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Physics in Session 2: I n Physics / Higher Physics 1B (PHYS1221/1231) n n n n n n n n Science, Advanced Science Engineering: Electrical, Photovoltaic,Telecom Double Degree: Science/Engineering 6 UOC Waves Physical Optics (light & interference) Introduction to Quantum Physics Solid State & Semiconductor Physics Physics in Session 2: II n Higher Physics 1B (Special) (PHYS1241) (6UOC) n Advanced Science Double Degree (Science/Engineering) n Credit or higher in Physics 1A n n n n n n n Waves: interference, diffraction, polarization Introduction to Quantum Mechanics Alternating Currents The Sun and the Planets Thermal Physics Special Relativity Much smaller classes! Physics in Session 2: II n Energy & Environmental Physics n n PHYS1211 (6UOC) / PHYS1249 (3UOC) A possible elective course? n Heat & Energy n Introductory quantum theory n Nuclear energy & radiation n n Photovoltaic energy Two kinds of electric charges n n n Topic 3 Electricity and Magnetism Revision Solar energy, alternative energy Electric Charges n Physics / Higher Physics 1A Coulomb’s Law n Like charges repel Unlike charges attract In vector form, F12 = ke Called positive and negative n n q1q2 rˆ r2 rˆ is a unit vector directed from q1 to q2 Like charges produce a repulsive force between them 1 The Superposition Principle n Electric Field The resultant force on q1 is the vector sum of all the forces exerted on it by other charges: F1 = F21 + F31 + F41 + … E= n Fe q = ke 2 rˆ qo r Continuous charge distribution ∆qi E = k e lim ∆qi →0 Electric Field Lines – Dipole n n The charges are equal and opposite The number of field lines leaving the positive charge equals the number of lines terminating on the negative charge n n i 2 rˆi = k e ∫ i dq rˆ r2 Electric Flux ∆ΦE = E i∆Ai cosθ i = E i ⋅ ∆A i Φ E = lim ∆Ai →0 ∑ E ⋅ ∆A i i = ∫ E ⋅ dA surface Field Due to a Plane of Charge Gauss’s Law ΦE = ∑r ∫ E ⋅ dA = q in The total charge in the surface is A Applying Gauss’s law σA σ ΦE = 2EA = , and E = ε0 2ε0 σ ε0 qin is the net charge inside the surface E represents the electric field at any point on the surface n n n Field uniform everywhere 2 Properties of a Conductor in Electrostatic Equilibrium Electric Potential Energy Electric field is zero everywhere inside conductor Charge resides on its surface of isolated conductor Electric field just outside a charged conductor is perpendicular to the surface with magnitude / o On an irregularly shaped conductor surface charge density is greatest where radius of curvature is smallest 1. 2. 3. σ 4. n n n B ∆U = UB − U A = −qo ∫ E ⋅ ds ε A Electric Potential, V n n Equipotential Surface The potential energy per unit charge, U/qo, is the electric potential ∆V = n B ∆U = − ∫ E ⋅ ds A qo The work performed on the charge is n W = ∆U = q ∆V n B B A A From ∆V = -E.ds = -Exdx n dV dx Along an equipotential surfaces ∆V = 0 n n Hence E ⊥ ds i.e. an equipotential surface is perpendicular to the electric field lines passing through it q r V Due to a Charged Conductor Finding E From V Ex = − Any surface consisting of a continuous distribution of points having the same electric potential For a point charge V = ke In a uniform field VB − VA = ∆V = − ∫ E ⋅ ds = −E ∫ ds = −Ed n Work done by electric field is F.ds = qoE.ds Potential energy of the charge-field system is changed by ∆U = -qoE.ds For a finite displacement of the charge from A to B, the change in potential energy is n n n n E · ds = 0 So, potential difference between A and B is zero Electric field is zero inside the conductor So, electric potential constant everywhere inside conductor and equal to value at the surface 3 Definition of Capacitance Cavity in a Conductor n n n n Assume an irregularly shaped cavity is inside a conductor Assume no charges are inside the cavity The electric field inside the conductor must be zero C= n n Capacitance – Parallel