Physics 2102 Gabriela González• Electric charge • Electric force on other electric charges • Electric field, and electric potential • Moving electric charges : current • Electronic circuit components: batteries, resistors, capacitors • Electric currents • Magnetic field • Magnetic force on moving charges • Time-varying magnetic field • Electric Field • More circuit components: inductors • All together: Maxwell’s equations • Electromagnetic waves • Matter wavesdA Magnetic Flux: B n “Lenz’s Law” A time varying magnetic flux creates an electric field, which induces an EMF (and a current if the edge of the surface is a conductor) € EMF = E ⋅ d s C∫= −dΦBdtNotice that the electric field has closed field lines, and is not pointing towards “lower” electric potential – this is only true for fields produced by electric charges.• Q: A long solenoid has a circular cross-section of radius R, carrying a current i clockwise. What’s the direction and magnitude of the magnetic field produced inside the solenoid? • The current through the solenoid is increasing at a steady rate di/dt. What will be the direction of the electric field lines produced? • Compute the variation of the electric field as a function of the distance r from the axis of the solenoid. Ans: from symmetry, we know the magnitude of E depends only on r. First, let’s look at r < R: Next, let’s look at r > R: € E (r) =µ0n2didtR2rmagnetic field lines R € Ans : B =µ0in, out of the page.electric field lines € EMF = E ⋅ d s C∫= −dΦBdt € E ⋅ d s C∫= −dΦBdtE (2πr) =ddtBπr2( )=πr2ddtµ0in( )E(r) =µ0n2didtrInductors are with respect to the magnetic field what capacitors are with respect to the electric field. They “pack a lot of field in a small region”. Also, the higher the current, the higher the magnetic field they produce. Capacitance → how much potential for a given charge: Q=CV Inductance → how much magnetic flux for a given current: Φ=Li Joseph Henry (1799-1878) Using Faraday’s law:• Solenoid of cross-sectional area A, length l, total number of turns N, turns per unit length n • Field inside solenoid = µ0 n i • Field outside ~ 0 i L = “inductance”• Set up a single loop series circuit with a battery, a resistor, a solenoid and a switch. • Describe what happens when the switch is closed. • Key processes to understand: – What happens JUST AFTER the switch is closed? – What happens a LONG TIME after switch has been closed? – What happens in between? Key insights: • If a circuit is not broken, one cannot change the CURRENT in an inductor instantaneously! • If you wait long enough, the current in an RL circuit stops changing!Loop rule: “Time constant” of RL circuit = L/R E/R i(t) Small L/R Large L/R iE/R Exponential discharge. i(t) i The switch is in a for a long time, until the inductor is charged. Then, the switch is closed to b. What is the current in the circuit? Loop rule around the new circuit:In an RC circuit, while charging, Q = CV and the loop rule mean: • charge increases from 0 to CE • current decreases from E/R to 0 • voltage across capacitor increases from 0 to E In an RL circuit, while charging, emf = Ldi/dt and the loop rule mean: • magnetic field increases from 0 to B • current increases from 0 to E/R • voltage across inductor decreases from -E to 0• Recall that capacitors store energy in an electric field • Inductors store energy in a magnetic field. Power delivered by battery = power dissipated by R + energy stored in L i• The current in a 10 H inductor is decreasing at a steady rate of 5 A/s. • If the current is as shown at some instant in time, what is the induced EMF? • Current is decreasing • Induced emf must be in a direction that OPPOSES this change. • So, induced emf must be in same direction as current • Magnitude = (10 H)(5 A/s) = 50 V (a) 50 V (b) 50 V iImmediately after the switch is closed, what is the potential difference across the inductor? (a) 0 V (b) 9 V (c) 0.9 V • Immediately after the switch, current in circuit = 0. • So, potential difference across the resistor = 0! • So, the potential difference across the inductor = E = 9 V! 10 Ω$10 H 9 V• Immediately after the switch is closed, what is the current i through the 10 Ω resistor? (a) 0.375 A (b) 0.3 A (c) 0 • Long after the switch has been closed, what is the current in the 40Ω resistor? (a) 0.375 A (b) 0.3 A (c) 0.075 A • Immediately after switch is closed, current through inductor = 0. • Hence, i = (3 V)/(10Ω) = 0.3 A • Long after switch is closed, potential across inductor = 0. • Hence, current through 40Ω resistor = (3 V)/(40Ω) = 0.075 A 40 Ω$10 H 3 V 10 Ω$• The switch has been in position “a” for a long time. • It is now moved to position “b” without breaking the circuit. • What is the total energy dissipated by the resistor until the circuit reaches equilibrium? • When switch has been in position “a” for long time, current through inductor = (9V)/(10Ω) = 0.9A. • Energy stored in inductor = (0.5)(10H)(0.9A)2 = 4.05 J • When inductor “discharges” through the resistor, all this stored energy is dissipated as heat = 4.05 J. 9 V 10 Ω$10 HOscillators are very useful in practical applications, for instance, to keep time, or to focus energy in a system. All oscillators operate along the same principle: they are systems that can store energy in more than one way and exchange it back and forth between the different storage possibilities. For instance, in pendulums (and swings) one exchanges energy between kinetic and potential form.Newton’s law F=ma!The magnetic field on the coil starts to collapse, which will start to recharge the capacitor. Finally, we reach the same state we started with (with opposite polarity) and the cycle restarts. Capacitor discharges completely, yet current keeps going. Energy is all in the inductor. Capacitor initially charged. Initially, current is zero, energy is all stored in the capacitor. A current gets going, energy gets split between the capacitor and the inductor.Compare with: Analogy between electrical and mechanical oscillations: (the loop rule!)The energy
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