the lesson, students should be able to:
i. State the meaning of gravitational potential and use the formula.
CONTENT
Gravitational potential - due to the field of the earth is defined as numerically equal to the work done in bringing a unit mass from infinity to a point in one field.
Unit is Jkg-1
potential V = work done
mass
v = mgR but g = GM
m R2
V = GMR V = -GM
Rx R
The negative sign is an indication that the potential at infinity zero is higher than potential close to the mass.
EVALUATION:
i. Explain briefly why gravitational potential is negative.
ii. Define gravitational potential at a point.
ASSIGNMENT:
1. The average radius of Jupiter's orbit round the sun of mass 2 x 1030kg is 7.8 x 1011m. if the mass of Jupiter is 1.9 x 1027kg, find the gravitational force the sun exerts on Jupiter.
[G = 6.67 x 10-11Nm2/kg2].
2. A mass of 100kg is on the earth surface. What is its gravitational potential due to the earth. Take mass of earth = 6.0 x 1024kg & radius of earth = 6.4 x 106m, G = 6.67 x 10-11Nm2kg-2).
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. State the Newton's law of universal gravitation.
ii. Define force of gravity
iii. Explain weightlessness and solve simple problems on gravitational potential.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: CONCEPT OF GRAVITATIONAL FIELD
Is a region of space surrounding the earth where object of masses experience force of gravity.
Newton's Law: states that every particle in the universe attracts every other particle with a force that is proportional to the product of their masses and inversely proportional to the square of their separation.
F α m1m2 = F = Gm1m2
r2 r2
note that G = 6.67 x 10-11 Nm2/kg2
example: what is the magnitude of force attraction between two 5kg ball, 50mm apart?
Relationship between "G" and "g" with earth
Fg = mg; and F = Gmm
R2
Mg = GMm
R2
G = GM
R2
At an altitude i.e. g (acc. At a point) is = g 1-2h
R
Which implies that the acceleration due to gravity decreases with increased rise in altitude.
Weightlessness occurs when
h = R
2
Also when a plane descends, or a body in a lift with i.e. a = g
Where R is reaction
W - R = ma
R = W- ma = mg - ma
M (g-a)
When R = 0, mg - ma =0
Mg = ma
And a=g
EVALUATION:
State Newton's law of universal gravitation.
ASSIGNMENT:
Two spheres of masses 100kg and 900kg respectively have their centres separated by a distance of 0.61m. Calculate the magnitude of the force of attraction between them [G = 6.67 x 10-11 Nm2kg-2]
SUB-TOPIC:
BEHAVIOURAL OBJECTIVES: At the end of the lesson, students should be able to:
i. State the meaning of gravitational potential and use the formula.
CONTENT
Gravitational potential - due to the field of the earth is defined as numerically equal to the work done in bringing a unit mass from infinity to a point in one field.
Unit is Jkg-1
potential V = work done
mass
v = mgR but g = GM
m R2
V = GMR V = -GM
Rx R
The negative sign is an indication that the potential at infinity zero is higher than potential close to the mass.
EVALUATION:
i. Explain briefly why gravitational potential is negative.
ii. Define gravitational potential at a point.
ASSIGNMENT:
1. The average radius of Jupiter's orbit round the sun of mass 2 x 1030kg is 7.8 x 1011m. if the mass of Jupiter is 1.9 x 1027kg, find the gravitational force the sun exerts on Jupiter.
[G = 6.67 x 10-11Nm2/kg2].
2. A mass of 100kg is on the earth surface. What is its gravitational potential due to the earth. Take mass of earth = 6.0 x 1024kg & radius of earth = 6.4 x 106m, G = 6.67 x 10-11Nm2kg-2).
WEEK 2
TOPIC: ESCAPE VELOCITY POTENTIAL IN GRAVITATIONAL FIELD
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Define escape velocity
ii. Solve simple questions on escape velocity
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: ESCAPE VELOCITY
This defined as the minimum velocity required for an astronomical body (e.g. satellite/rocket) to just escape/leave the gravitational influence or filed permanently.
Let m be mass of astronomical body and me be mass of earth, of distance r
W D m m = w = f x r
F = Gmme
R
W = Gmme -------- (1)
R
Thus W.D must be equal to the K.E
K.E = ½ mr2 = w
½ mr2 = gmme
R
Make V subject
Ve = √2Gme
r
Thus r is the Radius of the earth (R)
but we know that
ge = Gme making G subject
R2
G = Rg
Me
Ve = √ R2g me x 2 ÷ R
me
Ve = √ 2g R2
R
Ve = √ 2gR Q.E.D
EVALUATION: The teacher evaluates the lesson with the following questions:
Define escape velocity.
SUB-TOPIC:
BEHAVIOURAL OBJECTIVES: At the end of this lesson, the students should be able to :
i. Distinguish between orbital velocity and parking orbit.
ii. Solve simple problems on satellite body and velocity of escape.
CONTENT: ORBITAL VELOCITY
A satellite orbits a planet such as the earth with a velocity. It orbits a planet such as the earth with a velocity. It orbits just because the centripetal force on it is equal to the gravitational force on it.
Mv2 = GMm
r r2
V2 = Gm
r r2
V2 = Gmr
R
Hence V = √ Gm
r
where V is the orbital velocity.
PARKING ORBIT
This is the orbit of a satellite having the same period as that of the earth
T = 2ￗr
V
EVALUATION:
Distinguish between orbital velocity and parking orbit of an astronomical body.
ASSIGNMENT:
The mass of the moon is 1/80M and its 1/5R; where M and R are mass and radius of the earth respectively. Calculate the acceleration due to gravity on the surface of the moon. [g= 9.8m/s2]
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Define escape velocity
ii. Solve simple questions on escape velocity
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: ESCAPE VELOCITY
This defined as the minimum velocity required for an astronomical body (e.g. satellite/rocket) to just escape/leave the gravitational influence or filed permanently.
Let m be mass of astronomical body and me be mass of earth, of distance r
W D m m = w = f x r
F = Gmme
R
W = Gmme -------- (1)
R
Thus W.D must be equal to the K.E
K.E = ½ mr2 = w
½ mr2 = gmme
R
Make V subject
Ve = √2Gme
r
Thus r is the Radius of the earth (R)
but we know that
ge = Gme making G subject
R2
G = Rg
Me
Ve = √ R2g me x 2 ÷ R
me
Ve = √ 2g R2
R
Ve = √ 2gR Q.E.D
EVALUATION: The teacher evaluates the lesson with the following questions:
Define escape velocity.
