PHY 102 Electricity, Magnetism & Modern Physics PDF

PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse PHY 102: ELECTRICITY, MAGNETISM & MODERN PHYSICS CHAPTER ONE ELECTROSTATIC AND CAPACITORS 1.0 Introduction Electricity can either be static or current. Static electricity (Electrostatics) is the study of an electric charge at rest and the force that exist between them 1.1 Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. It is a familiar experience that, when very dry hair is combed with an ebonite or plastic comb, some cracking sound is experienced. The same is true when putting on or off dry cloth made of nylon material, at times it even accompanied with electric shocks. Also, when a pen is rubbed in the hair, it is found to attract bits of light materials. This phenomenon is called electrification and the material are said to be electrified. There are two types of electric charge, positive and negative charge. The net charge of a closed system never changes When two objects are rub together there will be a transfer of electron between one object to another. An object that loose an electron is said to be positively charge object, while object that gain electron is said to be negatively charge object. E.g Rubbing glass rod with silk, glass rod losses electron and becomes positively charge while silk gain electron and becomes negatively charge, similar process happen when ebonite rod is rub with fur. Fur becomes positively charge by losing electron while ebonite becomes negatively charge by gaining the electron. For the subatomic particles proton has a positive charge, and an electron has a negative charge. 1.1.1 Fundamental law of electrostatics: Fundamental law of electrostatics states that ―like charges repel each other, unlike charges attract‖. Page 1 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse Thus the electric force between two objects is repulsive if the objects carry ―like‖ charges, that is, if both are positively charged or negatively charged while attractive if the sign of the charges differ 1.1.2 Conductors and Insulators Conductors are materials that allow charge to move through them freely; examples all metals (eg copper, Aluminum), human body, tap water etc. Insulators (nonconductors) are materials that cannot allow charge to move through them freely; examples include rubber, plastic, glass, e.t.c Semiconductors are certain material whose conductivity lies between conductors and insulators e.g. Silicon, Germanium. 1.1.3 Charging a neutral body A neutral body is the one with equal number of positive and negative charge, so its net charge is zero. This body can be charge either by contact or by induction. 1.1.3.1 Charge by contact Charging a neutral body by contact method can be done by bringing a charge object in contact with the insulated or neutral object, so the same type of charge is transferred to the insulator. 1.1.3.2 Charge by induction Unlike charging by contact (friction) to charge the body by induction the two bodies is not necessary to be in contact. Thus only conducting substances can be charged by induction. Here, an electrically neutral body is imparting a charge by bringing a charged body near it. Experiment shows that the sign of the imparted charge is opposite that of the imparting charge. Thus a negatively charged ebonite rod will impart a positive charge on a previously uncharged metallic conductor. A conductor can be given a permanent charge by the process of induction as follows:  Consider a charged glass rod brought near a metallic conductor mounted on an insulating stand (fig.1.1a). The positive charges on the glass rod will attract free Page 2 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse electrons in the metallic conductor to the side facing the glass rod, and hence, leaving an equal positive charges on the opposite side.  Momentarily connect the far side of the conductor to the ground (fig. 1.1b). The positive charge of an electron will be neutralized by influx of electrons from the ground to which the conductor is connected.  Disconnect the earth and then remove the glass rod. The negative charges in the conductor distribute themselves over the entire surface (fig. 1.1c). Noted that if the glass rod is removed first before disconnecting the earth, electrons in the conductor would flow to the ground so that no charges would be obtained in the conductor. Example of the devices used for demonstration of charge by induction is the gold leave electroscope 1.2 Coulomb’s Law Coulomb’s Law states that “The force of attraction or repulsion between two charges and in a given medium is directly proportional to the product of the charges and inversely proportional to the square of the distance separating the charges 1.1 Where called a Coulomb constant, is the permittivity of free space equal to. Page 3 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse 1.2.1 Permittivity of the medium is defined as the degree to which a medium resist the flow of charge. If the charge is placed in a medium with permittivity other than free space we have to replace with. 1.2.2 Relative permittivity is define as the ratio of permittivity of the medium to the permittivity of free space i.e 1.2 Example 1: Two charges of and of are held fixed at position and respectively. Calculate the force acting on the negative charge Example 2 Suppose two equal charges repel one another with the force of when situated apart in vacuum. (a.) Calculate the magnitude of these charges b. What will be the magnitude of the force if they were situated in an insulating liquid whose permittivity is ten times that of vacuum Example 3: Three point charges each carrying a charge of are located at the corner of an equilateral triangle with side Calculate the force exerted on the charge Exercise 1.1 1. If the electric force of repulsion between two charges is , how far apart are they? 2. The electron and proton of a hydrogen atom are separated by a distance of about. Find the magnitudes of the electric force and the gravitational force that each particle exerts on the other.( , , ) 1.3 Electric line of forces Electric line of forces are lines or curve drawn such that tangent to it at any given point will give the direction of the electric field. Page 4 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse The line of forces associated with positive point charge radiate out ward from the source while that of negative point charge are directed toward the source, therefore field lines are originated from positive point charge and terminated at the negative point charge. Figure 1.3. Electric line of forces due to point charges 1.3.1 Electric Field Electric field is a region where an electric charge experiences a force. The electric field produced by a stationary charge can be investigated by measuring the force on a positive test charge placed at various point in space. Note that the direction of the electric field is away from positive charge while toward the negative charge as indicated below Figure 1.4 Direction of the electric field due to positive and negative charge If is very small, the Electric field strength is given as force per unit charge 1.3 is vector quantity having the same direction with the force. The unit of Electric field is From Coulomb law, = Therefore, Page 5 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse = 1.4 Example 1.3.1: (a) Find the electric field at a distance from a charge of. (b).What is the magnitude of force experience by proton at that point 1.3.3 Electric Flux Flux refers to the flow of some vectorial quantities through a given area. When an electric field pass through an area of a given surface then the magnitude of the electric field is proportional to the number of field lines of force in the area. Thus electric flux is defined as the number of field lines passing through a given surface or simply defines as the product of field magnitude and area. 1.5 Figure 1.5. Electric flux 1.4 Gauss’s theorem Gauss‘s theorem provide different way to express the relationship between electric charge and electric field. Gauss states that ―The total electric flux out of any close surface is equal to the total electric charge inside the surface divided by ‖ 1.6 combining eq. (1.5) and (1.6) we have ∫ 1.7 Eqn 1.7 is also form of Gauss law Page 6 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse Example 1.4.1: Calculate the electric field due to a uniformly charge infinite wire whose charge per unit length is λ Example 1.4.2: Use Gauss‘s law to verify that a charge Q uniformly distributed over the surface of a sphere is equivalent to the point charge at the Centre of the sphere 1.5 Electric potential When charge is placed in an electric field it experiences a force so as to pull a positive test charge from to in a direction opposite to that of , you must do positive work on it because the electric field tries to force it to the right. Figure 1.6. Electric potential Thus electric potential difference between and is defined as the work done per unit charge to move a positive test charge from A to B represented as 1.8a The unit of Electric potential