n Higher Physics 1B (Special) (PHYS1241) (6UOC) n Advanced Science n Double Degree (Science/Engineering) n Credit or higher in Physics 1A

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1 Physics in Session 2: I n Physics / Higher Physics 1B (PHYS1221/1231) n Science, dvanced Science n Engineering: Electrical, Photovoltaic,Telecom n Double Degree: Science/Engineering n 6 UOC n Waves n Physical Optics (light & interference) n Introduction to Quantum Physics n Solid State & Semiconductor Physics Physics in Session 2: II n Higher Physics 1B (Special) (PHYS1241) (6UOC) n dvanced Science n Double Degree (Science/Engineering) n Credit or higher in Physics 1 n Waves: interference, diffraction, polarization n Introduction to Quantum Mechanics n lternating Currents n The Sun and the Planets n Thermal Physics n Special Relativity Much smaller classes! Physics in Session 2: II n Energy & Environmental Physics n PHYS1211 (6UOC) / PHYS1249 (3UOC) n possible elective course? n Heat & Energy n Solar energy, alternative energy n Introductory quantum theory n Photovoltaic energy n Nuclear energy & radiation Physics / Higher Physics 1 Topic 3 Electricity and Magnetism Revision Electric Charges n Two kinds of electric charges n Called positive and negative n Like charges repel n Unlike charges attract Coulomb s Law n In vector form, F 12 =k e q 1 q 2 ˆ r r 2 n r ˆ is a unit vector directed from q 1 to q 2 n Like charges produce a repulsive force between them 1

2 The Superposition Principle n The resultant force on q 1 is the vector sum of all the forces exerted on it by other charges: F 1 = F 21 + F 31 + F 41 + Electric Field E F q e = = o q ke r r ˆ 2 n Continuous charge distribution q dq E = k lim rˆ = k rˆ i e 0 2 i e q 2 i i ri r Electric Field Lines Dipole n The charges are equal and opposite n The number of field lines leaving the positive charge equals the number of lines terminating on the negative charge Electric Flux Φ E =E i i cosθ i = E i i E Φ = lim E = d E i 0 i i surface Gauss s Law Φ E = E d = q in ε0 n q in is the net charge inside the surface n E represents the electric field at any point on the surface Field Due to a Plane of Charge n The total charge in the surface is σ n pplying Gauss s law Φ E = 2E = σ, and E = σ ε 0 2ε 0 n Field uniform everywhere 2

3 Properties of a Conductor in Electrostatic Equilibrium 1. Electric field is zero everywhere inside conductor 2. Charge resides on its surface of isolated conductor 3. Electric field just outside a charged conductor is perpendicular to the surface with magnitude σ /ε o 4. On an irregularly shaped conductor surface charge density is greatest where radius of curvature is smallest Electric Potential Energy n Work done by electric field is F. ds = q o E. ds n Potential energy of the charge-field system is changed by U = -q o E. ds n For a finite displacement of the charge from to B, the change in potential energy is U = U U = q E d s B o B Electric Potential, V n The potential energy per unit charge, U/q o, is the electric potential U B V = = d q E s o n The work performed on the charge is W = U = q V n In a uniform field B B V V = V = E ds = E ds = Ed B Equipotential Surface n ny surface consisting of a continuous distribution of points having the same electric potential n For a point charge q V = ke r Finding E From V n From V = -E.ds = -E x dx dv Ex = dx n long an equipotential surfaces V = 0 n Hence E ds n i.e. an equipotential surface is perpendicular to the electric field lines passing through it V Due to a Charged Conductor n E ds = 0 n So, potential difference between and B is zero n Electric field is zero inside the conductor n So, electric potential constant everywhere inside conductor and equal to value at the surface 3