SUB-TOPIC:
BEHAVIOURAL OBJECTIVES: At the end of this lesson, the students should be able to :
i. Distinguish between orbital velocity and parking orbit.
ii. Solve simple problems on satellite body and velocity of escape.
CONTENT: ORBITAL VELOCITY
A satellite orbits a planet such as the earth with a velocity. It orbits a planet such as the earth with a velocity. It orbits just because the centripetal force on it is equal to the gravitational force on it.
Mv2 = GMm
r r2
V2 = Gm
r r2
V2 = Gmr
R
Hence V = √ Gm
r
where V is the orbital velocity.
PARKING ORBIT
This is the orbit of a satellite having the same period as that of the earth
T = 2ￗr
V
EVALUATION:
Distinguish between orbital velocity and parking orbit of an astronomical body.
ASSIGNMENT:
The mass of the moon is 1/80M and its 1/5R; where M and R are mass and radius of the earth respectively. Calculate the acceleration due to gravity on the surface of the moon. [g= 9.8m/s2]
WEEK 3
TOPIC: ELECTRIC FIELD AND COULOMB’S LAW
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. State electric force field properties; calculate field strength and potential work done and energy in an electric field.
ii. Solve simple problems involving electric field and changes.
iii. State Coulomb’s Law.
CONTENT: ELECTRIC FORCE BETWEEN POINT CHARGE
Coulomb’s Law: the French physicists Charles Coulomb (1736-1806) postulated:
F α 1/r2; F α q1q2
This law states that the electric force between 2 point charges q1 and q2 separated by a distance r, is directly proportional to the product of the charges and inversely proportional to the distance between the charges.
Hence F = K q1q2 = K = 1
r2 4ￗ%
Definition of terms
Electric filed is a region where an electric force is experienced by a changed body.
Electrical lines of force: are imaginary lies drawn in such a way that the direction at any point is the same direction of the field at that point.
Electrical field intensity/strength: at any point is the force acting per unit change at that point.
E = f
q
Therefore: F = Eq
It implies that the force experienced by change is due to the presence of change q2.
Ex. 1
Two protons carrying positive charge each of + 1.6 x 10-19C and separated in free space at a distance 10-10m. Find the force of repulsion between the changes.
Ex.2
If a small positive charge q of +5 x 10-8C is positioned at point X of 2m away for a small negative charge q2 of -2 x 10-8C at y. if a pt A is located at 1.5m from A. find the electric field intensity ‘G’ at
1. Point A between X & Y
2. Point A if charge q is charged to positive charge.
EVALUATION: The teacher evaluates the lesson with the following questions:
State Coulomb’s law and write it out for two like charges at a distance of 0.1m apart.
ASSIGNMENT:
Calculate the electrical intensity and the force on a +5 x 10-6C charge placed mid way between two charges +30c and +11C; separated by a distance of 20cm apart.
SUB-TOPIC: ELECTRIC CURRENT & ELECTRIC POTENTIAL
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Define electric field potential
ii. Define capacitance and state basic principles about capacitors and use the information to solve simple problems.
CONTENT: CAPACITORS AND CAPACITANCES
A capacitor is a device for storing electricity (electric charges). It consists essentially of two conductors metal plates carrying opposite charges. The metal plates are separated by a small distance ‘d’ by an insulator.
Capacitance is defined as the ratio of the charge Q on either plates to the potential difference V between them.
C = Q
V
FACTORS
i. Size
ii. Shape
iii. Common area of plate
iv. Distance separating the two plates
v. Nature of material that separates them.
ENERGY STORED IN CAPACITOR
W.D in charging capacitor through p.d of 1/2V, is
w = ½ Vq = ½ qV
but since
C = q OR V= q
V c
Therefore:
W = ½ q2 C OR ½ CV2
APPLICATIONS OF CAPACITORS
i. Radio for tuning
ii. Ignition systems of motor vehicles
iii. The elimination of sparks when a containing inductance is suddenly operated.
EVALUATION: The teacher evaluates the lesson with the following questions:
Write out the expression for total energy of two capacitors identical in parallel.
ASSIGNMENT: Calculate
a. The value of capacitance in the diagram shown below
b. The p.d across each capacitor
c. The charge on each capacitor
d. Energy stored in total capacitance in the diagram below.
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. State electric force field properties; calculate field strength and potential work done and energy in an electric field.
ii. Solve simple problems involving electric field and changes.
iii. State Coulomb’s Law.
CONTENT: ELECTRIC FORCE BETWEEN POINT CHARGE
Coulomb’s Law: the French physicists Charles Coulomb (1736-1806) postulated:
F α 1/r2; F α q1q2
This law states that the electric force between 2 point charges q1 and q2 separated by a distance r, is directly proportional to the product of the charges and inversely proportional to the distance between the charges.
Hence F = K q1q2 = K = 1
r2 4ￗ%
Definition of terms
Electric filed is a region where an electric force is experienced by a changed body.
Electrical lines of force: are imaginary lies drawn in such a way that the direction at any point is the same direction of the field at that point.
Electrical field intensity/strength: at any point is the force acting per unit change at that point.
E = f
q
Therefore: F = Eq
It implies that the force experienced by change is due to the presence of change q2.
Ex. 1
Two protons carrying positive charge each of + 1.6 x 10-19C and separated in free space at a distance 10-10m. Find the force of repulsion between the changes.
Ex.2
If a small positive charge q of +5 x 10-8C is positioned at point X of 2m away for a small negative charge q2 of -2 x 10-8C at y. if a pt A is located at 1.5m from A. find the electric field intensity ‘G’ at
1. Point A between X & Y
2. Point A if charge q is charged to positive charge.
EVALUATION: The teacher evaluates the lesson with the following questions:
State Coulomb’s law and write it out for two like charges at a distance of 0.1m apart.
ASSIGNMENT:
Calculate the electrical intensity and the force on a +5 x 10-6C charge placed mid way between two charges +30c and +11C; separated by a distance of 20cm apart.
SUB-TOPIC: ELECTRIC CURRENT & ELECTRIC POTENTIAL
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Define electric field potential
ii. Define capacitance and state basic principles about capacitors and use the information to solve simple problems.
CONTENT: CAPACITORS AND CAPACITANCES
A capacitor is a device for storing electricity (electric charges). It consists essentially of two conductors metal plates carrying opposite charges. The metal plates are separated by a small distance ‘d’ by an insulator.
Capacitance is defined as the ratio of the charge Q on either plates to the potential difference V between them.
C = Q
V
FACTORS
i. Size
ii. Shape
iii. Common area of plate
iv. Distance separating the two plates
v. Nature of material that separates them.