is Volts(V) or JC-1 Similarly ( ) 1.8b Equation (1.8b) gives the expression for the potential difference between two points A and B. we can choose zero reference level for electric potential to be at infinity. Thus if the point B is at infinity then Therefore the potential at point A with respect to infinity is ( ) ( ) 1.8c Thus we can define absolute electric potential at a point as a work done in bringing a unit charge from infinity to that point. Page 7 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse The unit of Electric potential is Volts ( ) or 1.5.1 Relationship between electric field and electric potential By definition and We know that work done = force x distance. The negative comes as a result of the displacement is in opposite direction to the If is the potential difference between A and B Then 1.9 the electric field strength between two parallel conducting plates with a potential difference of and a plate separation of d is Example 1.5.1: Determine the absolute potential in air at a distance from a point charge of Example 1.5.2: Two point A and B are distant and from charge respectively. What is the potential difference between these point? Example 1.5.3: Determine the electric potential at point P due to two point charges and along a perpendicular bisector of the line joining the charges Page 8 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse CAPACITORS A capacitor is a device for storing electric charge. It consists of two metallic plates separated by an insulator e.g a plastic film, paper, air and water. The insulating material is called dielectric of the capacitor. The two plates have equal amount of opposite charges. There are different types of capacitors such as parallel plates, concentric spheres and cylindrical capacitors. Experiments shows that the amount of charge Q stored by a capacitor is directly proportional to the potential difference across the plates. Thus we can write. 1.20 where is constant of proportionality called the capacitance of the capacitor. We can therefore write If the charge stored is in Coulomb ( ) the potential different between the plates is in Volt (V), then capacitance is in Farad (F). One Farad is the capacitance of an extremely large capacitor. The capacitance of capacitors used in practical circuit such as radio receivers and audio amplifiers are expressed in microfarad (µF). There are smaller capacitors e.g pico farad (pF) and nanofarad (nF). Figure 1.7 Capacitor 1.6.1 Factors affecting the capacitance of a capacitor 1. The area of the plates: The capacitance increase as the common area A of the plates increase. 2. The nature and thickness of the dielectric: The capacitance of capacitor varies when different dielectrical materials are used. 3. The relative position of the plates: The capacitance of the capacitor decreases when the separation between the plates is increased Page 9 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse 4. The geometry of the plates: The capacitance of capacitor varies with the geometry of the plates 1.6.2 Parallel Plate Capacitor Figure 1.8. Parallel Plate Capacitor The electric field E due to a parallel plate capacitor is given by 1.21 Where is the surface density = where is the charge on each plates and is the area of plate. Therefore, the potential difference between the plates is given by; But, from equation 1.22 From equation (1.22) we see that the capacitance is; i. Directly proportional to the area of plates and ii. Inversely proportional to the distance between the plates Note The capacitance of a capacitor of an isolated conducting sphere of radius a and having a charge Q, located in a medium of permittivity is given by 1.2.3 Page 10 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse The capacitance of a capacitor of concentric spheres having radii a and b, where b > a is given b by 1.24 1.6.3 Capacitors with Dielectric Between the Plates The effect of introducing a dielectric between the plates is to increase the capacitance. The capacitance of the capacitor with dielectric is given in terms of a dielectric constant , We know that Where C = capacitance of the capacitor with dielectric, while is the capacitance without dielectric, , is the relative permittivity. For parallel plates capacitor Suppose , thus 1.25 1.6.4 Arrangement of capacitors Capacitors can be arranged either in series or in parallel 1.6.4.1 Series connection When capacitors are arranged in a consecutive manner then they are said to be connected in series in such a case the total potential difference across the series combination is the sum of the individual potential difference of each parallel plates. Therefore 1.26 But we know ( ) Thus Page 11 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse For the capacitor connected in series the reciprocal of their resultant is equal to the sum of the reciprocal of their individual capacitance 1.27 Figure 1.9 Capacitors arranged in series 1.6.4.2 Parallel connection Capacitors having all their left hand plates connected together and also their right hand plate connected, are said to be connected in parallel across the same potential difference, therefore the potential difference across each capacitor is the same. while, total charges is due to sum of the individual charges Figure 1.10 Capacitors arranged in Parallel 1.28 but Thus Since the potential across each capacitor is the same thus Therefore Page 12 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse ( ) Thus for n capacitors connected in parallel, the net capacitance is given by the sum of the individual capacitors 1.29 1.7 Energy stored in a capacitor Suppose that the capacitor is charge by increasing the potential difference from 0 to. the charge will increase from 0 to where. The potential difference ( ) between the plates at any moment will be ( ) ( ) 1.30 If an extra charge is transfer from the negative terminal to the positive terminals, the small amount of work require to carry out the transfer will be 1.31 if the process is continued till the capacitor is charge, the total work or energy stored in the capacitor is ∫ ∫ 1.32 equation (1.32) is for the energy store in the capacitor. Example 1.7.1: Find equivalent capacitance of and when connected in (a) Series (b) parallel Exercise 1. Three charges are located at the different corner of an equilateral triangle with side 10cm find (a) Total force on (b) total electric field at Page 13 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse 2. Three capacitors each of 100nF are charged to 500V and then connected in series, determine (a) The potential difference between the end plate (b) The charge on each capacitor (c) The energy store in a system Page 14 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse CHAPTER TWO CURRENT ELECTRICITY 2.1 Electric Current A current is a flow of charge. In metal wire many electrons are free to move, so that a current can flow in a metal wire as a flow of electrons, i.e. the current carriers (or charge carriers) are electrons. The unit for current is the Ampere (A), defined as: ―A current of one Ampere is flowing in a circuit when a charge of one Coulomb passes any point in the circuit in one second” Current (I) is related to charge (Q) moving through (entering and leaving) a wire in a time (t) seconds by (2.1) Equation 2.1 defines the Coulomb as 1As. The direction of current is taken as that of positive charge flow, i.e. opposite to that of electron flow. 2.2 Carrier Velocity (drift velocity) If carriers e.g. electrons in a metal wire, are moving with an average drift velocity, along the wire (Fig 2.1), then the current is (2.2) Where n is the carrier density (number of electrons per m3), A is the cross-section area of the wire (so that nA is the carriers per meter length of wire) and q is the charge of each carrier. Figure 2.1: current carrying conductor Page 15 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse The drift velocity of the electron is (2.3) 2.3 Current Density When a wire of uniform cross-sectional area A carries a current I, the current density, J, is current per unit cross-sectional area. Using equation (2.2) when the charge carriers are considered. (2.4) Current density may be used in charge carrier calculations. The unit for current density is ampere per metre square (Am-2). Example 2.1: How many electrons are passing through a wire per second if the current is 1.00 mA, given that the charge carried by each electron is 1.6 x 10-19 C. Solution: I=1.0 x 10-3A, q = 1.6 x 10-19C; let time t = 1s and number of electrons be n=? using I =nq/t we have Example 2.2: Calculate the mean velocity of electron flow (the drift velocity) in a wire where the free electron density is 5.0 x 1028 m-3 if the current is 1.0A and the wire has a uniform cross- sectional area of 1.0 mm2. (electron charge = -1.6 x 10-19 C). Solution I= 1A, n= 5.0 x 1028m-3, q=1.6 x 10-19C and A= 10-6m2 v=? Using I= nAqv Example2.3: A current of 4.5A flows through a car headlight. How many coulombs of charge flow through it in 1.0h? Solution I = 4.5A, t = 1h = 1 x 60 x 60s = 3600s, Q =? Using Q = It Q = 4.5 x 3600 = 16200C = 