4 Cavity in a Conductor n ssume an irregularly shaped cavity is inside a conductor n ssume no charges are inside the cavity n The electric field inside the conductor must be zero Definition of Capacitance n The capacitance, C, is ratio of the charge on either conductor to the potential difference between the conductors Q C = V n measure of the ability to store charge n The SI unit of capacitance is the farad (F) Capacitance Parallel Plates n Charge density σ = Q/ n Electric field E = σ/ε 0 (for conductor) n Uniform between plates, zero elsewhere C = Q V = Q Ed = Q ε = 0 Q ε 0 d d Capacitors in Parallel n Capacitors can be replaced with one capacitor with a capacitance of C eq n C eq = C 1 + C 2 Capacitors in Series n Potential differences add up to the battery voltage Q =Q 1 =Q 2 V = V 1 + V 2 V Q = V 1 Q 1 + V 2 Q 2 1 C = 1 C C 2 Energy of Capacitor n Work done in charging the capacitor appears as electric potential energy U: 2 Q 1 1 ( ) 2 U = = Q V = C V 2C 2 2 n Energy is stored in the electric field n Energy density (energy per unit volume) u E = U/Vol. = ½ ε o E 2 4

5 Capacitors with Dielectrics n dielectric is a nonconducting material that, when placed between the plates of a capacitor, increases the capacitance n For a parallel-plate capacitor C = κ C o = κ ε o (/d) Rewiring charged capacitors n Two capacitors, C 1 & C 2 charged to same potential difference, V i. n Capacitors removed from battery and plates connected with opposite polarity. n Switches S 1 & S 2 then closed. What is final potential difference, V f? Q 1i, Q 2i before; Q 1f, Q 2f after. Q 1i = C 1 V i ; Q 2i = -C 2 V i So Q=Q 1i +Q 2i =(C 1 -C 2 ) V i But Q= Q 1f +Q 2f (charge conserved) With Q 1f = C 1 V f ; Q 2f = C 2 V f hence Q 1f = C 1 /C 2 Q 2f So, Q=(C 1 /C 2 +1) Q 2f With some algebra, find Q 1f = QC 1 /(C 1 +C 2 ) & Q 2f = QC 2 /(C 1 +C 2 ) So V 1f = Q 1f / C 1 = Q / (C 1 +C 2 ) & V 2f = Q 2f / C 2 = Q / (C 1 +C 2 ) i.e. V 1f = V 2f = V f, as expected So V f = (C 1 - C 2 ) / (C 1 + C 2 ) V i, on substituting for Q Magnetic Poles n Every magnet has two poles n Called north and south poles n Poles exert forces on one another n Like poles repel n N-N or S-S n Unlike poles attract n N-S Magnetic Field Lines for a Bar Magnet n Compass can be used to trace the field lines n The lines outside the magnet point from the North pole to the South pole Direction n F B perpendicular to plane formed by v & B n Oppositely directed forces are exerted on charges of different signs n cause the particles to move in opposite directions 5

6 Direction given by Right-Hand Rule n Fingers point in the direction of v n (for positive charge; opposite direction if negative) n Curl fingers in the direction of B n Then thumb points in the direction of v x B; i.e. the direction of F B The Magnitude of F n The magnitude of the magnetic force on a charged particle is F B = q vb sin θ n θ is the angle between v and B n F B is zero when v and B are parallel n F B is a maximum when perpendicular Force on a Wire n F = I L x B n L is a vector that points in the direction of the current (i.e. of v D ) n Magnitude is the length L of the segment ni is the current = nqv D n B is the magnetic field Force on a Wire of rbitrary Shape n The force exerted segment ds is F = I ds x B n The total force is b d a F = I s B Force on Charged Particle n Equating the magnetic & centripetal forces: F =qvb = mv 2 r n Solving gives r = mv/qb Biot-Savart Law n db is the field created by the current in the length segment ds n Sum up contributions from all current elements I.ds B = µ 0 4π I ds ˆ r r 2 6