ENERGY STORED IN CAPACITOR
W.D in charging capacitor through p.d of 1/2V, is
w = ½ Vq = ½ qV
but since
C = q OR V= q
V c
Therefore:
W = ½ q2 C OR ½ CV2
APPLICATIONS OF CAPACITORS
i. Radio for tuning
ii. Ignition systems of motor vehicles
iii. The elimination of sparks when a containing inductance is suddenly operated.
EVALUATION: The teacher evaluates the lesson with the following questions:
Write out the expression for total energy of two capacitors identical in parallel.
ASSIGNMENT: Calculate
a. The value of capacitance in the diagram shown below
b. The p.d across each capacitor
c. The charge on each capacitor
d. Energy stored in total capacitance in the diagram below.
WEEK 4
TOPIC: ELECTRIC POTENTIAL AND CAPACITORS
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
State basic properties of simple cells, Daniel cell, Leclanche's cell and the lead acid accumulator.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: ELECTRIC CELLS
An electric cell is a device which supplies an electric current. It consists of electrolyte, and the electrodes as the basic components.
There are three types of cells: Simple, primary and secondary cell. The simple cell has a p.d of approximately 1V.
DEFECTS OF SIMPLE CELLS
i. Polarization: formation of H2 bubble at positive electrode.
ii. Local action
It cannot be left assembled as the dilute acid will react with the Zinc electrode.
Primary cells are those in which current is produced by an irreversible chemical change e.g Leclanche', Daniel and Nichel-Iron (Nife)
Secondary cells: these are those which can be recharged when they run down recharging is done by passing electric current through the cell.
i.e. discharging (+ve -ve, -ve +ve)
and recharging (+ve +ve, -ve -ve)
a good electric cell should have
i. A long life
ii. A minimum of polarization
iii. Cheap and safe electrodes
iv. Cheap and safe electrolytes
CELLS IN SERIES AND PARALLEL
Series: E1 = E1 + E2 + E3
Parallel: E = E1 = E2 = E3 in short the cell with the highest e.m.f
E.M.F AND TERMINAL P.D
The e.m.f of a cell is the p.d across the terminals of that cell when it is not delivering any current to an external circuit.
Potential difference: p.d is the e.m.f of a cell (or circuit) when it is delivering current to an external circuit.
V = E-v; V= terminal p.d supplied into R and v is the lost volt.
Therefore:
E = V +r V= E-r or E-Irs
Ex.
A battery made up of 2 cells joined in series supplies current to an external resistance of 5Ω. If the e.m.f and internal resistance of each cell is 0.6v and 3Ω respectively, calculate the current flowing in the external resistor, terminal p.d and lost voltage.
EVALUATION:
Two cells each having e.m.f of 1.5v and an internal resistance of 2Ω are connected in parallel. Find the current flowing when the cells are connected to a 11Ω resistor.
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
Describe the secondary cell and explain how it is charged.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: SECONDARY CELLS
These are cells that are rechargeable e.g. car battery. During discharge, the lead in the - ve plate and PbO2 are in +ve platesboth change into insoluble lead (II) sulphate (pbSO4)
EVALUATION:
Briefly describe how a car battery can be recharged.
ASSIGNMENT:
Calculate the
i. Total R
ii. P.d V across the 2 and 4Ω
iii. Current in each resistor
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
Explain cells in series and parallel and solve simple questions in cells.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: CELLS IN SERIES AND PARALLEL
Series: and parallel connections
Example:
If the terminal p.d across a circuit is 15V with internal resistance of 5Ω, resistance is 23Ω with p.d of 40V, calculate the terminal p.d.
Solution:
E = 15v(p.d across the whole circuit) (R +r)
r = 5Ω, R= 23Ω
from I = E = 15 = 5.357 x 10-1A
R + r 23 + 5
p.d across the load = IR = 12.3 x 10 volts
MEASUREMENT OF RESISTANCE
Voltmeter method:
Plot graph of v against I and obtain the slope of graph as resistance.
The disadvantage of this method is that some current is always directed through the voltmeter to make it work.
OTHER METHODS
i. Resistivity
ii. Wheatstone Bridge
iii. Metre Bridge
iv. Using measuring instruments
EVALUATION:
State Ohm's Law.
ASSIGNMENT:
i. Use diagram to describe and state Ohm's Law.
ii. Use diagram to describe cathode ray oscilloscope.
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
State basic properties of simple cells, Daniel cell, Leclanche's cell and the lead acid accumulator.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: ELECTRIC CELLS
An electric cell is a device which supplies an electric current. It consists of electrolyte, and the electrodes as the basic components.
There are three types of cells: Simple, primary and secondary cell. The simple cell has a p.d of approximately 1V.
DEFECTS OF SIMPLE CELLS
i. Polarization: formation of H2 bubble at positive electrode.
ii. Local action
It cannot be left assembled as the dilute acid will react with the Zinc electrode.
Primary cells are those in which current is produced by an irreversible chemical change e.g Leclanche', Daniel and Nichel-Iron (Nife)
Secondary cells: these are those which can be recharged when they run down recharging is done by passing electric current through the cell.
i.e. discharging (+ve -ve, -ve +ve)
and recharging (+ve +ve, -ve -ve)
a good electric cell should have
i. A long life
ii. A minimum of polarization
iii. Cheap and safe electrodes
iv. Cheap and safe electrolytes
CELLS IN SERIES AND PARALLEL
Series: E1 = E1 + E2 + E3
Parallel: E = E1 = E2 = E3 in short the cell with the highest e.m.f
E.M.F AND TERMINAL P.D
The e.m.f of a cell is the p.d across the terminals of that cell when it is not delivering any current to an external circuit.
Potential difference: p.d is the e.m.f of a cell (or circuit) when it is delivering current to an external circuit.
V = E-v; V= terminal p.d supplied into R and v is the lost volt.
Therefore:
E = V +r V= E-r or E-Irs
Ex.
A battery made up of 2 cells joined in series supplies current to an external resistance of 5Ω. If the e.m.f and internal resistance of each cell is 0.6v and 3Ω respectively, calculate the current flowing in the external resistor, terminal p.d and lost voltage.
EVALUATION:
Two cells each having e.m.f of 1.5v and an internal resistance of 2Ω are connected in parallel. Find the current flowing when the cells are connected to a 11Ω resistor.
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
Describe the secondary cell and explain how it is charged.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: SECONDARY CELLS
These are cells that are rechargeable e.g. car battery. During discharge, the lead in the - ve plate and PbO2 are in +ve platesboth change into insoluble lead (II) sulphate (pbSO4)
EVALUATION:
Briefly describe how a car battery can be recharged.
