Revision Compare Between. Application

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1 evision Compare etween Points of Comparison Series Connection Parallel Connection Drawing otal resistance ( ) = Potential Difference () = = = = 3 Electric Current (I) I = I = I = I 3 I = I + I + I 3 Subject Application Application. Series connection:. Parallel connection: 3. Ampere s right hand rule: 4. ight hand screw rule: 5. Flemming s left hand rule: Obtain a large resistance out of a bunch of small resistances. Obtain a small resistance out of a bunch of large resistances. It is used to determine the direction of the magnetic field when passing electric current in a straight wire. It is used to determine the direction of the magnetic field when passing electric current in a circular loop. It is used to determine the direction of motion of a straight wire when passing through it an electric current and placed normally to a magnetic field.

2 Graphs (olt) (Ohm) (esla) = I Slope = = I () esistance of wire (esla) I (A) Slope = L A L = A (ρ) esistivity of material (A) Area of wire (esla) L (m) I d Slope = = I d (μ) Permeability of medium (d) Distance between point and wire F wire (N) I (A) Slope = NI r r = NI (μ) Permeability of medium (N) Number of turns /r (I) Electric current intensity Slope = NI L L = μni (μ) Permeability of medium (N) Number of turns /L (I) Electric current intensity I (A) F wire = IL Sinθ F Slope = = LSinθ I () Magnetic flux density (L) Length of wire (θ) Angle between wire & field

3 τ (J) Sin θ τ = IAN Sinθ Slope = = IAN Sin () Magnetic flux density (I) Electric current intensity (A) Area of coil (N) Number of turns Subject Factors Affecting On (Depends On) Factors Affecting On (Depends On). esistance of the conductor: () (at same temperature). Magnetic flux density due to straight wire: () 3. Magnetic flux density due to circular coil: () 4. Magnetic flux density due to solenoid: () 5. Force acting on a wire in a magnetic field:(f) 6. orque of couple on a rectangular coil carrying current in a magnetic field: (τ) Length ( L) Area ( /A) Kind of material (ρ) Electric current intensity ( I) Distance between wire and point ( /d) Number of turns ( N) Electric current intensity ( I) adius of coil ( /r) Number of turns ( N) Electric current intensity ( I) Length of solenoid ( /L) Magnetic flux density (F ) Electric current intensity (F I) Length of wire (F L) Angle between wire and field (F Sin θ) Magnetic flux density (τ ) Electric current intensity (τ I) Area of coil (τ A) Number of turns (τ N) Angle between perpendicular to coil and field (τ Sin θ) 3

4 Definitions Scientific Concept Definitions Unit. Electric current intensity: (I). Potential difference between two points: () 3. Electromotive force: (E.M.F.) 4. esistance of material: () 5. Ohm s law: 6. esistivity of a material: (ρ e ) 7. he electrical conductivity: (σ) 8. Kirchhoff's Law: 9. Magnetic flux density: () 0. esla: (). Neutral point:. Sensitivity of the galvanometer: (S) 3. he electromagnetic induction: 4. Induced electric current: (I ind ) 5. Faraday s law: It is the quantity of charges passed a given cross section in one second. It is the work done to transfer unit charge from one point to another. It is the total work done to transfer unit charge throughout the whole electric circuit outside and inside the source O. It is the voltage difference across the source when the current ceases to flow in the circuit. he resistance of a material is its opposition to the flow of electric current. he current intensity in a conductor is directly proportional to the potential difference across its terminals at constant temperature. It is numerically equal to the resistance of a piece of material of length one meter and cross-sectional area m. It is the reciprocal of the resistivity of the material. he total voltage ( ), which is equal to the sum of the voltage differences across the resistors in the series circuit. It is the number of magnetic flux lines passing normally through a unit area. It is the magnetic flux density, which will exert a force (N) on a wire carrying current (A) of length (m) perpendicular to the field It is the point at which the total magnetic flux density is equal to zero. ( = 0) It is the scale deflection per unit current intensity passing through its coil. It is a phenomenon in which an induced e.m.f. and also an induced current are generated in the coil on plunging a magnet into or withdrawing out of it. It is the current generated in the conductor in a closed circuit due to the variation of the magnetic flux link with the conductor. he magnitude of the induced electromotive force is proportional to the rate by which the conductor cuts the lines of the magnetic flux linked with it and the number of turns of the conductor which cut the magnetic flux 4 Ampere olt olt Ohm.m Simon.m.m Weber/m esla N/A.m Degree/A Ampere

