Physics - Electricity & Magnetism Compendium

COMPENDIUM OF LECTURE NOTES S-PHY 14/S-SCI 15 (Electricity and Magnetism) Prepared By: CHRISTIAN DAVE A. DE LEON IV-19 BSE General Science

TABLE OF CONTENTS PART 1: The Electric Field 1.1 The Electric Charge 1.2 Coulomb’s Law 1.3 Electric Field and its Calculations 1.4 Lines of Force PART 2: Gauss’ Law 2.1 Electric Flux 2.2 Gauss’ Law 2.3 Electric Conductors (Sample Problems) PART 3: Electric Potential 3.1 Potential Difference 3.2 Potential of a system of charges 3.3 Equipotential Surfaces PART 4: Capacitance and Dielectrics 4.1 Capacitor 4.2 Capacitors in Series and Parallel 4.3 Energy of a charged capacitor 4.4 Dielectrics PART 5: Electric Current 5.1 Current and Motion of Charges 5.2 Ohm’s Law and Resistance 5.3 Electromotive Force: Energy in Electric Circuits PART 6: Direct-Current Circuits 6.1 Resistors in Series and Parallel 6.2 Kirchhoff’s Rules 6.3 Electric Meters

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PART 7: Magnetism and Magnetic Fields 7.1 Magnetism 7.2 Magnetic Field 7.3 Magnetic Flux 7.4 Motion of Charged Particles in a Magnetic Field 7.5 Magnetic Forces on Conductors PART 8: Electromagnetic Induction 8.1 Faraday’s Law and Lenz’s Law of Induction 8.2 Electromagnetic Devices 8.3 Transformers and Transmission of Power

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PART

1

a. Positively-charged protons and neutrally-charged neutrons comprise the nucleus b. Negatively-charged electrons surrounds the nucleus

Electric Field Electricity and Magnetism

1.1 The Electric Charge Electric = “electron” → amber (stone that could attract other objects) Electrostatics = the interactions between electric charges that are at rest. → “Like charges repels; unlike charges attract a. Plastic rods and for – two negative charges repel b. Glass rods and silk – two positive charges repel - The glass is positively charged because of electron deficiency - The electrons move from the glass to the silk c. Fur-rubbed rod and silk-rubbed glass – opposite charges attract

 Electric interactions are chiefly responsible for the structure of atoms, molecule, and solids.

Principle of Conservation of Charge: a. The algebraic sum of all the electric charges in a closed system is constant b. The magnitude of charge of the electron or proton is a natural unit of charge  Conductors are materials in which charge moves easily; in Insulators, charge doesn’t move easily.  Charging by conduction is when both objects acquire the same type of charge. If a negative object is used to charge a neutral object, then both objects become charged negatively.  Charging by induction is when a charged object brought near but not touched to a neutral conducting object. The presence of a charged object near a neutral conductor will induce electrons within the conductor to move. 1.2 Coulomb’s Law The magnitude of the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

𝑭=𝒌

|𝒒𝟏 𝒒𝟐 | 𝒓𝟐

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F = electric force 𝑞1 𝑎𝑛𝑑 𝑞2 = electric charges r = distance between charges k = Coulomb’s law constant

Problem 1.3 The charges are arranged as shown. Find the resultant force on q3 due to the others.

𝑘 = 8.988 𝑥 109 𝑁 • 𝑚2 /𝐶 2 1 𝑘= = 9.0𝑥109 𝑁 • 𝑚2 /𝐶 2 ; 4𝜋𝜀0 𝜀0 = 8.854 𝑥 10−12 𝐶 2 /𝑁 • 𝑚2 1 |𝒒𝟏 𝒒𝟐 | 𝑭= 4𝜋𝜀0 𝒓𝟐 Problem 1.1: Determine the magnitude and direction of the force on the electron of a hydrogen atom exerted by a single proton that is the atom’s nucleus. Assume that the average distance between them is 0.53 𝑥 10−10 𝑚. Given that the 𝑒 − and 𝑝+ has the same magnitude of charge 𝑞1 = 𝑞2 = 1.6 𝑥 10−19 𝐶. |𝑞1 𝑞2 | 𝑟2 (9.0 𝑥 109 𝑁 • 𝑚2 /𝐶 2 )|(1.6 𝑥 109 𝐶)(1.6 𝑥 109 𝐶)| = 0.53 𝑥 10−10 𝑚2 𝐹=𝑘

|𝒒𝟏 𝒒𝟑 | 𝒓𝟏 𝟐 6 (9𝑥10 )(8𝑥10−6 )(4𝑥10−6 ) = (0.05𝑚)2 = 115𝑁; 53.1°𝑆𝑊 |𝒒𝟐 𝒒𝟑 | 𝑭𝟐 = 𝒌 𝒓𝟐 𝟐 6 (9𝑥10 )(6𝑥10−6 )(4𝑥10−6 ) = (0.03𝑚)2 = 240𝑁; 𝑊𝑒𝑠𝑡 𝑭𝟏 = 𝒌

𝑭 = 𝟖. 𝟐 𝒙 𝟏𝟎−𝟖 𝑵

Principle of Superposition  the vector sum of the forces on the charges is the net force of all the charges present. Problem 1.2: A -6mC charge is placed 4cm from a +9mC charge. What is the resultant force on a -5mC charge located on the midway between the first 2 charges? (1𝑚𝐶 = 1 𝑥 10−6 𝐶)

𝐹1 = 𝑘

|𝑞1 𝑞3 | 𝑟1

2

=

𝐹1 = 675 N

𝐹2 = 𝑘

|𝑞2 𝑞3 | 𝑟2 2

𝐹2 = 1013 N

=

F1x = -(115 N) Cos 53.1o = - 69.2 N F1y = -(115 N) Sin 53.1o = - 92.1 N F2x = -240 N; F2y = 0 Rx = SFx ; Ry = SFy Rx = – 69.2 N – 240 N = -309 N Rx= -92.1 N Ry = -69.2 N – 0 = -69.2 N Ry= -240 N

(9 𝑥 106 )(6 𝑥 10−6 )(5 𝑥 10−6 ) (0.02𝑚)2

;

(9 𝑥 106 )(6 𝑥 10−6 )(5 𝑥 10−6 ) (0.02)𝑚2

;

The reference angle is: f = 77.40S of W Or, the polar angle q is: q = 1800 + 77.40 = 257.40 Resultant Force: R = 317 N, q = 257.40