ASSIGNMENT:
Calculate the
i. Total R
ii. P.d V across the 2 and 4Ω
iii. Current in each resistor
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
Explain cells in series and parallel and solve simple questions in cells.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: CELLS IN SERIES AND PARALLEL
Series: and parallel connections
Example:
If the terminal p.d across a circuit is 15V with internal resistance of 5Ω, resistance is 23Ω with p.d of 40V, calculate the terminal p.d.
Solution:
E = 15v(p.d across the whole circuit) (R +r)
r = 5Ω, R= 23Ω
from I = E = 15 = 5.357 x 10-1A
R + r 23 + 5
p.d across the load = IR = 12.3 x 10 volts
MEASUREMENT OF RESISTANCE
Voltmeter method:
Plot graph of v against I and obtain the slope of graph as resistance.
The disadvantage of this method is that some current is always directed through the voltmeter to make it work.
OTHER METHODS
i. Resistivity
ii. Wheatstone Bridge
iii. Metre Bridge
iv. Using measuring instruments
EVALUATION:
State Ohm's Law.
ASSIGNMENT:
i. Use diagram to describe and state Ohm's Law.
ii. Use diagram to describe cathode ray oscilloscope.
WEEK 5
TOPIC: ELECTRIC CELLS AND DEFECTS
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Mention at least 2 defects of simple cells.
ii. Describe the defects of simple cells
iii. Mention corrective technique for defects in simple cells.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: DEFECTS OF SIMPLE CELLS
1. Polarization
This is the release of hydrogen bubbles at the positive electrode. Thereby insulating the electrodes. This causes back e.m.f. polarization can be minimized by brushing the electrodes and also using depolarizers e.g. manganese dioxide.
2. Local Action:
Local action occurs due to improve zinc useage as electrodes. It can be minimized or corrected by cleaning zinc with H2SO4 and then rubbing or amalgamating with mercury.
DISADVANTAGE OF SIMPLE CELL OVER SECONDARY CELLS.
It does not allow for a steady supply of current.
It is not rechargeable after discharging.
EVALUATION:
List the defects of simple cells
ASSIGNMENT:
Use diagram to describe Leclanche' and lead-acid accumulator.
SUB-TOPIC: E.M.F AND TERMINAL P.D
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Solve simple problems involving cells and e.m.f of cell
ii. Distinguish between two points on a circuit.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: E.M.F OF A CELL
The e.m.f of a cell is the p.d across the terminals of a cell when it is in an open circuit; i.e. not supplying current to an external circuit.
E supplies current I to external R. therefore the p.d across R is called terminal p.d V while v is the p.d across internal resistance "r".
Therefore E= V + r
E = IR + Ir
E= I (R +r)
Therefore I= E
R + r
Where Ir is called lost volts.
Ex.
If the terminal p.d of a battery is 15v and the external resistance of 23Ω is connected and when the lost volt is 1.65v. Find the e.m.f and the internal of the cell.
Solution:
When p.d = 15v, R = 23Ω
Also p.d = 1.65v, R = 48Ω
From I = E, but E = V + v
R + r
(i) E.m.f is given by E = 15 + 1.65
E = 16.65V
(ii) R =?
Since terminal p.d across external resistor = 15V
Therefore: using Ohm's Law,
V = IR
Therefore: I = V = I = 15 = 0.652A
R 23
Using E = I (R + r)
Or E = I (R +r) gives
16.65 = 0.652 (23 + r)
Therefore: r = 16.65 - (0.652 x 23)
0.652
R = 1.654; r = 2.54Ω
0.652
EVALUATION:
Distinguish between e.m.f of a cell and terminal p.d of that cell.
ASSIGNMENT:
Calculate the e.m.f of a cell of an internal resistance of 15Ω connected to a 100Ω resistance. If the terminal p.d is 10V.
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Mention at least 2 defects of simple cells.
ii. Describe the defects of simple cells
iii. Mention corrective technique for defects in simple cells.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: DEFECTS OF SIMPLE CELLS
1. Polarization
This is the release of hydrogen bubbles at the positive electrode. Thereby insulating the electrodes. This causes back e.m.f. polarization can be minimized by brushing the electrodes and also using depolarizers e.g. manganese dioxide.
2. Local Action:
Local action occurs due to improve zinc useage as electrodes. It can be minimized or corrected by cleaning zinc with H2SO4 and then rubbing or amalgamating with mercury.
DISADVANTAGE OF SIMPLE CELL OVER SECONDARY CELLS.
It does not allow for a steady supply of current.
It is not rechargeable after discharging.
EVALUATION:
List the defects of simple cells
ASSIGNMENT:
Use diagram to describe Leclanche' and lead-acid accumulator.
SUB-TOPIC: E.M.F AND TERMINAL P.D
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Solve simple problems involving cells and e.m.f of cell
ii. Distinguish between two points on a circuit.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: E.M.F OF A CELL
The e.m.f of a cell is the p.d across the terminals of a cell when it is in an open circuit; i.e. not supplying current to an external circuit.
E supplies current I to external R. therefore the p.d across R is called terminal p.d V while v is the p.d across internal resistance "r".
Therefore E= V + r
E = IR + Ir
E= I (R +r)
Therefore I= E
R + r
Where Ir is called lost volts.
Ex.
If the terminal p.d of a battery is 15v and the external resistance of 23Ω is connected and when the lost volt is 1.65v. Find the e.m.f and the internal of the cell.
Solution:
When p.d = 15v, R = 23Ω
Also p.d = 1.65v, R = 48Ω
From I = E, but E = V + v
R + r
(i) E.m.f is given by E = 15 + 1.65
E = 16.65V
(ii) R =?
Since terminal p.d across external resistor = 15V
Therefore: using Ohm's Law,
V = IR
Therefore: I = V = I = 15 = 0.652A
R 23
Using E = I (R + r)
Or E = I (R +r) gives
16.65 = 0.652 (23 + r)
Therefore: r = 16.65 - (0.652 x 23)
0.652
R = 1.654; r = 2.54Ω
0.652
EVALUATION:
Distinguish between e.m.f of a cell and terminal p.d of that cell.
ASSIGNMENT:
Calculate the e.m.f of a cell of an internal resistance of 15Ω connected to a 100Ω resistance. If the terminal p.d is 10V.
WEEK 6
TOPIC: CONDUCTION OF ELECTRICITY THROUGH MATERIALS
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Mention at least two defects of simple cell.
ii. State Ohm's Law and apply it in calculations involving cells, p.d and e.m.f
iii. Verify Ohm's Law and apply Ohm's Law
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT:
Ohm's law states that the current passing a metallic conductor at constant temperature is directly proportional to the p.d between its ends.