1.62 x 104C Page 16 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse 2.4 Potential and Potential Difference The potential of a place may be thought of as its attractiveness for electrons or unattractiveness for positive charges. A place where there is a high concentration of electrons or which has a lot of electrons near it will have a low potential. The potential difference (PD) V between two places is defined as the work done per coulomb of charge moved from one place to the other. (2.5) where W is the work done (e.g. if positive charge Q moves from lower potential (-) to higher potential (+)) or energy obtainable from the movement (e.g. if negative charge Q goes from – to + place). The S I unit for Potential difference is the Volt (V). The potential of a place measured in volts is the Potential difference between the place concerned and some reference point, usually taken to be a place far away from any electric charges (i.e. at infinity), or otherwise the earth. In other words, either of these places may be taken as zero potential. Electric current flows spontaneously from a higher potential place (+) to a lower potential place (-) if the two places are joined by a conducting path. 2.5 Ohm’s law This law states that the current I through a given conductor is proportional to the potential difference between its ends, provided that its temperature does not change. (2.6) This law applies to metallic conductors and many others. 2.6 Ohmic and Non-Ohmic Conductors Ohm‘s law is obeyed by most conductors (metals). For example, copper and tungsten, used respectively in cables and lamp filaments obey Ohm‘s law. These are called ohmic conductors. Non-ohmic conductors are those which do not obey Ohm‘s law.. Many useful components in the electrical industry are non-ohmic; for example, a non-ohmic component is essential in a radio receiver circuit. Examples include diodes, LED, LDR, thermistor etc. 2.7 Resistance R of a conductor This is the opposition of the conductor to the current flow through it, and it is defined as the Potential difference needed across it (between its ends) per unit current: (2.7) Page 17 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse The unit for resistance is the Ohm (Ω). 2.8 Resistors These are devices for providing resistance to the flow of current. Resistors can be fixed, variable or potential divider. Some variable resistors are called rheostats. A Thermistor is a temperature sensitive resistor. An LDR is a light dependent resistor (photoconductor). 2.9 Resistivity (ρ) and Conductivity (σ) of a material It is found experimentally that the resistance R of a conductor is proportional to its length l, inversely proportional to its area of cross-section A, and dependent upon the nature of the material, described by its resistivity ρ, which is defined by the following equation: (2.8) So, (2.9) The unit for ρ is Ωm The term ‗conductivity’ σ of a material is used for the reciprocal of resistivity ρ so that (2.10) Example 2.4: Calculate the length of wire of 1.0 mm diameter and 5.0 μΩm resistivity that would have a resistance of 5.0 Ω. Solution Given that; So Page 18 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse Example 2.5: A rectangular copper block has dimensions 100cm x 0.5cm x 0.5cm. (a) What is the resistance between its two square ends? (b) What must be the diameter of a circular aluminum rod of length 200cm if its resistance is to be the same as that of the above rectangular copper? (Resistivity of copper at 200C = 1.72 x 10- 8 -m, and that of aluminum at 200C = 2.65 x 10-8-m). Solution (a) the cross-sectional area of the square end A is From equation (2.8) the resistance between the square ends is (b) So But the circular area is defined as √ √ 2.10 Electric circuit Fig 2.2: Circuit Elements Page 19 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse Often a current is produced by use of a voltaic cell or battery (two or more cells joined together). The cell creates and maintains a potential difference between its terminals. A current is obtained if these two terminals are joined by a conducting path, i.e. when a complete circuit is formed as shown in Fig above. The current obtained from a voltaic cell is direct current (DC) because its direction is constant. 2.11 Resistors in Series When two resistances R1 and R2 are connected as shown in fig. 2.4a they are in series and the total resistance is R, where (2.11) 2.12 Resistance In Parallel In this arrangement the resistance of the combination is given by (2.12) In a parallel combination the PD across one resistor is the same as that across the other, but the total circuit current in figure (b) is shared between the resistors. Example 2.6: Calculate the current through and the Potential difference across each of the resistors in the circuit shown in Figure below. Page 20 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse Solution The resistance of 3Ω in parallel with 6Ω is We see that the circuit can be regarded as 8.0Ω in series with 2.0Ω. Circuit resistance is Therefore, Note that we know only one PD, namely 6.0V, and to use ohm‘s law we must use V = 6.0V with the correct resistance. It is the 10Ω across which the PD is 6.0V. The current through the 8.0Ω resistor is I, which is 0.6A. PD across 8.0Ω (using Ohm‘s law) is given by To obtain answers for the 3.0Ω and 6.0Ω we can say either: PD across the 3.0Ω and 6.0Ω The current I3 through the 3.0Ω is And for the 6.0Ω the current I6 is Page 21 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse Example 2.7: Three cells of EMF 3V and internal resistance of 0.5Ω are connected in parallel in order to reduce the total potential difference across the circuit. Determine the effective internal resistance of the cell. Solution Let and For three resistors in series we have Example 2.8: A cell of e.m.f 12V and internal resistance 0.1Ω is connected to a 10Ω resistor. What is the voltage at the terminals of the cell? Solution E = 12V, r = 0.1Ω and R = 10Ω. V = ? 12V 0.1Ω V 10Ω E = terminal voltage + lost voltage terminal voltage (V) = E – lost voltage but ( ) ( ) or ( ) Example 2.9: Find the equivalent resistance in the circuit shown below (fig 2.5). Calculate the current in the circuit. Assuming R1=4 , R2=6 , R3=8 , R4=8 , R5=12 and R6=2 Page 22 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse E=10V R2 R1 R3 R4 R5 R6 Fig 2.5a Solution The circuit (Fig 2.5) can be reduced stage by stage using series-parallel equivalent method. Fig 2.5b: equivalent circuit diagram R3 and R4 are parallel. Thus R6 and R7 are series. The equivalent resistance Page 23 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse R8 and R5 are parallel. Thus Hence. The equivalent resistance R10 The total current in the circuit can be found using Ohm‘s law Therefore, 2.13 Electrical heating in a resistance When current flows through a resistance there is a PDV across the resistance and, for Q coulombs passing through, electrical potential energy is lost (work W is done), this becoming internal energy of the resisting material (its temperature has risen). Since ⁄ (2.13) Using equation (2.1) into (2.13) we have equation (2.14) (2.14) Equation (2.14) is the electrical heat produced in a resistance for which current flows. Substituting equation (2.7) into (2.14) we have equation (2.15) and (2.16) ⁄ (2.15) (2.16) The work done per second or heat produced per second is the electric power P given by equation (2.17). (2.17) The unit for power is watt (W). 1W=1Js-1. Page 24 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse The expression ‗power dissipated‘ (in a resistance) is often used. It means ―heat produced (per second)‖; but reminds us that the heat normally spreads and escapes from place where it is produced. The kilowatt-hour (product of KW and hour) is a unit for energy converted in 1 hour when the power is one kilowatt. 