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1.3 Electric Field and its Calculations  The electric force on a charged body is exerted by the electric field created by other field charges  Electric Field is the electric force per unit area (N/C) 𝑭𝒐 𝑬= 𝒒𝒐  The electric field produced at a point charge is directed radially away from the charge (if positive) or towards the charge (if negative)

𝐹𝑜 = 𝐸=

1 |𝑄𝑞| 4𝜋𝜖𝑜 𝑟 2 1 |𝑞| 4𝜋𝜖𝑜 𝑟 2

𝑬=𝒌

|𝒒| 𝒓𝟐

Problem 1.4 What is the magnitude of the electric field at a field point 2.0m from a point charge q=4.0 nC? (1nC=1x10-9 C) |𝒒| 𝑬=𝒌 𝟐 𝒓 (9.0 𝑥 109 𝑁 • 𝑚2 /𝐶2 )|(4.0 𝑥 10−9 𝐶)| = 2

1.4 Lines of Force  ELECTRIC FIELD LINE – is an imaginary line or curve drawn through a region of space so that its tangent at any point is in the direction of the electric-field vector at that point  The number of lines per unit area (perpendicular to that direction) is proportional to the magnitude of E at that point a. Lines of Force Start from positively charged particle and end on the negatively charged particle b. Lines of force repel each other sideways. They do not intersect nor break as they pass from one charge to another c. The greater the lines of force, the stronger is the electric field d. The direction of the electric field at a point is the same as the direction of the tangent to the line of force at that point

(2.0𝑚)

= 𝟗. 𝟎𝑵/𝑪

SUPERPOSITION OF ELECTRIC FIELDS  The total electric field at P is the vector sum of the fields at P due to each point charge in the charge distribution 𝑬 = 𝑬𝟏 + 𝑬𝟐 + 𝑬𝟑 + ⋯

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PART

2 Gauss’ Law Electricity and Magnetism

2.1 Electric Flux ELECTRIC FLUX – is the measure of the “flow” of electric field through a surface

𝜱𝑬 = 𝑬 ⊥ 𝑨 𝜱𝑬 = 𝑬𝑨𝒄𝒐𝒔𝜽 𝜱𝑬 = Electric Flux E = Perpendicular electric field A = Area  Positive charges cause outward flux, negative charges cause inward flux (closed surface, rectangular box) a. Whether there is a net outward or inward electric flux through a closed surface depends on the sign of the enclosed charge b. Charges outside the surface do not give a net electric flux through the surface c. The net electric flux is directly proportional to the net amount of charge enclosed within the surface, but it is otherwise independent to the size of the closed surfaces  These observations are qualitative statements of the Gauss Law

2.2 Gauss’ Law  States that the total electric flux summed over any closed surface is equal to the net charge (Q) enclosed by the surface divided by 𝜖𝑜 𝜮𝑬 ⊥ ∆𝑨 =

𝑸𝒆𝒏𝒄𝒍 𝝐𝒐

APPLICATIONS OF GAUSS LAW 1. Spherical symmetry a. Point charge on any spherical distribution with total charge Q, the field outside of the charge will be:

b. Spherical conductor with uniform surface charge density 𝜎, the field outside of the charge will be:

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c. And the field inside will be zero, since tha Gaussian surface contains no charge:

d. Spherical insulator with uniform charge density 𝜌, the field outside the charge will be:

c. And the field inside will be zero since that Gaussian surface contains no charge

d. Cylindrical insulator with uniform charge density 𝜌, the field outside the charge will be:

e. And inside the field will be: e. And inside the field will be:

2. Cylindrical Symmetry a. Line with uniform charge density λ: 3. Planar Symmetry a. Non-conducting plane of infinitesimal thickness with uniform surface charge density 𝜎:

b. Cylindrical conductor with uniform charge density 𝜎, the field outside the charge will be: b. Conducting plane of finite thickness with uniform surface charge density 𝜎:

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2.3 Electric Conductors (Sample Problems)

Table 2.1 Charge Distribution and Points in Electric Field

Problem 2.1 A disk or radius 0.10m is oriented at 30 degrees to a uniform electric field of 2.0x103 N/C

a. What is the electric flux through the disk?

b. What is the electric flux through the disk that 𝜱𝑬 is perpendicular to E?

c. What is the flux through the disk that 𝜱𝑬 is parallel to E?

Table 2.2 Charge Distribution and Electric Field Magnitude

Problem 2.2 A point charge is surrounded by an imaginary sphere of radius r = 0.20m centered on the charge Find the resulting electric flux through the sphere.

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PART

3 Electric Potential Electricity and Magnetism

ELECTRIC POTENTIAL AND POTENTIAL DIFFERENCE 3.1 Potential Difference ELECTRIC POTENTIAL ENERGY  The electric force caused by any collection of charges at rest is a conservative force. 𝑊 = 𝐹𝑑 𝑾 = 𝒒𝒐 𝑬𝒅; 𝐹 = 𝑞𝑜 𝐸

POTENTIAL (V) – is the electric potential energy per unit charge. Measured in Volts (V) or Joules per Coulumb (J/C) 𝑼 𝑽= 𝒒𝒐 𝑽=

 The work done by the electric force on a charged particle moving in an electric field can be represented by the change in potential energy 𝑊𝑎→𝑏 = 𝑈𝑎 − 𝑈𝑏  The electric potential energy for two-point charges q and qo depends on their separation r 1 𝑞𝑞𝑜 𝑈= 4𝜋𝜖𝑜 𝑟  The electric potential energy for a charge qo in the presence of a collection of charges (e.g. q1, q2, q3,) depends on the distance from qo to each of these other charges 𝑈=

𝑞𝑜 𝑞1 𝑞2 𝑞3 ( + + + ⋯) 4𝜋𝜖𝑜 𝑟1 𝑟2 𝑞3 Or 𝑞𝑜 𝑞𝑖 𝑈= ∑ 4𝜋𝜖𝑜 𝑟𝑖

𝟏 𝒒 𝟒𝝅𝝐𝒐 𝒓

𝑽=𝒌

𝒒 𝒓

 The potential difference between two points equals the amount of work done that would be required to move a unit positive test between those points 𝑉𝑏𝑎 = 𝑉𝑏 − 𝑉𝑎 𝑉𝑏𝑎 =