5 6. Lenz s law: 7. Mutual Induction: 8. Mutual induction coefficient: (M) 9. Self induction coefficient: (L) 0. Henry: (H). Eddy current: he induced current must be in a direction such as to oppose the change producing it. It is the electromagnetic interaction between two coils kept close to each other (or one inside the other). An electric current with time varying intensity passing in one coil (primary coil) will produce in the second one (secondary coil) an induced current in a direction such that to oppose the variations of the current intensity in the primary coil. It is the electromotive force induced in the secondary coil when the current passing through the primary coil changes at a rate equals one ampere per second. It is the electromotive force induced in the coil when the current passing through it changes at a rate equals one ampere per second. It is the self-inductance of a coil in which an e.m.f. of one volt is induced when the current passing through it changes at a rate one ampere per second. hey are induced currents that circulate in closed paths due to the change in magnetic flux through a solid conductor associating with heating effect. Give eason For.sec/A Ohm.sec Henry.sec/A Ohm.sec Henry () When the length of a wire increases its resistance also increases. L, resistance of wire is directly proportional with its length. A () he home electrical devices are not connected in series. As when connected in parallel, the current is divided on them, so when current is cutoff in one device, the other devices is still working, and do not be affected. Also parallel connection decreases the total resistance. (3) If the filament of a lamp in the house is cut off, the other lamps still lighten. As the lamps in the house are connected in parallel so the current flowing in each is different and if one of them is cutoff the others will be working. (4) When the total power of the electrical instruments used in houses exceeds a certain value the electric current intensity flowing through the fuse increases. As the power (P = I) and since the voltage of the electrical instruments in houses are constant so the total power is directly proportional to the current intensity. (5) If the electric circuit is switched off, the potential difference between the two poles of the electric source equals its E.M.F. = Ir, when circuit is switched off (I = 0), so =. 5

6 (6) A straight wire carrying a current placed in a magnetic field, wire did not move. ecause the straight wire is placed in parallel with the magnetic field. F = ILSin, = 0 so F = zero. (7) he magnetic flux density along the axis of a solenoid increases if we insert inside it a bar of soft iron. As it increases the permeability of the solenoid and concentrates the magnetic lines inside it, thus the magnetic flux density increases. (8) If an electric current flows through both a solenoid and a straight wire coinciding with the axis of the coil, there is no magnetic force acting on the wire. As the wire is placed at the axis of the solenoid parallel to the magnetic flux lines of the solenoid, so there is no force acting on the wire. F = ILSin, = 0 and F = zero (9) A rectangular coil carrying current is placed in a perpendicular position to a uniform magnetic field, it does not deviate. τ = IAN Sinθ, so when coil is perpendicular to the field, θ = 0, Sin 0 = 0 so τ = 0. (0) he couple acting on rectangular coil carrying a current between two magnetic poles, decreases as the coil starts rotating from the position in which the plane of the coil parallel to the magnetic field. τ = IAN Sinθ and when coil is parallel to the field, θ = 90, Sin 90 = so torque is maximum and as the coil rotates, sine the angle decreases so the torque decrease. 6

7 Proofs () Series Connection ut = = I = I = I 3 = I 3 I = I + I + I 3 3 () Parallel connection: I = I + I + I 3 And I, I, I, I (3) Force due to magnetic field acting on straight wire carrying current Magnetic flux density where (F ). Current intensity passing through the wire where (F I). Length of the wire where (F L). F IL F = constant x I L. (constant equal to unity) F IL 7

(4) he force between two parallel wires each carrying current Magnetic flux density of first wire: Force acting on second wire: Substituting with the magnetic flux density: I d I L F I F IL d I I L d

both (bc) and (ad) are parallel to the flux lines. he force acting on each wire is zero. As to both (ab) and (cd), they are perpendicular to the flux lines.