1kilowatt-hour=1000 watt x 60 x60 seconds = 3600 x 106J = 3.6 MJ. Exercise 1. Calculate the energy dissipated by a 100 watt lamp working for 1 day. Give the answer (a) In kilowatt-hour and (b) In joules 2. A current of 2mA flows in a radio resistor when PD of 4V is connected. What are the values of the resistance and conductance? G=I/V Unit Siemens or Mho 3. Show that for four resistors in parallel the current in one branch, for example the branch of R4, is related to the total current by ( ) 4. The resistivity of uniform wire of length 2.0m is 5.4 x 10-4Ωm. if the cross-sectional area of wire is 9.5 x 10-3cm2. Calculate the resistance of the wire. Page 25 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse 2.14 Electromotive force and internal resistance The PD between the terminals of a cell is caused by a chemical action which stops when the PD reaches a value characteristic of the type of the cell, called the EMF of the cell. EMF stands for electromotive force, although it is a voltage not a force. When cell is producing no current, i.e. it is an open circuit; the terminal PD (V) equals the EMF value E. ( ) (2.18) When a current is being produced, the PD falls from the EMF value E, the chemical action starts again and the terminal PD V that is maintained is less than E by an amount called the ‗lost volts‘. This drop E-V is a consequence of internal resistance r in the cell that hinders the cell‘s working. The lost volts equals Ir, so that (2.19) Either of the statements E =V when I = 0 or E-V = Ir may be used to define E, but a more satisfactory definition is given by equation (2.20) (2.20) According to equation (2.17) the total power dissipated in the circuit with resistance R and the internal resistance r is given by equation (2.21) (2.21) Substituting equation (2.21) into (2.20) we have equation (2.22) which agrees with equation (2.18) when there exists no current (i.e. I = 0). (2.22) For calculations a cell or other voltage source can be regarded as a cell of zero internal resistance with a separate internal resistance r in series with it, see Fig. 2.6. Fig 2.6: Internal resistance (r) and Electromotive force (E.M.F) in a circuit Page 26 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse A cell represented in this way is seen in the circuit of Fig.2.6. This circuit agrees with equation (2.19) and (2.20) and it is also seen that (2.23) 2.15 Maximum power When a cell or other voltage source, having internal resistance r, is connected to a ‗load‘ resistance (R in fig. 2.6) the current through R is given by E/(R+r), the PD across it is ER/(R+r) and the power dissipated in it is equal to the product of these. The current is at its largest when , the PD is large when and, it can be shown. The power is greatest when. Example 2.10: A cell of EMF 1.5V and internal resistance 1.0Ω is connected to a 5.0Ω resistor to form a complete circuit. (a) Calculate the current expected at the terminal PD and the power dissipated in the external circuit and in the cell. (b) What value would be needed for resistance R in order that the maximum power would be drawn from the cell? (c) Calculate the maximum power value. Solution (a) Using The power in the 5.0Ω is The power in the 1.0Ω internal resistance is (b) For a maximum power dissipation in resistance R this resistance must equal the internal resistance which is 1.0Ω (c) The total resistance of the circuit will then be 2.0Ω and the current will be E/R or 1.5/2.0 or 0.75A. Page 27 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse The power in the 1.0Ω is 2.16 Active and Passive Elements Active elements are voltage or current sources which are able to supply energy to the network. Resistors, inductors, and capacitors are passive elements which take energy from the sources and either convert it to another form or store it in an electric or magnetic field. 2.17 Circuit Laws An electric circuit or network consists of a number of interconnected single circuit elements of the type described above. The circuit will generally contain at least one voltage or current source. The arrangement of elements results in a new set of constraints between the currents and voltages. These new constraints and their corresponding equations, added to the current- voltage relationships of the individual elements, provide the solution of the network. The underlying purpose of defining the individual elements, connecting them in a network, and solving the equations is to analyse the performance of such electrical devices as motors, generators, transformers, electrical transducers, and a host of electronic devices. The solution generally answers necessary questions about the operation of the device under conditions applied by a source of energy. (i) Kirchhoff’s Voltage Law For any closed path in a network, Kirchhoff‘s voltage law (KVL) states that the algebraic sum of the voltages is zero. Some of the voltages will be sources, while others will result from current in passive elements creating a voltage, which is sometimes referred to as a voltage drop. The law applies equally to circuits driven by constant sources, DC, time variable sources, V(t) and I(t), and to circuits driven by sources. V  0 (2.24) Example 2.11 Write the KVL equation for the circuit shown in Fig.2.11 R1 Va i Vb R2 R3 Fig. 2.11: circuit diagram Page 28 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse Starting at the lower left corner of the circuit and using equation (2.24), for the current direction as shown in Fig (2.11), we have ( ) (ii) Kirchhoff’s Current Law The connection of two or more circuit elements creates a junction called a node. The junction between two elements is called a simple node and no division of current results. The junction of three or more elements is called a principal node, and here current division does take place. Kirchhoff‘s current law (KCL) states that the algebraic sum of the currents at a node is zero. It may be stated alternatively that the sum of the currents entering a node is equal to the sum of the currents leaving that node. The basis for the law is the conservation of electric charge equation (2.25). I 0 (2.25) Example 2.12: Write the KCL equation for the principal node shown in Fig. 2.12 i1 i3 i4 i2 i5 Fig 2.12: Kirchhoff‘s current law (KCL) for multiple currents Solution Example 2.13: In the circuit shown in figure 2.13 a 12-V power supply with unknown internal resistance r is connected to a run-down rechargeable battery with unknown emf and internal resistance 1Ω and to an indicator light bulb of resistance 3Ω carrying a current of 2A. The current through the run-down battery is 1A in the direction shown. Find the unknown current I, the internal resistance r, and the emf E. Page 29 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse 12V r (3)I E 1Ω (1)a b 1 (2) A 3 Ω 2A Figure 2.13: circuit diagram Solution First we apply the junction rule at point a, we find To determine r, we apply the loop rule to the outer loop labelled (1); we find To determine E, we apply the loop rule to loop (2) 2.18 Wheatstone bridge The Wheatstone bridge, devised in 1843, provides an alternative method of determining the resistance of an unknown resistor. The circuit is shown in Fig.2.14, four resistors are joined as shown, one of them being unknown resistor whose resistance is to be measured and one standard resistor. Assuming that R4 is the unknown. R3 can be varied and it is adjusted until no current flows in the galvanometer. In this condition also Page 30 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse Therefore (2.26) (2.27) From equation (2.26) and (2.27) we have (2.28) (2.28) Fig. 2.14: Wheatstone bridge 2.19 Potentiometer The potentiometer, devised by Poggendorf in 1850, is a very useful instrument for a number of measurements in electricity. In its simplest form it is simply a piece of resistance wire, usually a meter long, fixed between two points A and B with a cell of output voltage V connected between two ends (Fig.2.15). V A F B C l D G E Fig.2.15: potentiometer Page 31 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse If C is now connected to A the potential at junction AC becomes +V and that at D becomes (V- E). If D is now connected to the wire at a point F then if no current is to flow through the galvanometer the potential drops down the wire must be equal to that across the source and meter, that is p.d. across AF must equal p.d. across CD. Since we know the P.d. across AB then, assuming that it reduces uniformly down the wire, E can be found. Let length AF = l2 and Let length AB = l1 Thus (2.29) Notice that E must be less than V Example 2.13: A potentiometer is set up as shown in figure 2.16 and the balance point for the unknown e.m.f. V found at 74.5cm from the left-hand end of the meter wire. If the driver cell has an e.m.f. of 1.5V and negligible internal resistance find that of the unknown e.m.f. 1.5V G V Figure 2.16: circuit Diagram for example 2.14 Solution Page 32 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse 2.22 Thermoelectricity Thermoelectricity is a two-way process. It can refer either to the way a temperature difference between one side of a material and the other can produce electricity, or to the reverse: the way applying an electric current through a material can create a temperature difference between its two sides, which can be used to heat or cool things without combustion or moving parts. Exercise 1. A potentiometer wire carrying a steady current is 100cm long. When a standard cell of E.m.f 1.5V is connected to a balance length of 60cm was obtained. Calculate the E.m.f of a cell that gives a balance length of 80cm. 2. Consider a moving coil galvanometer with a resistance of 100Ω and a full scale deflection of 100µA. I. Convert to an ammeter reading of 5A II. Convert to a voltmeter reading of 5V Page 33 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse CHAPTER 3 MAGNETIC FIELD AND FORCES 3.1 Introduction Magnetic phenomena were first observed about 2500 years ago in fragments of magnetized iron ore called permanent magnets. These magnets are known to have fields around them called magnetic fields; such that when a magnetic material is brought in the vicinity of the fields, it will experience a magnetic force. Fundamentally, magnetism is experience only due to the interaction of mobile charges such as moment of electrons in a magnetic field. When a current-carrying conductor is brought into a magnetic field, it will experience magnetic force. Magnetism is known to concentrate at the ends of a magnet called magnetic poles. These poles exert attractive forces on each other when they are of opposite polarities (North and South poles attraction) and repel each other when they are of like polarities (North and North or South and South poles repulsion). Hence like poles attract each other while unlike poles repel each other. 