𝑈𝑏 − 𝑈𝑎 𝑞𝑜

𝑽𝒃𝒂 =

−𝑾𝒃𝒂 𝒒𝒐

 The Potential V due to a quantity of charge can be calculated by summing up if the charges is a collection of point charges 1 𝑞𝑖 𝑉= ∑ 4𝜋𝜖𝑜 𝑟𝑖 𝑖

𝑖

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Problem 3.2: A Proton (+e= 1.6x10-19C) moves a distance d=0.50m in a straight line between points a and b in a linear acceleration. The electric field is uniform along this line, with magnitude E=1.5x107 V/m (1Vm=1N/c) in the direction from A to B. Determine the: a. Force on the Proton  The potential difference between two points a and b where a uniform electric field E exists is given by: 𝑾 = 𝒒𝒐 𝑬𝒅 𝑊 = −𝑞(𝑉𝑏 − 𝑉𝑎 ) −𝑞𝑉𝑏𝑎 = 𝑞𝑜 𝐸𝑑 −𝑉𝑏𝑎 = 𝐸𝑑 𝑬=

b. Work done on it by the field

−𝑽𝒃𝒂 𝒅 c. Potential difference

3.2

Potential of a system of charges

Problem 3.3: An electric dipole consists of 2point charges q1=+12nC and q2-12nC placed 10.0 cm apart. Compute the electric potentials at points a, b and c.

Problem 3.1: A charge of 6.0x10-7 Cis transferred from infinity to point A. If the work done to do this is 1.2x10-5J, what is the potential energy at point A? −𝑾𝒃𝒂 −1.2x10−5 J 𝑽𝒃𝒂 = = = −𝟐𝟎𝑽 𝒒𝒐 6.0 x10−7 𝐶 ELECTRON VOLT (eV) – defined as the energy acquired by a particle carrying a charge whose magnitude equals that on the electron (q = e) as a result of moving through a potential difference of 1V 1𝑒𝑉 = 𝑞𝑉 = (1.6𝑥10−19 𝐶)(1.0𝑉) 𝟏𝒆𝑽 = (𝟏. 𝟔𝒙𝟏𝟎−𝟗 𝑱)

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Problem 3.4: Two parallel plates are charged to produce a potential difference of 50V. If the separation between the plates is 0.050m, calculate the magnitude of the electric field in the space between the plates −𝑽𝒃𝒂 50𝑉 𝑬= = = 1000𝑉/𝑚 𝒅 0.050𝑚 3.3 Equipotential Surfaces  An equipotential surface is a surface on which the potential difference has the same value at every point a. At a point where a field line cause an equipotential surface, the two are perpendicular b. When all of the charges are at rest, the surface of a conductor is always in equipotential surface and all points in the interior of a conductor are at the same potential c. When a cavity within a conductor contains no charge, the entire cavity is an equipotential region and there is no surface charge anywhere on the surface of the cavity

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PART

4

𝑪 = 𝝐𝒐

𝑨 𝒅

Where 𝜖𝑜 = 8.85𝑥10−12 𝐹/𝑚

Capacitance and Dielectrics Electricity and Magnetism

4.1 Capacitors CAPACITOR – any pair of conductors separated by an insulating material (or a vacuum)  In circuit diagrams, a capacitor is represented by:

 When the capacitor is charged, there are charges of equal magnitude Q and opposite sign the potential Vab of the positively charged conduction with respect to the negatively charged conductor is proportional top Q CAPACITANCE – the ratio of Q to Vab 𝑸 𝑪= 𝑽𝒂𝒃  The SI unit of capacitance is the Farad (F) 1𝐹 = 1𝐶/𝑉 PARALLEL-PLATE CONDUCTOR  Consists of two-parallel conducting plates, each with area A, separated by distance d  If they are separated by a vacuum, the capacitance depends only on A and d. 𝑄 𝐶= 𝑉𝑎𝑏 𝑄 1 𝑄𝑑 𝐸= 𝑎𝑛𝑑 𝑉𝑎𝑏 = 𝜖𝑜 𝐴 𝜖𝑜 𝐴 𝐴

 The capacitance of a parallel-plate conductor is affected by the following factors: a. Area of the plates – the bigger the area of the plates, the greater the capacitance b. Distance between the plates – the closer the plates to one another, the greater is the capacitance c. Insulating material – the dielectric material between them Problem 4.1: The parallel plates of a 1.0-F capacitor are 1.0 mm apart. What is their area? 𝐶𝑑 (1.0𝐹)(1.0𝑥10−3 𝑚) 𝐴= = 𝜖𝑜 8.85𝑥10−12 𝐹/𝑚 = 𝟏. 𝟏𝒙𝟏𝟎𝟖 𝒎𝟐

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Problem 4.2 The plates of a parallel-plate capacitor in vacuum are 5.00mm apart and 2.0m2 in area. A 10.0-kV potential difference is applied across the capacitor. Compute for the: a. Capacitance

b. Charge on the Capacitor

c. Electric field magnitude

2. PARALLEL CONNECTION

 When the capacitors are connected in parallel, the equivalent capacitance Ceq equals the sum of the individual capacitances 𝐶𝑒𝑞 = 𝐶1 + 𝐶2 + 𝐶3 + ⋯  In parallel connection. The equivalent capacitance is always greater than any individual capacitance Problem 4.3: Given a series and parallel connection of 2 capacitors. Let C1=6.0µF, C2=3.0 µF, and Vab=18V. Find the equivalent capacitance and the charge for each capacitor when it is connected in: a. Series Connection

4.2 Capacitors in Series and Parallel 1. SERIES CONNECTION b. Parallel Connection

 When capacitors with capacitances C1, C2, C3 … are connected in series, the reciprocal of the equivalent capacitance Ceq equals the sum of the reciprocals of the individual capacitors 1 1 1 1 = + + +⋯ 𝐶𝑒𝑞 𝐶1 𝐶2 𝐶3  In series connection. The equivalent capacitance is always less than any individual capacitances

4.3 Energy of a charged capacitor  The Energy U required to charge a capacitor C to a potential difference V and a charge Q is equal to the energy stored in the capacitor 1 𝑈 = 𝑄𝑉 2 1 2 𝑈 = 𝐶𝑉 ; 𝐶 = 𝑄𝑉 2 𝑼=