8 (4) he force between two parallel wires each carrying current Magnetic flux density of first wire: Force acting on second wire: Substituting with the magnetic flux density: I d I L F I F IL d I I L d F (5) Force and orque acting on a rectangular coil carrying current placed in a magnetic field If we have a rectangular coil (abcd), whose plane is parallel to the lines of a uniform magnetic flux, both (bc) and (ad) are parallel to the flux lines. he force acting on each wire is zero. As to both (ab) and (cd), they are perpendicular to the flux lines. hey will be acted upon by two forces equal in magnitude and opposite in direction according to Fleming L.H. and are parallel, each equal to F = I L, separated by a perpendicular distance (W). hus, the coil is acted upon by a torque which will cause the coil to rotate around its axis. he magnitude of the couple (torque) is equal to the force magnitude times the perpendicular spacing between the two equal forces: orque (couple of two forces) = force (one of them) x distance (between them) F W IL W ut A = L x W, where (A) is the area of the loop IA If the loop is replaced with a closely wound coil having (N) turns of wire IAN m d (N.m) Where md IAN is the magnetic dipole moment (A.m ) 8

9 In parallel connection: P Essay P ( is constant) he resistances connected in parallel, the smallest resistance consume largest power. In series connection: P = I P α (I is constant) he resistances connected in series, the largest resistance consume smallest power. ypes of Instruments () Analog multi-meter: hese instruments use pointers. () Digital multi-meter: hese instruments depend on reading numerals. Conclusion of Faraday s Experiments () Electromotive force induced in the conductor direction depends on the direction of motion of the conductor relative to the field. () he magnitude of the induced electromotive force is proportional to the rate by which the conductor cuts the lines of the magnetic flux linked with it i.e. emf t m, where (m) is the variation in the magnetic flux intercepted by the conductor through the time interval ( t). (3) he magnitude of the induced electromotive force is proportional to the number of turns (N) of the coil which cut (or link with) the magnetic flux. i.e. emf N Mutual induction between two coils o produce (emf) on secondary coil by mutual induction: () Plunge or take away the primary coil from inside the secondary coil. () Using rheostat, increase or decrease the intensity of the current in the primary coil. (3) Using switch, switch-on or switch-off the primary circuit. 9

Self induction When the switch is closed, backward induced emf is produced and the lamp will not glow. When the switch is opened, forward induced emf is produced and the lamp will glow.

10 Self induction When the switch is closed, backward induced emf is produced and the lamp will not glow. When the switch is opened, forward induced emf is produced and the lamp will glow. Eddy Currents Disadvantages of eddy currents: () Some of the electric energy is wasted and dissipated as heat. () he spoil of the coil surrounding it, due to this generated heat. Minimizing eddy currents: () he iron core made of high resistance. () he iron core is made up of thin sheets or wires with an insulating material between them to make the eddy current flow in smaller path and this decreases the heat effect. 0

11 What is meant by each of the following ) A current intensity in a conductor = 00 ma? It means that an amount of electric charge of 00 milli coulomb passes through the crosssection of the conductor in one second. ) he potential difference between two points =0 volts? he work done to transfer a charge of coulomb between the two points =0 Joules. 3) he electromotive force of a battery = volts? he total work done to transfer a charge of coulomb inside and outside the battery = Joules. 4) he resistance of a conductor = 60 Ω? he ratio between the potential difference across the conductor and the current intensity passing through it is 6 /A. 5) he work done to transfer coulomb between points in an electric circuit = 4 J. he potential difference between the two points = 4 volts. 6) he work done to transfer a charge of 3 coulombs between points = 8J. he potential difference between the two points = 6 volts. 7) he electric resistivity of copper = Ω.m? he resistance of a copper conductor of length m and cross-section area m is Ω.m. 8) he electric conductivity-of silver = 6 l0 7 simon.m -? he reciprocal of resistivity of silver = simon.m. O: he resistance of a silver conductor of unit length and unit cross-section area = 6x0 7 9) A wire of length m and cross-section area m has a resistance = Ω? he resistivity of the wire material = Ω.m. 0) he ratio between the potential difference across a conductor and the current intensity passing through it is 0 /A? he resistance of this conductor =0 Ω.