3.2 Magnetic fields Magnetic field is the region in space around a magnet where magnetic effect or magnetism can be experienced by a magnetic material or another magnet. This field can be produced in two ways 1. When electrically charged particles flow in a conductor to produce electromagnet and 2. By means of elementary particles such as electrons which are known to have intrinsic magnetic field around them. Similar to electric field, magnetic field is a vector field. It is denoted ad ⃗ with the direction of the field defined in the direction in which the north pole of compass needle will tend to point when placed in the field. Magnetic field around a charge particle is defined in terms of the magnetic force and is defined as ⃗ ⃗ (3.1) Where is the charged particle ⃗ ⃗ defines the vector product between velocity vector ⃗ and magnetic field vector ⃗. From eqn. 3.1, we define the magnitude of the force as | | (3.2) Page 34 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse Where is the angle between ⃗ ⃗. Deductions from eqn. 3.2 reveals that the maximum value of the force is obtain when ⃗ and ⃗ are perpendicular to each other. The magnitude of the field perpendicular to the velocity vector is given form eqn. 3.2 as | | (3.3) Magnetic field is measured in Tesla T and is equivalent to or. Another unit of B is called Gauss ( ). Example: A beam of protons ) moves at speed of in a uniform magnetic field directed in positive z-axis experience a magnetic force of magnitude. If the velocity of each proton lies in x-y plane at to positive z-axis, find the magnitude of the field responsible for this force. Solution | | ( )( ) Example: A uniform magnetic field of magnitude 1.2mT is directed vertically upward throughout the volume of a laboratory chamber. If a proton moves with kinetic energy of 5.3Mev horizontally from south-north, what is the magnitude of the force that acts on the proton? Solution Note the direction of the velocity and field are such that they are perpendicular to each other. So | | | | ( ⁄ ) 3.3 Magnetic field lines and flux 3.3.1 Magnetic field lines Magnetic field is represented using magnetic field lines. These are imaginary lines showing the direction and strength of magnetic field. They are drawn such that- 1. The lines through any point are tangent to the magnetic field vector ⃗ at that point. 2. The spacing of the lines represents the magnitude of the field at that point. i.e the field is stronger at regions where the lines are closely packed and vice versa. 3. The field lines emerge from the north poles of permanent magnet and enter through south poles thereby always forming loops. Page 35 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse Fig. 3.1 magnetic field lines of magnetic sources 3.3.2 Magnetic flux and Gauss’s law for magnetism If the surface area of a magnetic material is divided into smaller elements , we can determine the component of magnetic field perpendicular to each element as shown in fig. 3.2 below BT B φ BH dA Fig. 3.2: magnetic field through an element From the figure, the component perpendicular to the element is defined as , where is the angle between the direction of the field and the a line perpendicular to the surface. We define the magnetic flux as (3.4) By integrating eqn. 3.4 gives the total magnetic flux over the entire surface as: ∫ ∫ ∫⃗ (3.5) Page 36 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse Magnetic flux is a scalar quantity and from eqn. 3.5, the maximum value of magnetic flux in the surface will be when the magnetic field ⃗⃗ is perpendicular to the surface Hence (3.6) It has S.I unit of Weber (Wb) defined as. What will be the total magnetic flux in a closed surface? The answer to this question is a statement of Gauss law. It states that “the total magnetic flux in an enclosed surface is equal to the total charges enclosed by the surface”. This is however the general statement of the law. In reality, it is not possible to isolate a magnetic pole (monopole) but instead, we have only dipoles. As a result Gauss law for magnetic flux state that the total magnetic flux of a closed surface is equal to zero. The reason is that the poles cancel out. We therefore have Gauss law as ∮⃗ (3.7) 3.4 Magnetic materials Materials respond differently to the force of a magnetic field. There are three main classifications of magnetic materials. A magnet will strongly attract ferromagnetic materials, weakly attract paramagnetic materials and weakly repel diamagnetic materials. The orientation of the spin of the electrons in an atom, the orientation of the atoms in a molecule or alloy, and the ability of domains of atoms or molecules to line up are the factors that determine how a material responds to a magnetic field. Ferromagnetic materials have the most magnetic uses. Diamagnetic materials are mainly used in magnetic levitation. i. Ferromagnetic materials Ferromagnetic materials are strongly attracted by a magnetic force. Ferromagnetic materials include iron (Fe), nickel (Ni), cobalt (Co) and Gadolinium (Gd). The reasons these metals are strongly attracted are because their individual atoms have a slightly higher degree of magnetism due to their configuration of electrons, their atoms readily line up in the same magnetic direction and the magnetic domains or groups of atoms line up more readily. Alloys of iron, nickel, cobalt, gadolinium and certain ceramic materials can become "permanent" magnets, such that they retain their magnetism for a long time. Temperature effect affects strongly ferromagnetic materials like nickel or steel. They lose all their magnetic properties if they are heated to a high enough temperature. The temperature at which a metal loses its magnetism is called the Curie temperature, and it is different for every metal. The Curie temperature for nickel, for example, is about 350°C. ii. Paramagnetic materials Page 37 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse Paramagnetic materials are metals that are weakly attracted to magnets. Aluminum and copper are such metals. These materials can become very weak magnets, but their attractive force can only be measured with sensitive instruments. Temperature can affect the magnetic properties of a material. Paramagnetic materials like aluminum, uranium and platinum become more magnetic when they are very cold. The force of a ferromagnetic magnet is about a million times that of a magnet made with a paramagnetic material. Since the attractive force is so small, paramagnetic materials are typically considered nonmagnetic. iii. Diamagnetic materials Certain materials are diamagnetic, which means that when they are exposed to a strong magnetic field, they induce a weak magnetic field in the opposite direction. In other words, they weakly repel a strong magnet. Some have been used in simple levitation demonstrations. Bismuth and carbon graphite are the strongest diamagnetic materials. They are about eight times stronger than mercury and silver. Other weaker diamagnetic materials include water, diamonds, wood and living tissue. Note that the last three items are carbon-based. The electrons in a diamagnetic material rearrange their orbits slightly creating small persistent currents, which oppose the external magnetic field. 3.5 Sources of magnetic field 3.5.1 Magnetic field of a moving charge A mobile charge at constant velocity ⃗ has a magnetic field around it. The magnitude of the field is proportional to the product of the charge | |, the velocity , and inversely proportional to the square of the distance to the point of examination. Mathematically we have | | (3.8) | | (3.9) Where is the constant of proportionality and is called permeability of free space. Note that the magnitude of the field at all points lying on a line parallel to the plane of ⃗ is zero since while the maximum field will be experienced on a line perpendicular to the plane of the velocity ⃗. At this line, | | (3.10) Page 38 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse 3.5.2 Magnetic field of a current-carrying element (Biot and Savart law) The principle of superposition of magnetic fields states that the total magnetic field created by several moving charges (current) is the vector sum of the fields of individual charges. Using this idea, we shall determine the magnetic field due to current in a conductor. Consider a small segment of a current carrying conductor with small volume (where A is the cross sectional area) as shown in the figure below P I dl Fig. 3.3: Magnetic field due to Current-carrying element dl of a conductor at point P. Let be the number of charges in the small element, each of charge. Then for