𝑸𝟐 𝟐𝑪

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 The energy density (u) is the electric energy stored per unit volume that is proportional to the square of the electric field magnitude 1 𝑈 = 𝐶𝑉 2 2 𝐴 𝐶 = 𝜖𝑜 𝑎𝑛𝑑 𝑉 = 𝐸𝑑 𝑑 1 𝐴 𝑈 = (𝜖𝑜 ) (𝐸 2 𝑑2 ) 2 𝑑 1 2 𝑈 2 (𝜖𝑜 𝐸 𝐴𝑑) 𝑢= = 𝑉 𝐸𝑑 𝒖=

𝟏 (𝝐 𝑬𝟐 ) 𝟐 𝒐

Problem 4.4: A camera unit stores energy in a 150µF capacitor at 200V. How much electric energy can be stored? 1 1 𝑈 = 𝐶𝑉 2 = (150𝜇𝐹)(200𝑉)2 = 𝟑. 𝟎𝑱 2 2 4.4 Dielectrics DIELECTRIC – a non-conducting material in between the conducting plates of capacitors

Table 4.1 Values of Dielectric Constants MATERIAL K Vacuum 1 Air (1atm) 1.00059 Air (100atm) 1.0548 Teflon 2.1 Polyethylene 2.25 Benzene 2.28 Mica 3-6 Mylar 3.1 Polyvinyl Chloride 3.18 Glass 5-10 Rubber, Neoprene 6.70 Water 80.4 Glycerin 42.5 Strontium Titanate 310 Germanium 16  The quantity 𝜖 = 𝑘𝜖𝑜 is called the permittivity of the dielectric 𝐴 𝐶 = 𝑘𝜖𝑜 𝑑 𝑪=𝝐

𝑨 𝒅

DIELECTRIC STRENGTH – refers to the maximum electric field before breakdown (charge flow) occurs Table 4.2. Dielectric Strengths (V/m) MATERIAL V/m Air (1atm) 3 x106 Paraffin 10 x106 Polystyrene 24 x106 Vinyl (Plastic) 50 x106 Paper 15 x106 Quartz 8 x106 Oil 12 x106 Glass (Pyrex) 14 x106 Rubber, Neoprene 12 x106 Porcelain 5 x106 Mica 150 x106 Strontium Titanate 8 x106

DIELECTRIC CONSTANT (K) – property of dielectrics in the increase of capacitance 𝐶 𝐾= 𝐶𝑜 S-PHY 14/S-SCI 15: Electricity and Magnetism (De Leon, Christian Dave A. SciEd)

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PART

5 Electric Current Electricity and Magnetism

5.1 Current and Motion of Charges CURRENT – any motion of charge from one region to another; the amount of charge flowing through a specified area, per unit time ∆𝑄 𝐼= ∆𝑡  The SI unit for current is the Ampere (A) 1𝐴 = 1𝐶/𝑠 CURRENT DENSITY – the current per cross-sectional area 𝐼 𝐽= 𝐴  The units for current density is Amperes per Square meter (A/m2) Problem 5.1 The electric current in a wire is 6A. How many electrons flow past a given point within 3 seconds? ∆𝑄 = 𝐼∆𝑡 = (6𝐴)(3𝑠) = 18 𝐶 1𝑒 18𝐶 ( ) 1.6 𝑥 10−19 𝐶 = 𝟏. 𝟏𝟐𝟓 𝒙 𝟏𝟎𝟐𝟎 𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏𝒔 Problem 5.2 An 18-guage copper wire, with a diameter of 1.02 mm carries a constant electric current of 1.67 A. Find the: a. Cross-sectional area of the wire 𝜋𝑑2 (3.14)(1.02 𝑥 10−3 𝑚) 𝐴= = 4 4 = 𝟖. 𝟏𝟕 𝒙 𝟏𝟎−𝟕 𝒎𝟐

b. Current Density 𝐼 1.67 𝐴 𝐽= = = 2.04 𝑥 106 𝐴/𝑚2 𝐴 8.17 𝑥10−7 𝑚2  Current is usually described in terms of a flow of positive charge, even when the charges are actually negative or of both charges 5.2 Ohm’s Law and Resistance OHM’S LAW – states that the current I through a given conductor is directly proportional to the potential difference V between its end points 𝐼∝𝑉  Resistance is the opposition certain material offers to the flow of current 𝑉 𝑅= 𝐼  The SI unit for resistance is Ohms (𝛺) 1 𝛺 = 1 𝑉/𝐴  The resistance R of a wire is inversely proportional to its cross-sectional area A, and directly to its length l and to a property called resistivity (𝜌) 𝜌𝐿 𝑅= 𝐴  The resistivity of a material is the ratio of the magnitudes of electric field and current density 𝐸 𝜌= 𝐽

Table 5.1 Resistivities at 20oC Material 𝝆(𝜴 ∙ 𝒎) Silver 1.47x10-8 Copper 1.72 x10-8 Gold 2.44 x10-8 Aluminum 2.75 x10-8

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Tungsten Steel Lead Mercury Manganin Constantan Nichrome Graphite Silicon Material Amber Germanium Glass Lucite Mica Quartz Sulfur Teflon Wood

5.25 x10-8 20 x10-8 22 x10-8 95 x10-8 44 x10-8 49 x10-8 100 x10-8 3.65 x10-5 2300 𝝆(𝜴 ∙ 𝒎) 5 x1014 0.60 10 10 - 1014 > x1013 x1013 75 x1016 1015 > 1013 108 - 1011

Problem 5.3 An 18-gauge Cu wire has a cross-sectional area of 8.20x10-7 m2. It carries a current of 1.67A. Find the: a. Electric field magnitude in the wire ρcu= 1.72x10-8 Ω m 𝜌𝐼 (1.72𝑥10−8 )(1.67𝐴) 𝐸 = 𝜌𝐽 = = 𝐴 8.20𝑥10−7 = 𝟎. 𝟎𝟑𝟓𝟎 𝑽/𝒎 b. Potential difference between two points in the wire 50.0m apart 0.0350𝑉 𝑉 = 𝐸𝐿 = ( ) (50.0𝑚) = 𝟏. 𝟕𝟓 𝑽 𝑚 c. Resistance of a 50.0m length of this wire 𝜌𝐿 (1.72𝑥10−8) (50.0𝑚) 𝑅= = = 𝟏. 𝟎𝟓 𝜴 𝐴 8.20𝑥10−7 𝑅=