12 ) he potential difference across an electric cell when no current flows = 9? he electromotive force of this cell = 9 volts. ) Magnetic flux cutting an area = 0 webers? he total number of magnetic flux lines cutting perpendicular this area =0. 3) Magnetic flux density at a point = 0.esla he number of magnetic flux lines passing normally through unit cross section area 4) he magnetic flux density = 0.5 N/A. m a force acting on a wire of length m carrying a current of intensity amp placed to that M.F = 0.5N 5) Ampere s right and rule It is used to determine the direction of the magnetic field when passing electric current in a straight wire. 6) ight hand screw rule It is used to determine the direction of the magnetic field when passing electric current in a circular loop. 7) Fleming s left hand rule It is used to determine the direction of motion of a straight wire when passing through it an electric current and placed normally to a magnetic field. 8) he coefficient of mutual induction between two coils is 0. Henry 0.olt is the electromotive force induced in the secondary coil when the current passing through the primary coil changes at a rate equals one ampere per second. 9) he coefficient of self induction of a coil is 0.5 Henry 0.5olt is the electromotive force induced in the coil when the current passing through it changes at a rate equals one ampere per second.

13 Solved Problems ) When a current of Ampere flows through an unknown resistor, the potential difference across it will be l0volts. Calculate the resistivity and the electrical conductivity of the material of such resistor if its length is meters and its crosssectional area is 0.cm. = I. 0 = = 5Ω 4 A 5 0.x0 e ρ e = Ω.m L σ = Ω -.m - 5 e.5x0 ) A student made two resistors from the same material; the second is double the first in length and its diameter is equal to the radius of the first. Calculate the ratio of the second resistance to first one. ρ = ρ L = L L L r d = r r = r r L L r r 8 8 3

14 3) In show figure, find the change in the reading of the ammeter and the voltmeters in the circuit when the variable resistance ( ) increases. (ariable resistance called rheostat) I r I he reading of ammeter (A) decreases When resistance ( ), the current (I) he reading of the voltmeter ( ) decreases = I. r When the current (I), while () constant, the voltage () he reading of the voltmeter ( ) increases = I. When the current (I), while (), the voltage () Ex: If = = r = 0 Ω. So + + r = 30 Ω And if ( ) increases 4 times where ( ) = 40 Ω. So + + r = 60 Ω When the total resistance increases times, the current (I) decreases to ½ its value he reading of the voltmeter ( 3 ) increases 3 = I.r When the current (I), while () constant, (r) constant, the voltage (3) 4) A cell of E.M.F. volts and internal resistance of 0.Ω is connected to a circuit of external resistance 3.9Ω. Calculate the current flowing through the circuit. I r I I = 0.5A

15 5) hree resistors of 5Ω, 70Ω and 85Ω are connected in series to a 45volt battery of negligible internal resistance. Calculate the current flowing in each resistor and the terminal potential difference of each. = = 80Ω I r 0 45 I = 0.5A 80 0 I battery = I 5Ω = I 70Ω = I 85Ω 5Ω = I. = 0.5 x 5 = Ω = I. = 0.5 x 70 = Ω = I. = 0.5 x 85 =.5 = 5Ω + 70Ω + 85Ω = = 45 6) If the resistors in the previous problem are connected in parallel to the same battery. Calculate: (a) he total resistance (b) he current through the circuit (c) he current flowing in each resistor a) b) I = 5.4Ω r 45 I I =.97A 5.4 c) battery = 5Ω = 70Ω = 85Ω 45 I5 =.8A 5 45 I70 = 0.643A I85 = 0.59A 85 I = I 5Ω + I 70Ω + I 85Ω = =.97A 5

16 7) In the circuit shown in figure. Calculate: (a) he total resistance. (b) he current flowing through the circuit. (c) he current flowing through each of (A) and (). 3 6 a) 3, = + 7 = 9Ω b) I r 8 I A 9 0 c) 3Ω = 6Ω = I. (3,6) 3Ω = 6Ω = x = 4 3 I3 4 I A 6 I6 4 I A O I I I A 3 6 I I I A 3 6 8) Find the current through battery. (Internal resistance of each battery = 3Ω) 8Ω 9Ω 6 4Ω 4Ω 7Ω 5Ω 6