V α I
Therefore: V = IR
V = p.d, I = current flowing between ends of conductor and R is the constant (resistance)
R = V
I
VERIFICATION OF OHM'S LAW
In an experiment to verify Ohm's Law, the slope of graph of V against I will give the resistance.
DEFECTS OF CELLS
i. Polarization
ii. Local Action
Ex.
A circuit consists of 1Ω wire in series with a parallel arrangement of 6Ω and 3Ω and a p.d of 12V is connected across the whole circuit.
Calculate:
i. Total circuit resistance
ii. Current in each resistor
iii. P.d across each resistor
EVALUATION:
State Ohm's law
ASSIGNMENT:
As shown in the diagram, calculate:
i. Effective resistance
ii. Current in each resistor
iii. P.d across each resistor.
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Distinguish between electrical energy and electrical power.
ii. Solve simple problems on electrical energy and power.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: ELECTRICAL ENERGY AND POWER
Electrical energy is when quantity charge moves between two points, which have potential difference, while electrical power is the rate of energy transferred.
i.e.
Power = Energy transferred
Time taken
Energy (W) = QV put Q - It
W = IVt, put I = V
R
W = V2, put V = IR,
R2
W = I2Rt
Unit [Joules]
From power = Energy
Time
P = IV or r2t - t = v2
R R R
Or
P = I2Rt = I2R
T
ILLUSTRATION:
How much heat is produced in 2minutes by a heater using 500mA and a potential difference of 60V.
Solution:
+ = 2min = 2 x60 = 120s, V= 60V, I = 500mA = 5 x 10 -1
Using heat energy =IVt,
= it = 0.5 x 60 x 120 = 3600J
ENERGY CONSUMPTION
Commercial power is consumed in kilowatt hour (kwh) meaning energy supplied by a rate of working of 1000watts for 1 hour.
Illustration:
A house consumed 400w for lighting and 200w for refrigerator for 24hours. Calculate the total of the tariff is 50k/unit.
EVALUATION:
Distinguish between electrical energy and electrical power.
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Mention at least two defects of simple cell.
ii. State Ohm's Law and apply it in calculations involving cells, p.d and e.m.f
iii. Verify Ohm's Law and apply Ohm's Law
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT:
Ohm's law states that the current passing a metallic conductor at constant temperature is directly proportional to the p.d between its ends.
V α I
Therefore: V = IR
V = p.d, I = current flowing between ends of conductor and R is the constant (resistance)
R = V
I
VERIFICATION OF OHM'S LAW
In an experiment to verify Ohm's Law, the slope of graph of V against I will give the resistance.
DEFECTS OF CELLS
i. Polarization
ii. Local Action
Ex.
A circuit consists of 1Ω wire in series with a parallel arrangement of 6Ω and 3Ω and a p.d of 12V is connected across the whole circuit.
Calculate:
i. Total circuit resistance
ii. Current in each resistor
iii. P.d across each resistor
EVALUATION:
State Ohm's law
ASSIGNMENT:
As shown in the diagram, calculate:
i. Effective resistance
ii. Current in each resistor
iii. P.d across each resistor.
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Distinguish between electrical energy and electrical power.
ii. Solve simple problems on electrical energy and power.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: ELECTRICAL ENERGY AND POWER
Electrical energy is when quantity charge moves between two points, which have potential difference, while electrical power is the rate of energy transferred.
i.e.
Power = Energy transferred
Time taken
Energy (W) = QV put Q - It
W = IVt, put I = V
R
W = V2, put V = IR,
R2
W = I2Rt
Unit [Joules]
From power = Energy
Time
P = IV or r2t - t = v2
R R R
Or
P = I2Rt = I2R
T
ILLUSTRATION:
How much heat is produced in 2minutes by a heater using 500mA and a potential difference of 60V.
Solution:
+ = 2min = 2 x60 = 120s, V= 60V, I = 500mA = 5 x 10 -1
Using heat energy =IVt,
= it = 0.5 x 60 x 120 = 3600J
ENERGY CONSUMPTION
Commercial power is consumed in kilowatt hour (kwh) meaning energy supplied by a rate of working of 1000watts for 1 hour.
Illustration:
A house consumed 400w for lighting and 200w for refrigerator for 24hours. Calculate the total of the tariff is 50k/unit.
EVALUATION:
Distinguish between electrical energy and electrical power.
WEEK 7
TOPIC: ELECTROLYSIS AND ITS APPLICATIONS
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Explain the conditions under which liquids and gases conduct electricity.
ii. Explain the behaviour of charge carriers in liquids and gases.
iii. State Faraday's laws and calculate the charge transferred in the process.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: CONDUCTION OF ELECTRICITY THROUGH LIQUIDS
Electrolysis is the chemical change in a liquid due to the flow of electric current through the liquid.
Voltametre is used for studying the flow of current through a liquid.
IONIC THEORY
An electrolyte is a liquid/molten substance that conducts electric current.
In an electrolyte, there are positively and negatively charged particles called ions. The molecules that constitute the electrolyte are split in solution into these ions.
EXAMPLES OF ELECTROLYSIS
1. Electrolysis of acidulated water
2. Electrolysis of CUSO4
APPLICATIONS OF ELECTROLYSIS
1. Electroplating: this is the process of coating a metallic object with another metal in order to improve its appearance or to safe guard it against corrosion.
2. Calibration of an Ammetre: the copper is weighed before the current is switched on. Let mass of copper be M and the e.c.e of copper be Z(i.e. electrochemical)
The electrochemical equivalent (e.c.e) of a substance is the mass of that substance deposited by one coulomb of electrolysis (unit = g/coulomb)
3. Extraction of metals such as aluminum and sodium.
4. Manufacture of electrolytic capacitors used in electronic appliances.
5. Manufacture of some pure metals e.g. pure cooper which is used in the manufacture of cables.
EVALUATION:
Define electrolysis and electrolytes
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. State Faraday's Law of electrolysis
ii. Solve simple problems on electrolysis and
iii. Describe conduction of electricity through gases.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: FARADAY'S LAWS OF ELECTROLYSIS
1st law: states that the mass m of a substance librated in electrolysis is directly proportional to the quantity of electricity Q, which has passed through the electrolyte.
2nd Law: states that the masses of different substances deposited by the same quantity of electricity are directly proportional to the chemical equivalent of the substance.
M α Q - 1
M α It
And m = Zit
Where Z is electrochemical equivalent.