𝑉 1.75𝑉 = = 𝟏. 𝟎𝟓 𝜴 𝐼 1.67𝐴

 Resistivity usually increases with temperature; for small temperature changes with variations, with (𝛼) as the temperature coefficient of resistivity 𝜌(𝑇) = 𝜌𝑜 [1 + 𝛼(𝑇 − 𝑇𝑜 )]

 The following are the factors that affect resistance 1. Length – greater length; higher resistance 2. Cross-sectional area – greater area; less resistance 3. Type of material – resistivity of a material 4. Temperature – given by the following equation 𝑅(𝑇) = 𝑅𝑜 [1 + 𝛼(𝑇 − 𝑇𝑜 )] Table 5.2 α Temperature Coefficients of Resistivity Material 𝜶 [(𝑪𝒐 )−𝟏 ] Aluminum 0.0039 Brass 0.0020 Carbon -0.005 Constantan 1.00x10-5 Copper 3.93 x10-3 Iron 5.0 x10-3 Lead 4.3 x10-3 Manganin 0 Mercury 8.8 x10-4 Nichrome 4.0 x10-4 Silver 3.8 x10-3 Tunsgten 4.5 x10-3 Problem 5.4 Suppose the resistance of a copper wire is 1.05Ω at 20oC. Find the resistance at: a. 0oC 𝑅(𝑇) = 𝑅𝑜 [1 + 𝛼(𝑇 − 𝑇𝑜 )] 𝑅(𝑇) = 1.05[1 + 0.00393(0 − 20)] = 𝟎. 𝟗𝟕𝜴 b. 100oC 𝑅(𝑇) = 𝑅𝑜 [1 + 𝛼(𝑇 − 𝑇𝑜 )] 𝑅(𝑇) = 1.05[1 + 0.00393(100 − 20)] = 𝟏. 𝟑𝟖𝜴

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 A Resistor is a circuit device that is made to have a specific value of resistance between its ends. Usually, a resistor has 3 or 4 colored bands: a. First 2 bands – first 2 digits of the resistance b. Third band – multiplier to power of ten c. Fourth band – represents the tolerance (if present) 1. Silver band – +/- 10% 2. Gold band – +/- 5% 3. No band – +/- 20%

5.3 Electromotive Force: Energy in Electric Circuits ELECTROMOTIVE FORCE (emf) – is the influence that makes the current flow from lower potential to higher potential  It is the voltage across terminals of a cell when no current flows  It is symbolized by 𝜀 and it is measured in Volts (V)  A source of emf is a device that converts another type of energy into electrical energy a. Electric Cells – chemical to electrical b. Generators – mechanical to electrical c. Photovoltaic Cells – light to electrical  The source of emf provides the voltage to force the electrons through electric resistance  A COMPLETE CIRCUIT has a continuous carrying path

Table 5.3 Color bands for Resistors Value as Color Multiplier Digit Black 0 1 Brown 1 101 Red 2 102 Orange 3 103 Yellow 4 104 Green 5 105 Blue 6 106 Violet 7 107 Gray 8 108 White 9 109 Problem 5.5 A resistor has these color bands; red, blue, brown, and gold. What is its: a. Resistance 1st band: Red = 2 2nd band: Blue = 6 3rd band: Brown = 101 4th band: Gold = +/- 5%  26 x 101 = 260 Ω b. Tolerance 5% of 260 = +/- 13 Ω

Table 5.4 Symbols for Circuit Diagrams Symbol Device Conducting path Resistor Source of emf Source of Emf with internal resistance Voltmeter Ammeter TERMINAL VOLTAGE (Vab) – is the potential difference across the terminals of the cells when there is current flowing through it  Because a cell has internal resistance (r), terminal voltage is less than electromotive force 𝑽𝒂𝒃 = 𝜺 − 𝑰𝒓

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 The current in the external circuit connected to the source terminals a and ab, and is still determined by Vab = IR; then: 𝑉𝑎𝑏 = 𝜀 − 𝐼𝑟 𝐼𝑅 = 𝜀 − 𝐼𝑟 𝑰=

𝜺 𝑹+𝒓

Problem 5.6 A 65.0-Ω resistor is connected to the terminals of a battery whose emf is 12.0V and whose internal resistance r is 0.5Ω. Calculate the: a. Current in the circuit 𝑉𝑎𝑏 = 𝜀 − 𝐼𝑟 𝐼𝑅 = 𝜀 − 𝐼𝑟 𝑰=

𝜺 12.0 𝑉 = = 𝟎. 𝟏𝟖𝟑𝑨 𝑹 + 𝒓 65.0𝛺 + 0.5𝛺

b. Terminal voltage in the battery 𝑉𝑎𝑏 = 𝜀 − 𝐼𝑟 𝑉𝑎𝑏 = (12.0𝑉) − (0.183𝐴)(0.5𝛺) 𝑽𝒂𝒃 = 𝟏𝟏. 𝟗 𝑽

ELECTRIC POWER  The rate at which energy is transformed in a resistance R from electric to other forms of energy is equal to the product of current and voltage 𝑄𝑉 𝑃= 𝑡 𝑷 = 𝑰𝑽; 𝐼 = 𝑉/𝑡

c. Power in the Resistor R 𝑷 = 𝑰𝟐 𝑹 2 (65.0𝛺) = (0.183𝐴) = 𝟐. 𝟏𝟖 𝑾 d. Power in the battery’s internal resistance 𝑷 = 𝑰𝟐 𝑹 = (0.183𝐴)2 (0.5𝛺) = 𝟎. 𝟎𝟐 𝑾

 Power in resistors can be written as 𝑃 = 𝐼𝑉 = 𝐼(𝐼𝑅) 𝑷 = 𝑰𝟐 𝑹 𝑉 𝑃 = 𝐼𝑉 = ( ) 𝑉 𝑅 𝐕𝟐 𝐏= 𝐑  The SI unit for power is Watts (1 W = 1 J/s)

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PART

6 Direct-Current Circuits Electricity and Magnetism

 The current is divided between the resistors a. The reciprocal of the total resistance is the sum of the reciprocals of the individual resistances 1 1 1 𝑅𝑇 = + + +⋯ 𝑅1 𝑅 2 𝑅 3 b. The main current entering the combination of resistors divides and port of it goes through each resistor 𝐼𝑇 = 𝐼1 + 𝐼2 + 𝐼3 = ⋯ c. Voltages across the resistors are equal 𝑉𝑇 = 𝑉1 = 𝑉2 = 𝑉3 = ⋯