the entire volume, we have. The behaviour of this charges in the element is equivalent to that of a single charge moving with a drift velocity (i.e, ). Using eqn. 3.9 we write for an element | | | | (3.11) The quantity | | ( ) (3.12) Equation 3.12 is the famous Biot-Savart law. It is used in finding the total magnetic field at any point in space due to a current element. The total magnetic field due to this element is obtained from the integral of equation 3.12 as ⃗⃗⃗⃗ ∫ (3.13) Note: For a current carrying conductor the field is approximately given by (3.14) Example: If a very long wire carries a current of 1.0A, at what distance from the wire is the magnetic field produced by the current equal to ? Page 39 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse Solution ( )( ) ( ) 3.5.3 Ampere’s law So far, we have used Biot-Savart law to determine magnetic fields of different sources. Another approach is to use Ampere‘s law to find magnetic flux around a current carrying conductor of any shape. Ampere‘s law state that for a continuous closed line or loop drawn round one or more current carrying conductors, the line integral of magnetic field is equal to the product of the enclosed current and permeability of free space. ∮ (3.15) Applications of Ampere’s law to long straight wire Fig. 3.4: Consider a long straight conductor carry current as shown in fig. 3.6 above, the flux lines are circles centred at the conductor as shown. At every point on the enclosed line, B is tangent to a small element dl such that we can write (3.16) Taking the radius to be r, then the total length of the circle is Applying Ampere‘s law we have as follow ∮ Page 40 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse (3.17) 2. Ampere’s law for a solenoid Consider a circular line enclosed by a solenoid of N turns as shown in fig. 3.5. The field is same every at any point taken on the circle of total length L. Fig. 3.5 Using Ampere‘s law, ∮ ∮ (3.18) is the number of turns per unit length. 3.6 Magnetic force on a moving charge particle in a magnetic field A charge particle moving in a magnetic field B will experience a magnetic force. This force has four basic characteristics namely- 1. The magnitude of the force is proportional to the charge | | 2. The magnitude of the force is also proportional to the strength of the field B 3. The force is dependent on the velocity of the charge in the field and 4. The direction of the force is always perpendicular to the direction of the field and velocity of the charge as stated in Fleming Right Hand Rule. Fleming Right Hand Rule state that when the thumb and the fore fingers of the right hand are arranged at right angle to each other and the fore fingers are curl in the direction of the rotation of ⃗ to ⃗⃗ , the thumb will point in the direction of the force. Page 41 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse These four properties can be put together into an equation describing the force experienced by a charged particle in a magnetic field as | |⃗ ⃗ (3.19) Where ⃗ ⃗ defines the cross product of the two vector defined as ⃗ ⃗ The magnitude of the force is therefore given as (3.20) are magnitude of the charge, velocity, and field respectively. is the angle measured in the direction of ⃗ and ⃗. The implications of eqn. 3.20 is that when ; ⃗ and ⃗ are parallel so no force act on the charge and The charge experience the maximum force in the field with value as (3.21) 3.6.1 Motion of charge particle in a magnetic field From the four characteristics of force on a charge in a magnetic field, it is worthy of mention that if both the field and the velocity vectors are perpendicular to each other and the force always been perpendicular to the plane containing the two vectors of ⃗ and ⃗ , the charge particle will exhibit circular motion at constant radius R as shown in fig.3. 6. This motion can be describe using Newton‘s 2nd law of motion as Fig. 3.6 Page 42 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse | | (3.22) | | | | (3.23) Where R is the radius of the circular path, P is the momentum, m is the mass of the charge V is the velocity and | | is the magnitude of the charge. Example: A magnetron in a microwave oven emits electromagnetic waves with angular frequency. What magnetic field strength is needed for the electrons to move in a circular path? Solution From eqn. 3.23, ( )( ) | | | | 3.6.2 Magnetic force on a current carrying conductor Current is simply charges in motion in a conductor. When a conductor carrying current I is placed in uniform magnetic field, it will experience a force of magnitude F. This force is due to the resultant effects of the constituent charges flowing in the conductor of length and cross sectional area with a drift velocity ⃗⃗⃗⃗ as shown below + + + Fig. 3.7: magnetic force on a current carrying conductor The resultant force on each charge is given by | |⃗ ⃗ and magnitude Page 43 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse The total force on number of charges in the conductor is given by the force on the entire volume. The quantity of charges in the volume is as such the magnitude of the force on the conductor is ( ) (3.24) Where. If the field is not perpendicular to the conductor, the force will be given by ⃗ , with magnitude as (3.25) Note: For a case of non-straight conductor, the force is obtain by taking the integral of the force due to infinitesimal segment of the conductor as ∫ ⃗ ∫ (3.26) Exercises 1. A wire 36 m long carries a current of 22 A from east to west. If the magnetic force on the wire due to Earth‘s magnetic field is downward (toward Earth) and has a magnitude of , find the magnitude and direction of the magnetic field at this location. 2. The magnetic force on a straight 0.15 m segment of wire carrying a current of 4.5 A is 1.0 N. What is the magnitude of the component of the magnetic field that is perpendicular to the wire? Page 44 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse CHAPTER 4 ELETROMAGNETIC INDUCTION 4.1 Introduction Based on the fact that electric current produces magnetic field in a circuit or in a current carrying conductor, and the fact that electric circuits uses either electric power supply or a battery to create a potential difference within the circuit and the electric field associated with that potential difference causes charges to moves through the circuit and create a current. This leads one to think whether it is possible to produce an electric current using only wires and no battery or power supply and whether magnetic field will produce currents. Electromagnetic induction is the process of producing[inducing] a current in closed electric circuit by changing the magnetic field that passes through the circuit when a magnet and the circuit move relative to each other. 4.2 Faraday’s Law of Magnetic Induction States that “the induced EMF [E] in a circuit is directly proportional to the time rate of change of magnetic flux through the circuit” Mathematically written as (4.1) Where is the induced emf, is the magnetic flux, t is the time. For that pass through a finite area A we have (4.2) where is the angle between. The total magnetic flux through a finite area is the integral of this expression over the area A; ∫ (4.3) If B is uniform over a flat area A, then (4.4) If for example the magnetic field B has a constant magnitude and is everywhere perpendicular to the surface area of the coil, then equation 4.4 becomes (4.5) Now from the definition of Faraday‘s law, the mathematical expression of this law will be given by Page 45 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse (4.6) (4.7) where C is a positive constant. Explanation beyond the scope of this book revealed that the constant C in equation has numerical value equal to 1. Therefore, equation can also be written as (4.8) This is the equation representing Faraday‘s law of induction. Thus, Faraday‘s law can be stated as follows: The induced e.m.f in a circuit is equal to the negative of the time rate at which the magnetic flux through the circuit is changing. If the magnetic flux is in weber (Wb) or Tesla meter square (T-m2), the e.m.f will be in volts. The negative sign appears due to the direction of the induced EMF, which was not explained by Faradays, but latter explained by Lenz using law of conservation of energy principle (Lenz‘s law). 