CONDUCTIONS OF ELECTRICITY THROUGH GAS
The conduction of electricity through gases is studied using discharge tubes. Gases conduct electricity are:
1. Low pressure (0.01mmHg)
2. High voltage (1000V); they break into ions
NATURE AND PROPERTIES OF CATHODE RAYS
1. They consist of streams of fast moving particles of negative electricity (ELECTRON)
2. They cause glass and other materials to fluoresce or glow in the green colour.
3. They travel in straight lines.
4. They are deflated by magnetic and electric fields.
5. They can ionize a gas.
6. They will turn a light paddle wheel in tube to move.
7. They produce intense heat.
8. They affect photographic plates.
9. They produce x-rays
10. They have a high penetrating power.
THERMONIC EMISSION:
This is the emission of electrons from the surface of a hot metal.
The cathode ray oscilloscope is used for studying all types of wave forms especially the a.c waveform and to measure frequency, amplification of voltages of electronic devices.
EVALUATION: The teacher evaluates the lesson with the following questions:
State Faraday's Laws of electrolysis.
ASSIGNMENT:
A piece of metal has an area of 20cm2 and it takes one hour to deposit a layer of silver. What current is needed to do this? (Assume that 1cm2 of silver has a mass of 10.5g and that 1 coulomb of electricity deposits 0.0012g of silver.
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Explain the conditions under which liquids and gases conduct electricity.
ii. Explain the behaviour of charge carriers in liquids and gases.
iii. State Faraday's laws and calculate the charge transferred in the process.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: CONDUCTION OF ELECTRICITY THROUGH LIQUIDS
Electrolysis is the chemical change in a liquid due to the flow of electric current through the liquid.
Voltametre is used for studying the flow of current through a liquid.
IONIC THEORY
An electrolyte is a liquid/molten substance that conducts electric current.
In an electrolyte, there are positively and negatively charged particles called ions. The molecules that constitute the electrolyte are split in solution into these ions.
EXAMPLES OF ELECTROLYSIS
1. Electrolysis of acidulated water
2. Electrolysis of CUSO4
APPLICATIONS OF ELECTROLYSIS
1. Electroplating: this is the process of coating a metallic object with another metal in order to improve its appearance or to safe guard it against corrosion.
2. Calibration of an Ammetre: the copper is weighed before the current is switched on. Let mass of copper be M and the e.c.e of copper be Z(i.e. electrochemical)
The electrochemical equivalent (e.c.e) of a substance is the mass of that substance deposited by one coulomb of electrolysis (unit = g/coulomb)
3. Extraction of metals such as aluminum and sodium.
4. Manufacture of electrolytic capacitors used in electronic appliances.
5. Manufacture of some pure metals e.g. pure cooper which is used in the manufacture of cables.
EVALUATION:
Define electrolysis and electrolytes
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. State Faraday's Law of electrolysis
ii. Solve simple problems on electrolysis and
iii. Describe conduction of electricity through gases.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: FARADAY'S LAWS OF ELECTROLYSIS
1st law: states that the mass m of a substance librated in electrolysis is directly proportional to the quantity of electricity Q, which has passed through the electrolyte.
2nd Law: states that the masses of different substances deposited by the same quantity of electricity are directly proportional to the chemical equivalent of the substance.
M α Q - 1
M α It
And m = Zit
Where Z is electrochemical equivalent.
CONDUCTIONS OF ELECTRICITY THROUGH GAS
The conduction of electricity through gases is studied using discharge tubes. Gases conduct electricity are:
1. Low pressure (0.01mmHg)
2. High voltage (1000V); they break into ions
NATURE AND PROPERTIES OF CATHODE RAYS
1. They consist of streams of fast moving particles of negative electricity (ELECTRON)
2. They cause glass and other materials to fluoresce or glow in the green colour.
3. They travel in straight lines.
4. They are deflated by magnetic and electric fields.
5. They can ionize a gas.
6. They will turn a light paddle wheel in tube to move.
7. They produce intense heat.
8. They affect photographic plates.
9. They produce x-rays
10. They have a high penetrating power.
THERMONIC EMISSION:
This is the emission of electrons from the surface of a hot metal.
The cathode ray oscilloscope is used for studying all types of wave forms especially the a.c waveform and to measure frequency, amplification of voltages of electronic devices.
EVALUATION: The teacher evaluates the lesson with the following questions:
State Faraday's Laws of electrolysis.
ASSIGNMENT:
A piece of metal has an area of 20cm2 and it takes one hour to deposit a layer of silver. What current is needed to do this? (Assume that 1cm2 of silver has a mass of 10.5g and that 1 coulomb of electricity deposits 0.0012g of silver.
WEEK 8
TOPIC: MAGNETIC FIELD AND MAGNETISM
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Plot magnetic field around magnets and magnetic substances.
ii. Describe the working principle of electric bell and the telephone earpiece.
iii. State the relation between magnetic force and the motion of a charge in a magnetic field.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: MAGNETIC FIELD
The area around a magnet in which it can attract or repel objects or in which a magnetic force can be detected is called magnetic field of the magnet.
Magnet lines of force:
These are many lines along which a free- North pole would tend to move if placed in the field. A line of force may also be considered as a line such that the tangent to it at any point gives the direction of the field at that point.
PATTERNS OF MAGNETIC FIELD
Magnetic field around a straight conductor carrying current.
The direction of the field depends on the direction of flow of current. Such a direction can always be obtained by applying the R-H Grip or CLENCHED FIST RULE. this states that if the straight wire is grasped with the right hand so that the thumb points in the direction of the current, then the direction in which the fingers are curled indicates the direction of the magnetic field.
MAGNETIC CORKSCREW RULE B
This rule states that if a right handed corkscrew is turned so that its tip travels along the direction of current, the direction of rotation of the corkscrew gives the direction of the magnetic field or magnetic line of force.
MAGNETIC FIELD AROUND A SOLEMOID
A solemoid is a long cyclindrical coil of wire whose turns are usually wound close together.
METHODS OF MAKING MAGNETS
1. Electrical method: The magnetic effect of current on magnets . polarity of magnet depends on direction of flow of current in solemoid. If current flows anti clockwise in coil when one side is viewed, then the specimen at that end has a N pole.
2. By constant method:
i. Single touch
ii. Double touch
At the end of each repeated stroking, each end of the specimen will be found to have opposite polarity to that of the stroking pole.
3. Demagnetization:
i. Slant method (a.c current)
ii. Mechanical method : pointing in E-w direction.
iii. Heating method
MAGNETIC PROPERTIES OF IRON AND STEEL
1. Iron is easily magnetized than steel but also loses its magnetism more quickly than steel.
2. Iron produces a stronger magnet than steel, thus steel
i. Steel is used for making p.magnets
ii. Iron is used for making + magnets
iii. Iron nails are often used for lab experiments on magnetization and demagnetization.