INTRODUCTION: Types of Circuits based on direction of currents 1. Direct-Current (DC) – the direction of the current does not change with time Ex: Flashlights and automobile wiring systems 2. Alternating-Circuits (AC) – the current oscillates back and forth Ex: Household electrical power

Problem 6.1 The Equivalent resistance of the network

6.1 Resistors in Series and Parallel 1. Series Circuit  Contains more than one piece of electrical apparatus (resistors) connected one after another in a single line  The current flows in a single path a. Total Resistance is the sum of the individual resistances 𝑅𝑇 = 𝑅1 + 𝑅2 + 𝑅3 + ⋯ b. The Current is the same in all resistors 𝐼𝑇 = 𝐼1 = 𝐼2 = 𝐼3 = ⋯ c. The total voltage across the combination is the sum of the voltages across each resistor 𝑉𝑇 = 𝑉1 + 𝑉2 + 𝑉3 + ⋯ 2. Parallel Circuit  Two or more pieces of electrical apparatus (resistors) are connected side by side

Problem 6.2 2 identical light bulbs (R=2Ω) are connected to a source of (𝜀 = 8𝑉). Find the current through each bulb, potential difference across each bulb, and the power delivered to each bulb and to the entire network if the bulbs are connected to: a. Series connection

𝑅𝑒𝑞 = 𝑅1 + 𝑅2 = (2Ω + 2Ω) = 𝟒𝛀

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𝑉𝑎𝑐 8 𝑉 𝐼= = = 𝟐𝑨 𝑅𝑒𝑞 4 Ω

 Based on the conservation of charges 𝜮𝑰 = 𝟎

𝑉𝑎𝑏 = 𝑉𝑏𝑐 𝑉 = 𝐼𝑅 = (2𝐴)(2Ω) = 𝟒 𝑽 𝑃 = 𝐼 2 𝑅 = (2𝐴)2 (2Ω) = 𝟖 𝐖 b. Parallel connection

𝑅𝑒𝑞 =

2. Kirchhoff’s Loop Rule – the algebraic sum of the potential differences in any loop must be equal to zero Based on the conservation of energy and the conservative nature of electrostatic fields 𝜮𝑽 = 𝟎

1 1 2 4Ω + = = = 𝟐𝛀 2Ω 2Ω 4Ω 2 𝐼=

𝑉𝑑𝑒 8 𝑉 = = 𝟒𝑨 𝑅𝑒𝑞 2Ω

𝑉𝑑𝑒 = 𝜀 = 𝟖𝑽 𝑃 = 𝐼 2 𝑅 = (4𝐴)2 (2Ω) = 𝟑𝟐 𝐖 6.2. Kirchhoff’s Rules  JUNCTION – the point where 3 or more conductors meet  LOOP – any closed conducting path

Problem 6.3 Find the current in each resistor and the equivalent resistance of the five resistors

Kirchhoff’s Loop Rule

1. Kirchhoff’s Junction Rule – the algebraic sum of the currents into any junction is zero S-PHY 14/S-SCI 15: Electricity and Magnetism (De Leon, Christian Dave A. SciEd)

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I2 = I1+I3 13 V = I1 (2 Ω) – I3 (2 Ω) 13 V = I1 (3 Ω) + I3 (5Ω)

VOLTMETER – a device that measures the emf or potential difference between two points in a current; a galvanometer with high resistance in series

78 V = I1 (13 Ω); I1 = 6A 13 V = 6A (2 Ω) – I3 (1 Ω); I3 = -2A I2 = I1 + I3; I2 = 5A I1 + I2 = 11A 𝑅𝑒𝑞 =

𝑉𝑎𝑏 13 𝑉 = = 𝟏. 𝟐𝛀 𝐼 11 𝐴

6.3 Electric Meters GALVANOMETER – an electrical device that is primarily used to detect the presence of a current AMMETER – a galvanometer with low resistance, called the shunt resistor, connected parallel to it used to measure current

𝑽𝒗 = 𝑰𝒇𝒔 (𝑹𝒄 + 𝑹𝒔 ) Vv = full-scale reading Ifs = full-scale current Rc = coil resistance Rs = Series resistance Problem 6.5 What series resistance is required to make the 1.00-mA, 20.0 Ω meter described above into a voltmeter with a range of 0 to 10.0 V?

𝑰𝒇𝒔 𝑹𝒄 = (𝑰𝒂 − 𝑰𝒇𝒔 )𝑹𝒔𝒉 Rsh = Shunt Resistance Ifs = Full-scale current Rc = Coil Resistance Ia = Full-scale reading Problem 6.4 What shunt resistance is required to make the 1.00-mA, 20.0 Ω meter described above into an ammeter with a range of 0 to 50.0 mA?

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PART

7 Magnetism and Magnetic Fields Electricity and Magnetism

7.1 Magnetism  A magnet has two poles, north and south. The North Pole is the end which points towards the geographic north  Like poles repel; unlike poles attract

 In 1820, Oersted ran experiments with conducting wires run near a sensitive compass. The orientation of the wire and the direction of the flow both moved the compass needle

 Ampere/Faraday/Henry – moving a magnet near a conducting loop can induce a current  The magnetic forces between two bodies are due to the interaction between moving electrons in the atoms  Inside a magnetized body, there is a coordinated motion of certain atomic electrons 7.2 Magnetic Field  ELECTRIC FIELD 1. Distribution of electric charge at ret creates the electric field E in the surrounding place 2. The electric field exerts a force FE=qE on any other charges in the presence of that field  MAGNETIC FIELD 1. A moving charge or current creates magnetic field in the surrounding space 2. The magnetic field B exerts a force on any other moving charge or current present in that field 𝑭𝒎 = |𝒒|𝒗𝑩 or 𝑭𝒎 = |𝒒|𝒗𝑩𝒔𝒊𝒏𝜱  The moving charge interacts with the field magnet. The force between them is at a maximum when the velocity of the charge is perpendicular to the magnetic field

Problem 7.1 A proton having a speed of 5.0x106 m/s in a magnetic field feels a force of 8.0x10-14 N toward the west when it moves vertically upward. Find the magnitude of the magnetic field S-PHY 14/S-SCI 15: Electricity and Magnetism (De Leon, Christian Dave A. SciEd)