4.2.1 Lenz’s Law Lenz‘s law states that “An induced current has a direction such that the magnetic field due to the current opposes the change in the magnetic flux that induces the current”. For a coil of N turns, an e.m.f will appear in every loop and these e.m.fs are to be added. If the loops in the coil are assumed to have the same area then the flux through each turn will be the same. Thus, the induced e.m.f in such coils is given by ( ) , (4.9) It should be noted that the induced e.m.f, would be zero if the rate of change of magnetic flux is zero. Example 4.1. A coil with 25 turns of wire is wrapped around a hollow tube with an area of 1.8m2. Each turn has the same Area as the tube. A uniform magnetic field is applied at a right angle to the plane of the coil. If the field increase uniformly from 0.00T to 0.55T in 0.85s, find the magnitude of the induce EMF in the coil. If the resistance in the coil is 2.5Ω, find the magnitude of the induce current in the coil. Exercises 1. Circular coil of copper wire has 750tuns and radius of 3.00cm is placed between N and S pole of an electro magnet. Consider that the magnetic field is uniform and at an angle of 450 with plan of the coil. If the magnitudes of the magnetic field decrease at rate of 0.25Ts-1 determine the magnitude of the resulting induce EMF. Page 46 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse 2. A coil with 205turns of wires, a total resistance of 23Ω and across sectional area of 0.25m2 is positioned with its plane perpendicular to the field of a powerful electromagnet. What average current is induced in the coil during the 0.25s that the magnetic field drops from 1.6T to 0.0T? 4.3 The Transformer Is a device whose operation depends on Faraday‘s law of magnetic induction. It transforms voltages [alternative emf] from one value to another. A transformer consists of two coils of wire wound around a core of soft Iron. The coil on the left has Np turns and is connected to the A.C. source is called a primary coil and the coil on the right has Ns turns and is connected to the output voltages is called the secondary coil as shown below. source Load Vp Np Ns Vs Fig 4.3 transformer When an alternating current flows through the primary coil it gives rise to a change in magnetic flux around the primary coil and hence around the secondary coil. The changes of magnetic flux in the secondary coil then give rise to an induced EMF in the secondary coil. The induced EMF in the two coils may be obtained from Faraday‘s law as; (4.10) where Ns is the number of turns in the secondary coil. Page 47 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse Consider that there is no power loss in the primary winding, i.e. the resistance of the coil is negligible, then the change in magnetic flux will also produce a back e.m.f in the primary winding which is equal in magnitude to the applied alternating voltage. This is given by (4.11) where Np is the number of turns in the primary windings. Now dividing equation 4.10 by 4.11 yields (4.12) Thus, a transformer increases or decreases alternating voltages according to its turns ratio given by secondary voltage secondary windings primary voltage primary windings We can conclude that the EMF of a transformer depends on the number of turns in the primary and secondary coil because the strength of the magnetic field in the iron core and the cross sectional area of the core are the same for the two coils. However when Ns is greater than Np the secondary EMF is greater than that of the primary and the transformer is called step-up transformer. When Ns is less than Np the secondary EMF is less than that of the primary and the transformer is called the step-down transformer. For an ideal transformer we assume that the internal resistance is negligible that is there is no loss of energy. Thus the electrical output power is equal to the input power and the efficiency is nearly 100% then. Example 4.2: A step-up transformer is used on a 120V line to provide a potential difference of 2400V. if the primary coil has 75tuns how many turns must the secondary coil has? Example 4.3: A portable X-ray unit has a step-up transformer, the 120V input of which is transformed to the 100KV output needed by the X-ray tube. The primary has 50loops and draws a current of 10.00A when in use. a)What is the number of loops in the secondary? b)Find the current output in the secondary. Exercises 1. A step down transformer providing electricity for a residential neighborhood has exactly 2680 turn in its primary when the potential difference across the primary is 58sovs the potential difference at the secondary is 120v. How many turns are in the secondary? 2. A step up transformer for long range transmission of electric power is used to create a potential difference of 119340v across the secondary. If the potential difference across the primary is 1172v and the number of turns in the secondary is 25,500, what is the number of turns in the primary? Page 48 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse 4.4 Electric Generators These are machines which convert mechanical energy to electrical energy. There exist two types of generators: A.C generators and D.C generators i A.C Generator (Alternator) An alternator consists of a rectangular coil of wire (armature) placed between two poles of electromagnet in such a way that it can rotate around the horizontal axis so that it cuts the magnetic lines of force and develop a sinusoidally varying emf. This is taken off for use with the help of graphite blocks called brushes, which press against two copper slip-rings. ii D.C Generator (Dynamo) A Dynamo is similar to the alternator except that a commutator is used instead of the slip-rings. The commutator consists of ring of brass or copper which has been split into two semicircular segments carefully insulated from each other, the terminal of the armature coil are each soldered to one segment The carbon brushes are arranged so that the change-over of contact from one slit ring to the other occurs when the coil is vertical. In this, the induced emf in the coil reverses and so one brush is always positive and the other negative. The emf at the brushes is unidirectional and so it is a D.C voltage. 4.5 Electric Motors Is a device that converts electrical energy into a rotational kinetic energy by the action of the force on a coil pivoted in a magnetic field. It can be an A.C. motor or a D.C. motor. A motor is always identical in construction to a D.C generator. The coil of wire is mounted on a rotating shaft and is positioned between the poles of a magnet. Brushes makes contact with the commutator which alternates the current in the coil. This alternation of the current causes the magnetic field produced by the current to regularly reverse and thus always repelled by the fixed magnetic field, thus the coil and the shaft are kept in continuous rotational motion. 4.6 Eddy Current Eddy currents are induced electric currents within the conductor (larger than a piece of wire) used as a very effective electromagnetic brake. These circulating eddy currents have inductance and thus induce magnetic fields. These fields can cause repulsion, attraction, propulsion, drag and heating effects. The stronger the applied magnetic field, the greater the electrical conductivity of the conductor and the faster the field changes the greater the currents that are developed and the greater the field produced. Page 49 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse Energy is lost in the form of heat in the iron core due to eddy current such loss can be reduced by laminating the core. Lamination of the core reduces eddy currents by breaking up their path of flow. 4.7 Self Inductance Is the property of a conductor which enables it to induce an emf when the current in the circuit changes. Consider a coil with N turns carrying a current I as shown below; L I K E Fig. 4.7 self inductance coil Suppose a current I is passing through a coil or solenoid containing N loops. Because of this current, a magnetic flux Ф passes through each turn. If the current is changing the resulting magnetic flux will also be changing and so an emf will be induced in the coil itself. Such an emf is called self-induced emf or the back emf. The electromagnetic induction process is called self-induction and the coil is called an inductor with self-inductance “L”. According the Faraday‘s law, the induced e.m.f produced in each loop is given by (4.13) where is the rate at which the magnetic flux through the loop is changing. This is called back e.m.f. If the magnetic flux is the same through all the N loops, the total back e.m.f is given by ( ) (4.14) The quantity is the total flux enclosed by the circuit and is called flux linkage. This quantity for a given coil must be proportional to the current I provided no magnetic material such as iron is near it. So that Page 50 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse (4.15) (4.16) where L, the proportionality constant, is called the self-inductance of the coil. Therefore, self- inductance L is defined as ‗flux linkage in the coil per unit current in the same coil.’ (4.17) Substituting equation 4.17 into equation 4.14 will yeild ( ) (4.18) Thus, for coils of any shapes and sizes, whether they are closed packed or not and whether iron or any magnetic materials is nearby or not, the self - inductance is defined from equation 4.18 as (4.19) We can define self-inductance of a coil or circuit as the magnitude of the self-induced e.m.f per unit rate of change of current. If nor iron or similar magnetic materials are nearby, the magnitude of L depends only on the geometry of the coil. The important characteristic in an inductor is the presence of a magnetic field analogous to the presence of an electric field in a capacitor. The unit of inductance will be volt – seconds per ampere or ohm – seconds. The S.I. unit of inductance is Henry, H. Therefore, 1 henry = 1 volt-seconds/ampere = 1 ohm-seconds The unit milli Henry 1mH 10 3 H , and micro Henry 1 H 10 6 H are commonly used. 4.8 Mutual Inductance Consider two coils, coil 1 and coil 2 arranged as shown below I N1 N2 A. C Page 51 of 74 Coil 1 Coil 2 Fig 4.8 mutual inductance coil PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse The a.c. current in coil 1 set up time varying magnetic flux, some of which passes through coil 2 and induced an emf in coil 2. From Faraday‘s law (4.20) Since B is proportional to I, then is also proportional to I1. Thus (4.21) Therefore (4.22) where M21 is the mutual inductance of the two coils. And has similar unit to that of self- inductance. Differentiating eqn 4.23 above yields ( ) (4.23) Therefore applying Faraday‘s law we have (4.24) Page 52 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse CHAPTER 5 ALTERNATING CURRENT CIRCUITS 5.1 ALTERNATING CURRENT CIRCUITS An A.C Current/Voltage is the one which varies with time about a mean value. i.e an oscillating function of time. The usual diagram of an A.C source is shown in fig 4.90 fig. 5.1 A.C source The emf that is produced/induced when a coil of wire rotates with constant angular speed ω in a uniform magnetic field and angular frequency f is a sinusoidal alternating voltage. Alternating voltages produces alternating currents, which has the same graph as that of voltage as shown in fig 5.1 The instantaneous voltage V and current developed has the form (5.1) Where V and Io are the peak/maximum values called the amplitude of the voltage in Volts and current in Amperes and is the angular speed and rad/s, and f is the frequency in Hertz. Not only rotating coils do produce alternating voltages but also electronic devices called oscillators are common sources of alternating voltages. crest r.m.s value Vr.m.s trough Peak value Fig. 5.2 sine wave diagram Page 53 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse The diagram in figure 5.2 represents the variation of voltage with time, the voltage is not constant but alternates sinusoidally between two extremes. NB: Terms used in describing Alternating Sources i. The Period T is the time taken for one complete oscillation. ii. The Frequency f is the inverse of the period (1/T) iii. The average value of the current or voltage is zero as much as the curve lies below the central line as above it. iv. The Mean Value or Half Cycle Average this is the average value taken over half a cycle. v. The Peak Value is the distance from the average value to the crest or trough of the wave. This is clearly also the amplitude of the wave. vi. The Root Mean Square Value is the value of the alternating current which is equal to that value of a steady/direct current which has the same heating effect as the alternating current. OR The square root of the mean value of the square of the current taken over a whole cycle. Suppose an alternating voltage is applied across a resistor of resistance R thereby carrying alternating current and a direct current (d.c) applied through another resistor of the same resistance. Assuming that the two equal resistors are dissipating the same power, P as heat. The root mean square value of the alternating current I ms , is said to be equal to direct current. The power dissipated by the direct current and hence the a.c. circuit would respectively be 2 P Idc R (5.2) and 2 P I rms R Also if I is the value of the a.c. current at any instant, the power delivered to the resistance R at that instant is I 2R. Thus, average power delivered will be given by P average value of I 2R R average value of I 2 But the resistance R is constant. Therefore, 2 P I rms R R average value of I 2 2 I rms average value of I 2 But average value of I2 is given by 2 I Average value of I 2 2 The above substitution gives 2 Io2 I rms Average value of I 2 2 I 02 I0 I rms 0.707I 0 (5.3) 2 2 Page 54 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse Similarly, the root mean square value of alternating voltage can be found to be V Vrms 0.707V (5.4) 2 5.2 Ohm’s Law in Alternating Current Circuit (i) A Circuit containing a Pure Resistance Suppose a sinusoidal alternating current passes through a circuit of pure resistor as shown in fig. 4.92. R I Fig. 4.92 resistor circuit The current I passing through the circuit is (5.5) From Ohm‘s law the voltage becomes (5.6) From above V0 I 0R is the maximum voltage or peak voltage and Io the maximum current. Example 5.1: An A.C. voltage of 4V peak (maximum) is connected to a 100Ω resistor R. (a)What is the phase of the current and voltage? (b) Calculate the current in R in mA? (c) What is the power in R in mW? Solution (a) The current and voltage are in phase √ (b) √ (c) 5.3 A.C Circuit containing a capacitor; Suppose a sinusoidal alternating voltage source is connected across a capacitor of capacitance C as shown in fig. 5.3. A. C Page 55 of 74 Fig. 5.3: AC circuit containing a capacitor PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse The output voltage of the A.C source is given by V V0 sin t (5.7) At the instant A.C source is connected to the capacitor, the capacitor begins to charge. Before the capacitor is fully charged, the alternating voltage reverses its direction thereby discharging the capacitor. The continuous charging and discharging of this capacitor results in the flow of an alternating current in the circuit. The instantaneous voltage across the capacitor is given by Q (5.8) VC C Then the charge on the capacitance is Q CVC (5.9) And the current flowing through the capacitance is dQ (5.10) I dt Differentiating 4.64 with respect to time yields ( ) (5.11) dQ But is the current I and from trigonometry identities dt ( ) Therefore, ( ) ( ) (5.12) Where I 0 CV0 is the peak value of the current. It should be noted that the current flowing into the capacitor leads the voltage across it by 90 ; this is the same as saying that the peak current arrive a quarter period earlier than when the voltage reaches its maximum value. 1 V0 I C 0 1 Vo the proportionality constant has to do with the resistance is called capacitive C Io reactance, XC of the capacitor Thus, and therefore, Page 56 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse V0 I 0XC (5.13) Equation 5.13 is of the same form as Ohm‘s law V IR. Therefore the reactance XC in alternating current circuit plays the same role as resistance in d.c circuit, that is, it impede the flow of current in the circuit. It is defined as the peak a.c. voltage divided by the peak a.c. current. Example : An alternating voltage of peak value 220V is applied across 4 F capacitor. (i) Determine the capacitive reactance of the circuit if the frequency f of the source is 50Hz. (ii) What is the peak current flowing into the capacitor. Example: A capacitor C of 1µF is used in radio circuit where the frequency is 1000Hz and the current flowing is 2mA(R.M.S.). (i) Calculate the voltage across C. (ii) What current flows when an a.c. voltage of 20V (R.M.S.), is connected to this capacitor? NB: REACTANCE is the effective resistance of a component to a.c i.e the measure of opposition of a component to an a.c. Amplitude of the voltage across a component Denoted by X given as X Amplitude of the current flowing through it 5.4 Alternating Voltage across an Inductor A. C Fig.5.4 circuit containing an inductor Assuming that the alternating voltage source is now connected across an inductor of inductance L as shown in figure 5.4 from the figure immediately the a.c voltage source is connected to the inductor, current will start to flow through the inductor. This will give rise to an induced e.m.f in the coil. The current I through the inductor L is I I 0 sin t Page 57 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse From self inductance the back emf ϵ is ( ) (5.14) From trigonometric identities cos A sin A 90 We can write 5.14 as ( ) (5.15) The inductor has zero resistance, for current to flow in the circuit the voltage must be equal and opposite to ϵ. Therefore ( ) (5.16) but ( ) (5.17) where V0 is the maximum current in the inductor. Equation 4.47 shows that the current I I0 L lags the applied voltage by 90 or the voltage leads the current by 90. From eqn. 5.17 has to do with the resistance and is called the inductive reactance of the inductor. It also opposes the flow of current in the circuit. It is measured in Ohms Example: An inductor of 2H and negligible resistance is connected to a 12V (R.M.S.) mains supply, f = 50Hz. Find the current flows when the inductance is changed to 6H? Solution Given: When the inductance is increase to 6H, its reactance is increased three (3) times since Page 58 of 74 PHY 102: Electricity, Magnetism and Modern Physics @ Physics Dept, Federal University Dutse For a given frequency. So the current is reduced to one-third (1/3) of its value. So 5.5 LR Series Circuit Suppose an inductor of inductance L be in series with resistance R and the combination connected across an alternating voltage source of RMS voltage V and frequency f as shown in figure 5.5 VR VL R L VL V θ A.C VR

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