ELECTROMAGNETS AND ITS APPLICATIONS OF ELECTRIC FIELD
They are soft iron core in a current carrying solemoid.
Electromagnets are used for:
i. Producing intense magnetic fields(e.g. electric motors)
ii. For lifting and transporting heavy pieces of iron and steel.
iii. Used in separating iron from mix of non magnetic substances.
iv. In construction of electromagnetic devices.
THE ELECTRIC BELL
The instrument works on a "make and break" device. The making of the contact between spring and screw at c causes the clapper to move and strike gong and consequently breaking of the contact causes the clapper to leave the gong.
TELEPHONE EAR-PIECE
The variation of electric current due to changes in the speech energy causes a variation in the magnetism of the electromagnet. Vibration of the diaphragm creates sound waves of the same frequency in a speech current.
THE EARTH'S MAGNETIC FIELD
The earth acts like a magnet. The N-geographical pole has south magnetic polarity since it attracts North pole of a magnet, the south geographical pole has North magnetic polarity.
Angle of Declination/variation:
This is the angle between North and geographical north direction at the place concerned.
Angle of Dip/inclination:
Is the angle between the direction of the earth's resultant magnetic field at the horizontal.
The angle of dip at points - component of Earth's magnetic field
Is given by tanθ = V
H
Patterns of Bar magnet in Earth's field
At the neutral point, the magnet force due to the magnetic field is equal and opposite to the force due to the magnetic force due to the field of the bar magnet.
Magnetic force on a charge moving in a magnetic field
Magnetic field exerts a force on a charge moving in the field. If electrons want to move with average velocity v, and the wires lie in the magnetic field b, with force F, on each electron, then
F = Bevsinθ
But moving q
F = qvBsinθ, becomes
F = vvB i.e. v cross B
If angle between B and v is 90∘, then sinθ = 1
EVALUATION:
Distinguish between angle of dip and declination.
ASSIGNMENT:
A beam of ions passes undeflated through two parallel horizontal plates 2.5m apart having a p.d of 2500v when the perpendicular magnetic field is 2 x 10-2 T, and in the magnetic field along radius 3m. calculate the charge per mass ratio of the ions.
SUB-TOPIC: CALCULATIONS ON ELECTRIC FIELD AND E.M.F
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Solve simple problems on force on a material in a magnetic field.
ii. Show a relationship between electric and magnetic field, charge and motion.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: CLASS WORK
1. A charge of 1.6 x 10-1C enters a magnetic field of 10T with and of 2.5 x 102m/s. calculate the angle between the magnetic field and the direction of the motion at a force of 10 x 10-12N.
2. Obtain the magnetic force on an electron of charge q = 1.6 x 10-19C in a field of flux 20 tesla, with a speed of 2 x 107m/s in the direction (a) 90∘ (b) 60∘ (c) parallel to the magnetic field.
ELECTROMAGNETIC FIELD
In a field representing the joint interaction of electric and magnetic forces. It exerts a force on a charged particle; the force in a charge " moving with velocity " is given by
F= q (E + V + B)
Where f = force that occurs in both fields
q= charge, E = Electric field
B= magnetic field in tesla, or weber/m2
It can occur in an electric motor moving coil galvanometer, electromagnetic induction, generator, transformers etc.
EVALUATION:
Define electromagnetic field and give the relationship between force, charge, electric and magnetic field and motion.
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Plot magnetic field around magnets and magnetic substances.
ii. Describe the working principle of electric bell and the telephone earpiece.
iii. State the relation between magnetic force and the motion of a charge in a magnetic field.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: MAGNETIC FIELD
The area around a magnet in which it can attract or repel objects or in which a magnetic force can be detected is called magnetic field of the magnet.
Magnet lines of force:
These are many lines along which a free- North pole would tend to move if placed in the field. A line of force may also be considered as a line such that the tangent to it at any point gives the direction of the field at that point.
PATTERNS OF MAGNETIC FIELD
Magnetic field around a straight conductor carrying current.
The direction of the field depends on the direction of flow of current. Such a direction can always be obtained by applying the R-H Grip or CLENCHED FIST RULE. this states that if the straight wire is grasped with the right hand so that the thumb points in the direction of the current, then the direction in which the fingers are curled indicates the direction of the magnetic field.
MAGNETIC CORKSCREW RULE B
This rule states that if a right handed corkscrew is turned so that its tip travels along the direction of current, the direction of rotation of the corkscrew gives the direction of the magnetic field or magnetic line of force.
MAGNETIC FIELD AROUND A SOLEMOID
A solemoid is a long cyclindrical coil of wire whose turns are usually wound close together.
METHODS OF MAKING MAGNETS
1. Electrical method: The magnetic effect of current on magnets . polarity of magnet depends on direction of flow of current in solemoid. If current flows anti clockwise in coil when one side is viewed, then the specimen at that end has a N pole.
2. By constant method:
i. Single touch
ii. Double touch
At the end of each repeated stroking, each end of the specimen will be found to have opposite polarity to that of the stroking pole.
3. Demagnetization:
i. Slant method (a.c current)
ii. Mechanical method : pointing in E-w direction.
iii. Heating method
MAGNETIC PROPERTIES OF IRON AND STEEL
1. Iron is easily magnetized than steel but also loses its magnetism more quickly than steel.
2. Iron produces a stronger magnet than steel, thus steel
i. Steel is used for making p.magnets
ii. Iron is used for making + magnets
iii. Iron nails are often used for lab experiments on magnetization and demagnetization.
ELECTROMAGNETS AND ITS APPLICATIONS OF ELECTRIC FIELD
They are soft iron core in a current carrying solemoid.
Electromagnets are used for:
i. Producing intense magnetic fields(e.g. electric motors)
ii. For lifting and transporting heavy pieces of iron and steel.
iii. Used in separating iron from mix of non magnetic substances.
iv. In construction of electromagnetic devices.
THE ELECTRIC BELL
The instrument works on a "make and break" device. The making of the contact between spring and screw at c causes the clapper to move and strike gong and consequently breaking of the contact causes the clapper to leave the gong.
TELEPHONE EAR-PIECE
The variation of electric current due to changes in the speech energy causes a variation in the magnetism of the electromagnet. Vibration of the diaphragm creates sound waves of the same frequency in a speech current.
THE EARTH'S MAGNETIC FIELD
The earth acts like a magnet. The N-geographical pole has south magnetic polarity since it attracts North pole of a magnet, the south geographical pole has North magnetic polarity.