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𝐵=

𝐹 = 𝑞𝑣

8.0𝑥10−14 𝑁 5.0𝑥106 𝑚 (1.6𝑥10−19 𝐶)( ) 𝑠 = 𝟎. 𝟏𝟎 𝑻

 The unit for magnetic field is Tesla (T) or Gauss (G) 1 𝑇 = 1𝑁/𝐴 ∙ 𝑚 1 𝐺 = 10−4 𝑇 7.3 Magnetic Flux MAGNETIC FLUX – the number of lines of force; equals to the product of the area of the loop times the perpendicular component of the magnetic field strength 𝜱𝑩 = 𝑩 ⊥ 𝑨 𝜱𝑩 = 𝑩𝑨𝒄𝒐𝒔𝜽 Problem 7.2 A square loop of wire 10.0cm on a side is in a 1.25T magnetic field. What is the maximum value of the flux that can pass through the loop (θ=0) 𝛷𝐵 = 𝐵𝐴𝑐𝑜𝑠𝜃 = (1.25𝑇)(0.100𝑚)(𝑐𝑜𝑠0) = 𝟎. 𝟎𝟏𝟐𝟓 𝑾𝒃  The unit for magnetic flux is Weber (Wb) 1 𝑊𝑏 = 1 𝑇 ∙ 𝑚2 7.4 Motion of Charged Particles in a Magnetic Field  Motion of charged particle under the action of magnetic field alone is always in motion with constant speed 𝛴𝐹 = 𝑚𝑎 𝑣2 𝑞𝑣𝐵 = 𝑚( ) 𝑅 𝒎𝒗 𝑹= 𝒒𝑩 Problem 7.3 An electron travels 2.0x107 m/s in a plane perpendicular to a uniform 0.010T magnetic field. Describe the path quantitatively 2.0𝑥107 𝑚 −31 ) 𝑚𝑣 (9.1𝑥10 𝑘𝑔)( 𝑠 𝑅= = −19 𝑞𝐵 (1.6𝑥10 𝐶)(0.010 𝑇) = 𝟏. 𝟏 𝒄𝒎

7.5 Magnetic Forces on Conductors  A Magnetic field exerts a force on an electric current. For a straight wire of length l with a current I, the force has a magnitude of 𝑭 = 𝑰𝒍𝑩𝒔𝒊𝒏𝜽

Where θ is the angle between magnetic field B and the current I Problem 7.4 A wire carrying a 30 A current has a length l=12cm between the pole faces of a magnet at angle of 60o. The magnetic field is approximately uniform at 0.90T. What is the magnitude of the force on the wire? 𝑭 = 𝑰𝒍𝑩𝒔𝒊𝒏𝜽 𝐹 = (30.0𝐴)(0.12𝑚)(0.90𝑇)(𝑠𝑖𝑛60) = 𝟐. 𝟖 𝑵  The torque on a loop or coil of N turns and current I in a magnetic field B at an area A at known angle θ is 𝝉 = 𝑵𝑰𝑨𝑩𝒔𝒊𝒏𝜽  The quantity NIA is the magnetic-dipole moment of the wire 𝑀 = 𝑁𝐼𝐴 Problem 7.5 A circular coil of wire has a diameter of 20.0c, and contains 10 loops. The current in each loop is 3.00A and the coil is placed in a 2.00T external magnetic field. Determine the maximum torque at 90o A = πr2 = 3.14(0.1000)2 = 3.14x10-2m 𝝉 = 𝑵𝑰𝑨𝑩𝒔𝒊𝒏𝜽 𝝉 = (10)(3.00𝐴)(3.14𝑥10−2 𝑚)(2.00𝑇) = 𝟏. 𝟖𝟖 𝑵 ∙ 𝒎

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 The magnetic field due to a long-straight wire is directly proportional to the current and inversely proportional to the distance from the wire 𝐼 𝐵∝ 𝑟 𝝁𝒐 𝑰 𝑩= 𝟐𝝅 𝒓  The constant µo is the permeability of free space 𝜇𝑜 = 4𝜋 𝑥 107 𝑇 ∙ 𝑚/𝐴 Problem 7.6 An electric wire in the wall of the building carries a dc current of 25A vertically upward. What is the magnetic field due to this current at a point 10cm due north? 𝝁𝒐 𝑰 (4𝜋 𝑥 107 𝑇 ∙ 𝑚/𝐴)(25𝐴) 𝑩= = 𝟐𝝅 𝒓 2(3.14)(0.10𝑚) = 𝟓. 𝟎 𝒙 𝟏𝟎−𝟓 𝑻  In two-parallel wires, the magnetic field produced at the position of wire 2 due to the current in wire 1 is 𝝁𝒐 𝑰𝟏 𝑩𝟏 = 𝟐𝝅 𝒅  The force in this field exerts on a length l2 of wire 2 is 𝝁𝒐 𝑰𝟏 𝑰𝟐 𝑭𝟐 = 𝒍 𝟐𝝅 𝒅 𝟐

Problem 7.7 The two wires of a 2.0m long appliance cord are 3.00mm apart and carry a current of 8.0A dc. Calculate the force one wire exerts on the other wire 𝝁𝒐 𝑰𝟏 𝑰𝟐 𝑭𝟐 = 𝒍 𝟐𝝅 𝒅 𝟐

𝐹2 =

(4𝜋 𝑥 107 𝑇 ∙ 𝑚/𝐴) (8.0𝐴)(8.0𝐴) 2.0𝑚 2𝜋 3.0 𝑥 10−3 𝑁 = 8.5 𝑥 103 𝑁

 A long coil of wire consisting of many loops (or turns) of wire is called a solenoid; if it is tightly wrapped, the magnetic field in s interior is almost uniform 𝝁𝒐 𝑰𝑵 𝑩= 𝒍

AMPERE’S LAW – states that around any chosen closed loop path segment ∆𝑙 times the component of magnetic field B parallel to the segment equals 𝜇𝑜 times the current I enclosed by the closed path 𝜮𝑩||∆𝒍 = 𝝁𝒐 𝑰𝒆𝒏𝒄𝒍