7 B for a Long, Straight Conductor B = µ 0 I 2πa B for a Long, Straight Conductor, Direction n Magnetic field lines are circles concentric with the wire n Field lines lie in planes perpendicular to to wire n Magnitude of B is constant on any circle of radius a n The right-hand rule for determining the direction of B is shown n Grasp wire with thumb in direction of current. Fingers wrap in direction of B. Magnetic Force Between Two Parallel Conductors F 1 = µ 0 I 1 I 2 2πa l n Parallel conductors carrying currents in the same direction attract each other n Parallel conductors carrying currents in opposite directions repel each other Definition of the mpere n The force between two parallel wires can be used to define the ampere F 1 l = µ 0 I 1 I 2 2πa with µ 0 = 4π 10 7 T m -1 n When the magnitude of the force per unit length between two long parallel wires that carry identical currents and are separated by 1 m is 2 x 10-7 N/m, the current in each wire is defined to be 1 mpere s Law n The line integral of B. ds around any closed path equals µ o I, where I is the total steady current passing through any surface bounded by the closed path. B ds = µ 0 I Field in interior of a Solenoid n pply mpere s law n The side of length l inside the solenoid contributes to the field n Path 1 in the diagram B ds = B ds = B ds = Bl path1 path 1 B = µ 0 N l I = µ 0 ni 7

8 mpere s vs. Gauss s Law B ds = µ 0 I E d = q ε 0 n Integrals around closed path vs. closed surface. n i.e. 2D vs. 3D geometrical figures n Integrals related to fundamental constant x source of the field. n Concept of Flux the flow of field lines through a surface. Gauss Law in Magnetism n Magnetic fields do not begin or end at any point n i.e. they form closed loops, with the number of lines entering a surface equaling the number of lines leaving that surface n Gauss law in magnetism says: Φ B = B.d = 0 Faraday s Law of Induction n The emf induced in a circuit is directly proportional to the rate of change of the magnetic flux through that circuit ε = N dφ B dt QuickTime and a Cinepak decompressor are needed to see this picture. Ways of Inducing an emf ε = d dt ( Bcosθ) n Magnitude of B can change with time n rea enclosed,, can change with time n ngle θ can change with time n ny combination of the above can occur Motional emf n Motional emf induced in a conductor moving through a constant magnetic field n Electrons in conductor experience a force, F B = qv x B that is directed along l n In equilibrium, qe = qvb or E = vb Sliding Conducting Bar n Magnetic flux is n The induced emf is ε = dφ B = d ( Blx)= Bl dx dt dt dt = Blv n Thus the current is I = ε R =Blv R Φ B =Blx 8

9 Induced emf & Electric Fields n changing magnetic flux induces an emf and a current in a conducting loop n n electric field is created in a conductor by a changing magnetic flux n Faraday s law can be written in a general form: ε = E.ds = dφ B dt n Not an electrostatic field because the line integral of E. ds is not zero. Generators n Electric generators take in energy by work and transfer it out by electrical transmission n The C generator consists of a loop of wire rotated by some external means in a magnetic field Rotating Loop n ssume a loop with N turns, all of the same area, rotating in a magnetic field n The flux through one loop at any time t is: Φ B = B cos θ = B cos ωt Motors n Motors are devices into which energy is transferred by electrical transmission while energy is transferred out by work n motor is a generator operating in reverse n current is supplied to the coil by a battery and the torque acting on the current-carrying coil causes it to rotate ε = N dφ B dt = NB d ( cosωt)=nbω sinωt dt Eddy Currents n Circulating currents called eddy currents are induced in bulk pieces of metal moving through a magnetic field n From Lenz s law, their direction is to oppose the change that causes them. n The eddy currents are in opposite directions as the plate enters or leaves the field Equations for Self-Inductance n Induced emf proportional to the rate of change of the current ε L = L di dt n L is a constant of proportionality called the inductance of the coil. 9

10 Inductance of a Solenoid n Uniformly wound solenoid having N turns and length l. Then we have: B = µ 0 ni = µ 0 N l I N Φ B =B = µ 0 I l L = NΦ B I = µ 0N 2 l Energy in a Magnetic Field n Rate at which the energy is stored is du di = LI dt dt I U =L IdI = 1 2 LI 2 0 n Magnetic energy density, u B, is u B = U l = B2 2µ 0 RL Circuit I = ε R 1 e Rt L = ε R 1 e t τ n Time constant, τ = L / R, for the circuit n τ is the time required for current to reach 63.2% of its max value 10