17 98 9, (5,7) = = Ω 4, (4,6,8) = = 8Ω I I r 6 r r A 9) Calculate the current flowing through each resistance and also the potential difference across each of them , (60,0) = = 80Ω I r 40 I 3A 80 0 (80) = I = 3 x 80 = 40 (0) = I = 3 x 0 = 60 (60) = I = 3 x 60 = 80 0) Determine the magnetic flux density at a distance of 0cm in air from the center of a long wire carrying a current of 0A (air = 4πx0-7Wb/Amp.m) 7 I 4x0 0 = x0-5 esla d Ω 75 Ω 300 Ω 7

18 ) he given figure represents cross-section of two straight parallel conductors carrying currents as shown in the fig. Calculate the total magnetic flux density at the points (a, b and c). ( = 4πx0-7 Weber/Amp.meter). At point (a): 7 I 7 0 x0 x0 = x0-4 esla d I 7 30 x0 x0 = 5x0-5 esla d 0. = = x0-4 5x0-5 At point (b): =.5x0-4 7 I 7 0 x0 x0 = x0-4 esla d I 7 30 x0 x0 = x0-4 esla d 0.06 = + = x0-4 + x0-4 At point (c): = x0-4 7 I 7 0 x0 x0 = 3.077x0-5 d I 7 30 x0 x0 = x0-4 esla d 0.03 = = x x0-5 =.7x0-4 ) Determine the magnetic flux density at the center of a circular loop of radius cm carrying a current.4amp. If the wire consists of a coil having 0turns and air = 4πx0-7 Wb/A.m. 7 NI 4x0 0.4 = 6x0-5 esla r 0. 8

19 3) A long solenoid has 800turns. he current flowing through the wire is 0.7Arnpere. Find the magnetic flux density in the interior of the solenoid. Knowing that is length is 0cm. NI 4x0 L = 3.5x0-3 esla 4) A solenoid is constructed by winding 800 turns of wire on a 0cm iron core. What current is required to produce a flux density of 0.85esla in the center of the solenoid? he permeability of iron is.63x0 - Wb/amp.m. NI L.63x0 800 I I = 0.05A 5) A circular coil of radius 0cm, of 50turns carrying a current of A, Calculate the magnetic flux density at its center? If the turns of the coil is a parted from each other regularly such that its length become l00cm. Calculate the magnetic flux density at the axis of the coil? If a bar of iron of permeability (0.0Wb/A.m) is linked with the coil, what is the change in the magnetic flux density at its axis? coil solenoid solenoid NI r x0 50 = 6.83x0-4 esla 7 NI 4x0 50 =.56x0-4 esla L NI = esla L 6) A wire 30cm long supports a current of 4amperes in a direction perpendicular to a magnetic field. If the force on the wire is 6Newton. Find the magnetic flux density. 9

20 F = IL Sinθ 6 = x 4 x 0.3 x Sin90 = 5esla 7) If the wire in the previous problem makes an angle of 30 with respect to the field. Find the force acting on the wire. F = IL Sinθ = 5 x 4 x 0.3 x Sin30 = 3N 8) A rectangular loop ( x 0) cm of 40turns carrying a current of A. Calculate the torque acting on the loop if it is placed in a magnetic field of density 3 such that: a) he plane of the coil is parallel to the magnetic flux lines. b) he plane of the coil is perpendicular to the magnetic flux lines. c) he plane of the coil makes an angle 60 with the magnetic flux lines. d) he perpendicular to its plane makes an angle 60 with the magnetic flux lines. a) τ = IAN Sinθ = Sin90 =.88 Joule b) τ = IAN Sinθ = Sin0 = zero c) τ = IAN Sinθ = Sin30 =.44 Joule d) τ = IAN Sinθ = Sin60 =.494Joule 0

1. An isolated stationary point charge produces around it. a) An electric field only. b) A magnetic field only. c) Electric as well magnetic fields.

1. An isolated stationary point charge produces around it. a) An electric field only. b) A magnetic field only. c) Electric as well magnetic fields. 2. An isolated moving point charge produces around it.