Plates n n Charge density = Q/A Electric field E = σ/ε0 (for conductor) σ n Capacitors in Series n Potential differences add up to the battery voltage A measure of the ability to store charge The SI unit of capacitance is the farad (F) n n Capacitors can be replaced with one capacitor with a capacitance of Ceq Ceq = C1 + C2 Energy of Capacitor n Q = Q1 = Q2 ∴ ∆V ∆V1 ∆V2 = + Q1 Q2 Q 1 1 1 = + C C1 C2 Work done in charging the capacitor appears as electric potential energy U: U= ∆V = ∆V1 + ∆V2 ∴ Q ∆V Capacitors in Parallel Uniform between plates, zero elsewhere Q Q Q εA = = = 0 C= Q ∆V Ed d d ε0 A The capacitance, C, is ratio of the charge on either conductor to the potential difference between the conductors n n 1 Q2 1 = Q ∆V = C ( ∆V )2 2C 2 2 Energy is stored in the electric field Energy density (energy per unit volume) uE = U/Vol. = ½ εoE2 4 Capacitors with Dielectrics n n A dielectric is a nonconducting material that, when placed between the plates of a capacitor, increases the capacitance For a parallel-plate capacitor C = Co = o(A/d) κ κ Rewiring charged capacitors n n n ε Two capacitors, C1 & C2 charged to same potential difference, ∆Vi. Capacitors removed from battery and plates connected with opposite polarity. Switches S1 & S2 then closed. What is final potential difference, ∆Vf? Q1i, Q2i before; Q1f, Q2f after. Q1i = C1∆Vi; Q2i = -C2∆Vi Magnetic Poles So Q=Q1i+Q2i=(C1-C2)∆Vi But Q= Q1f+Q2f (charge conserved) n Every magnet has two poles With Q1f = C1∆Vf; Q2f = C2∆Vf hence Q1f = C1/C2 Q2f n So, Q=(C1/C2+1) Q2f n With some algebra, find Q1f = QC1/(C1+C2) & Q2f = QC2/(C1+C2) Called north and south poles Poles exert forces on one another n Like poles repel n Unlike poles attract So ∆V1f = Q1f / C1 = Q / (C1+C2) & ∆V2f = Q2f / C2 = Q / (C1+C2) n i.e. ∆V1f = ∆V2f = ∆Vf, as expected n So ∆Vf = (C1 - C2) / (C1 + C2) ∆Vi, on substituting for Q Magnetic Field Lines for a Bar Magnet n n Compass can be used to trace the field lines The lines outside the magnet point from the North pole to the South pole N-N or S-S N-S Direction n n FB perpendicular to plane formed by v & B Oppositely directed forces are exerted on charges of different signs n cause the particles to move in opposite directions 5 Direction given by Right-Hand Rule n Fingers point in the direction of v n n n The Magnitude of F n (for positive charge; opposite direction if negative) n n Curl fingers in the direction of B Then thumb points in the direction of v x B; i.e. the direction of FB n n F=ILxB n n n Magnitude is the length L of the segment n Equating the magnetic & centripetal forces: mv 2 F = qvB = r Solving gives r = mv/qB The total force is b F = I ∫ ds × B I is the current = nqAvD B is the magnetic field a Force on Charged Particle n The force exerted segment ds is F = I ds x B L is a vector that points in the direction of the current (i.e. of vD) n n θ is the angle between v and B FB is zero when v and B are parallel FB is a maximum when perpendicular Force on a Wire of Arbitrary Shape Force on a Wire n The magnitude of the magnetic force on a charged particle is FB = |q| vB sin θ Biot-Savart Law n n dB is the field created by the current in the length segment ds Sum up contributions from all current elements I.ds B= µ0 4π ∫ I ds × rˆ r2 6 B for a Long, Straight Conductor B for a Long, Straight Conductor, Direction Magnetic field lines are circles concentric with the wire Field lines lie in planes perpendicular to to wire Magnitude of B is constant on any circle of radius a The right-hand rule for determining the direction of B is shown n n B= µ0 I 2πa n n n Magnetic Force Between Two Parallel Conductors F1 = n n Definition of the Ampere n µ 0 I1I 2 l 2πa The force between two parallel wires can be used to define the ampere F1 µ0 I1I2 = with µ0 = 4 π •10−7 T m A-1 l 2πa Parallel conductors carrying currents in the same direction attract each other Parallel conductors carrying