Angle of Declination/variation:
This is the angle between North and geographical north direction at the place concerned.
Angle of Dip/inclination:
Is the angle between the direction of the earth's resultant magnetic field at the horizontal.
The angle of dip at points - component of Earth's magnetic field
Is given by tanθ = V
H
Patterns of Bar magnet in Earth's field
At the neutral point, the magnet force due to the magnetic field is equal and opposite to the force due to the magnetic force due to the field of the bar magnet.
Magnetic force on a charge moving in a magnetic field
Magnetic field exerts a force on a charge moving in the field. If electrons want to move with average velocity v, and the wires lie in the magnetic field b, with force F, on each electron, then
F = Bevsinθ
But moving q
F = qvBsinθ, becomes
F = vvB i.e. v cross B
If angle between B and v is 90∘, then sinθ = 1
EVALUATION:
Distinguish between angle of dip and declination.
ASSIGNMENT:
A beam of ions passes undeflated through two parallel horizontal plates 2.5m apart having a p.d of 2500v when the perpendicular magnetic field is 2 x 10-2 T, and in the magnetic field along radius 3m. calculate the charge per mass ratio of the ions.
SUB-TOPIC: CALCULATIONS ON ELECTRIC FIELD AND E.M.F
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Solve simple problems on force on a material in a magnetic field.
ii. Show a relationship between electric and magnetic field, charge and motion.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: CLASS WORK
1. A charge of 1.6 x 10-1C enters a magnetic field of 10T with and of 2.5 x 102m/s. calculate the angle between the magnetic field and the direction of the motion at a force of 10 x 10-12N.
2. Obtain the magnetic force on an electron of charge q = 1.6 x 10-19C in a field of flux 20 tesla, with a speed of 2 x 107m/s in the direction (a) 90∘ (b) 60∘ (c) parallel to the magnetic field.
ELECTROMAGNETIC FIELD
In a field representing the joint interaction of electric and magnetic forces. It exerts a force on a charged particle; the force in a charge " moving with velocity " is given by
F= q (E + V + B)
Where f = force that occurs in both fields
q= charge, E = Electric field
B= magnetic field in tesla, or weber/m2
It can occur in an electric motor moving coil galvanometer, electromagnetic induction, generator, transformers etc.
EVALUATION:
Define electromagnetic field and give the relationship between force, charge, electric and magnetic field and motion.
WEEK 9
TOPIC: TYPES OF RADIATIONS
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Distinguish between electromagnetic waves and mechanical waves.
ii. Compare and contrast in terms of frequency, wavelength, six radiations in electromagnetic spectrum.
iii. State some uses of electromagnetic waves.
iv. Apply the formula v= f⋋ to solve simple problems relating to electromagnetic waves.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: ELECTROMAGNETIC WAVES
Mechanical waves and electromagnetic waves.
Electromagnetic waves are those waves that do not require a material medium for their propagation. They arise from the vibration of electric and magnetic field. The combination of electric and magnetic field waves is called ELECTROMAGNETIC WAVES.
Mechanical waves are waves that require a material medium for their propagation E.m travel at a sped less than that of light.
Mechanical waves may be transverse or longitudinal, but e.m is always transverse.
Equation:
V = f⋋ or
C= f⋋
e.g. estimate the wavelength of a radio station (Ray power) whose frequency is transmission is 100.5MHZ. [Take c = 3 x 108m/s]
EVALUATION:
Distinguish between mechanical and electromagnetic waves.
ASSIGNMENT:
List out 5 uses of electromagnetic waves.
SUB-TOPIC: USES OF E.M WAVE
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. List out some types of radiation
ii. Outline the various uses of electromagnetic waves.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: USES OF ELECTROMAGNETIC WAVES
Uses of e.m waves are numerous since there are different kinds. Hence the use of any one type is paramount important for the uses of electromagnetic waves.
1. Gamma waves: they are waves of ⋋10-8cm. with very great penetration. About x 10 of x-rays. They are used in medical field to sterilize supplies. They are also used to treat cancer and tumours.
2. Radio and TV waves: these are e.m waves with ⋋ - 1mm a few km. they are used in radio and television transmissions
- Radar detection around 1cm(⋋)
3. X-rays: (⋋ between 10-10 - 10-7cm). since they are blocked/stopped by some solids such as isomers and metals, they are equally used in hospitals to make pictures of bones and radiographs of internal body organs.
- Used in detecting flaws in metal industry.
- Used to study couptal structures.
- In air ports used as scanners to detect
EVALUATION:
List 5 uses of e-m waves.
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Distinguish between electromagnetic waves and mechanical waves.
ii. Compare and contrast in terms of frequency, wavelength, six radiations in electromagnetic spectrum.
iii. State some uses of electromagnetic waves.
iv. Apply the formula v= f⋋ to solve simple problems relating to electromagnetic waves.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: ELECTROMAGNETIC WAVES
Mechanical waves and electromagnetic waves.
Electromagnetic waves are those waves that do not require a material medium for their propagation. They arise from the vibration of electric and magnetic field. The combination of electric and magnetic field waves is called ELECTROMAGNETIC WAVES.
Mechanical waves are waves that require a material medium for their propagation E.m travel at a sped less than that of light.
Mechanical waves may be transverse or longitudinal, but e.m is always transverse.
Equation:
V = f⋋ or
C= f⋋
e.g. estimate the wavelength of a radio station (Ray power) whose frequency is transmission is 100.5MHZ. [Take c = 3 x 108m/s]
EVALUATION:
Distinguish between mechanical and electromagnetic waves.
ASSIGNMENT:
List out 5 uses of electromagnetic waves.
SUB-TOPIC: USES OF E.M WAVE
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. List out some types of radiation
ii. Outline the various uses of electromagnetic waves.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: USES OF ELECTROMAGNETIC WAVES
Uses of e.m waves are numerous since there are different kinds. Hence the use of any one type is paramount important for the uses of electromagnetic waves.
1. Gamma waves: they are waves of ⋋10-8cm. with very great penetration. About x 10 of x-rays. They are used in medical field to sterilize supplies. They are also used to treat cancer and tumours.
2. Radio and TV waves: these are e.m waves with ⋋ - 1mm a few km. they are used in radio and television transmissions
- Radar detection around 1cm(⋋)
3. X-rays: (⋋ between 10-10 - 10-7cm). since they are blocked/stopped by some solids such as isomers and metals, they are equally used in hospitals to make pictures of bones and radiographs of internal body organs.
- Used in detecting flaws in metal industry.
- Used to study couptal structures.
- In air ports used as scanners to detect
EVALUATION:
List 5 uses of e-m waves.