 Parallel current attract; anti-parallel current repel

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PART

8 Electromagnetic Induction Electricity and Magnetism

8.1 Faraday’s and Lenz’s Laws of Induction  ELECTROMAGNETIC INDUCTION – the process by which a changing magnetic flux produces electric current a. Induced Current – produced current b. Induced emf – produced emf  When a conductor moves across flux lines, magnetic forces on the free electrons induce electric currents.  INDUCED EMF: Faraday’s Observation 1. Relative motion induces emf 2. Direcion of emf depends on the direction of the motion 3. Emf is proportional to the rate at which the lines are cut (V) 4. Emf is proportional to the number of turns

∆𝜱𝑩 ∆𝒕 Where the negative sign means that the E opposes its cause 𝜺 = −𝑵

Problem 8.1 A single coil of wire is situated in a 0.5 T uniform magnetic field. The area of the coil is 2.0m2. a. What is the magnetic flux at an angle of 60o? 𝜱𝑩 = 𝑩𝑨𝒄𝒐𝒔𝜽 𝛷𝐵 = (0.5𝑇)(2.0)(cos 60) = 0.5 𝑊𝑏 b. Magnetic flux when the field is parallel to the normal after 5 seconds? 𝜱𝑩 = 𝑩𝑨𝒄𝒐𝒔𝜽 𝛷𝐵 = (0.5𝑇)(2.0)(cos 0) = 1 𝑊𝑏 ∆𝜱𝑩 = 1 𝑊𝑏 − 0.5 𝑊𝑏 = 𝟎. 𝟓 𝑾𝒃 c. What is the induced emf? ∆𝜱𝑩 𝜺 = −𝑵 ∆𝒕 −𝟏(𝟎. 𝟓 𝑾𝒃) 𝜺= = −𝟎. 𝟏 𝑽 𝟓𝒔 LENZ’S LAW – a current produced by an induced emf moves in a direction so that its magnetic field opposes the original change in flux  Flux increasing to the left induces flux to the right  Flux increasing to the right induces flux to the left Problem 8.2 What is the direction of the induced current in the following circular loops due to the currents?

FARADAY’S LAW OF INDUCTION – the magnitude of the induced emf equals time rate of change of the magnetic flux through the loop times the number of loops in the coil

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a. Counter-clockwise b. Clockwise c. Zero d. Clockwise MOTIONAL EMF  Faraday’s Law also tells that a changing magnetic field produces and electric field  A straight wire of length l moving with a speed v perpendicular to a magnetic field of strength B has an emf induced between its equals ends equal to: ∆𝜱𝑩 𝜺= ∆𝒕 𝐵∆𝐴 𝜺= ∆𝑡 𝐵𝑙𝑣∆𝑡 𝜺= ∆𝑡 𝜺 = 𝑩𝒍𝒗 Problem 8.3 An airplane travels 280 m/s in a region where the Earth’s magnetic field is 5.0 x 10-5 T and is nearly vertical. What is the potential difference induced between the wing tips that are 70 m apart? 𝜺 = 𝑩𝒍𝒗 280𝑚 𝜀 = (5.0𝑥10−5 𝑇)(70𝑚) ( ) = 𝟏. 𝟎𝑽 𝑠

c. Brushes – usually made up of carbon; two contacts that rub against the commutator and keep the coil connected to the battery d. Permanent Magnet – provides the magnetic field  The turning effect of the coil can be increased by: a. Increasing the current b. Increasing the number of turns on the coil c. Increasing the area of the coil – longer coil fives higher forces due to magnetic field; wider coil gives more leverages on the forces 2. Electric Generator – converts mechanical energy to electrical energy  Parts of a generator a. Armature – coil of wire that rotates b. Slip rings – attached to the ends of the wire in the coil c. Brushes – allows the loop to turn freely d. Permanent Magnet – provides the magnetic field AC Generator

8.2 Electromagnetic Devices 1. Electric Motor – converts electrical energy to mechanical energy  Parts of an electric motor  Also called as alternators  Current flows through the fixed coils and the magnet is field  The magnet’s north and south poles cause the current to the flow in opposite directions, producing an alternating current  Used to power small motors and electrical appliances

a. Armature – a coil of wire mounted on a rotating shaft b. Commutator – a split-ring fixed to the coil and rotates with it which reverses the flow of current through the motor

DC Generator  Contains a commutator, a connector that reverses a connection from the loop to the outside circuit of each half turn, instead of slip rings  The coil through which the current flows rotates in a fixed field

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Problem 8.4 A transformer for home-use radio reduces 120-V ac to 9.0-V ac. The secondary coil contains 30 turns and the radio draws 400 mA. Calculate: a. The number of turns in the primary coil 𝑁𝑠𝑉𝑝 (30)(120𝑉) 𝑁𝑠 = = = 400 𝑡𝑢𝑟𝑛𝑠 𝑉𝑠 9.0 𝑉  The commutator balances the charges leaving and returning to the generator, resulting in a current that does not change directions  Used to power very large electric motors and provide efficient energy supply that can charge the banks of batteries 8.3 Transformers and Transmission of Power TRANSFORMER  A device that changes the magnitude of an AC voltage  Consists of a primary coil and a secondary coil  The changing flux due to an AC voltage in the primary coil induces an AC voltage in the secondary coil

b. The current in the primary 𝐼𝑠𝑁𝑠 (0.40 𝐴)(30) 𝐼𝑝 = = = 0.030 𝐴 𝑁𝑝 400 c. The power transformed 𝑃 = 𝐼𝑠𝑉𝑠 = (0.40𝐴)(9.0𝑉) = 3.6 𝑊 𝑃 = 𝐼𝑝𝑉𝑝 = (0.030𝐴)(120𝑉) = 3.6𝑊  Transformers are also used in the transmission of electrical energy from the power plants to our home

Types of Transformers 1. Step-Up Transformer  Raises the voltage  Secondary coil has more loops than the primary coil 2. Step-down transformer  Lowers the voltage  Primary coil has more loops than the secondary coil  In a 100% efficient transformer, the ratio of output to input voltages (Vs/Vp) equals the ratio of the turns Ns in the secondary coil to the number Np to the primary 𝑉𝑠 𝑁𝑆 = 𝑉𝑃 𝑁𝑃  The ratio of the secondary current to primary current is in the ratio of the turns 𝐼𝑆 𝑁𝑃 = 𝐼𝑃 𝑁𝑆

S-PHY 14/S-SCI 15: Electricity and Magnetism (De Leon, Christian Dave A. SciEd)

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