currents in opposite directions repel each other 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 A Field in interior of a Solenoid Ampere’s Law n n Grasp wire with thumb in direction of current. Fingers wrap in direction of B. The line integral of B . ds around any closed path equals µoI, where I is the total steady current passing through any surface bounded by the closed path. ∫ B ⋅ ds = µ I 0 n Apply Ampere’s law The side of length inside the solenoid contributes to the field ℓ n Path 1 in the diagram ∫ B ⋅ ds = ∫ B ⋅ ds = B ∫ ds = Bl path 1 B = µ0 path 1 N I = µ0 nI l 7 Ampere’s vs. Gauss’s Law ∫ B ⋅ ds = µ I ∫ E ⋅ dA = q 0 n n ε0 Magnetic fields do not begin or end at any point n n Integrals around closed path vs. closed surface. n n Gauss’ Law in Magnetism i.e. 2D vs. 3D geometrical figures Gauss’ law in magnetism says: n Integrals related to fundamental constant x source of the field. Concept of “Flux” – the flow of field lines through a surface. ΦB = Faraday’s Law of Induction n ∫ B.dA = 0 Ways of Inducing an emf ε=− The emf induced in a circuit is directly proportional to the rate of change of the magnetic flux through that circuit ε = −N i.e. they form closed loops, with the number of lines entering a surface equaling the number of lines leaving that surface n dΦB dt n n n d (BAcosθ ) dt Magnitude of B can change with time Area enclosed, A, can change with time Angle θ can change with time Any combination of the above can occur QuickTime™ and a Cinepak decompressor are needed to see this picture. Sliding Conducting Bar Motional emf n n Motional emf induced in a conductor moving through a constant magnetic field Electrons in conductor experience a force, FB = qv x B that is directed along In equilibrium, qE = qvB or E = vB n Magnetic flux is ΦB = Blx The induced emf is dΦ d dx ε = − B = − (Blx ) = −Bl = −Blv dt dt dt n Thus the current is n ℓ n I= ε R = Blv R 8 Induced emf & Electric Fields n n n A changing magnetic flux induces an emf and a current in a conducting loop An electric field is created in a conductor by a changing magnetic flux Faraday’s law can be written in a general form: ε= n Generators ∫ E.ds = − n dΦ B dt Not an electrostatic field because the line integral of E.ds is not zero. Rotating Loop Assume a loop with N turns, all of the same area, rotating in a magnetic field The flux through one loop at any time t is: Φ B = BA cos θ = BA cos ωt n n ∴ ε = −N n Circulating currents called eddy currents are induced in bulk pieces of metal moving through a magnetic field From Lenz’s law, their direction is to oppose the change that causes them. n Motors n Motors are devices into which energy is transferred by electrical transmission while energy is transferred out by work A motor is a generator operating in reverse n n A current is supplied to the coil by a battery and the torque acting on the current-carrying coil causes it to rotate dΦ B d = −NAB (cosωt ) = NABω sin ωt dt dt Eddy Currents n Electric generators take in energy by work and transfer it out by electrical transmission The AC generator consists of a loop of wire rotated by some external means in a magnetic field n The eddy currents are in opposite directions as the plate enters or leaves the field Equations for Self-Inductance n n Induced emf proportional to the rate of change of the current dI εL = −L dt L is a constant of proportionality called the inductance of the coil. 9 Inductance of a Solenoid n Uniformly wound solenoid having N turns and length . Then we have: Energy in a Magnetic Field n ℓ N I l NA ΦB = BA = µ0 I l NΦ B µ0N 2 A ∴L = = I l Rate at which the energy is stored is dU dI =LI dt dt I B = µ0 nI = µ0 n U = L ∫ IdI = 12 LI 2 0 Magnetic energy density, uB, is uB = U B2 = Al 2µ0 RL Circuit − Rt −t I = ε R 1− e L = ε R 1− e τ n n Time constant, τ = L / R, for the circuit τ is the time required for current to reach 63.2% of its max value 10