Our Goal: Maxwell’s Equations Gauss's law ( electric ) : q E A d ⋅ = S∫ εo Gauss's law in (magnetic): 0 ∫ B ⋅ dA = S Faraday's law: d ΦB − ∫ E ⋅ ds = dt Ampere-Maxwell law ∫ B ⋅ ds = μo I + μo d ΦE dt Electric Force and Field q1q2 F12 = k 2 rˆ r dq = dE k ∫ 2 rˆ E ∫= r 1 Φ E= ∫ E ⋅ dA = ε0 ∫ ρ dV= qin ε0 Magnetic Force and Field F = qv × B Φ B= ∫ B ⋅ dA = 0 µo I ds × rˆ B= 2 ∫ r 4π ∫ B ⋅ ds= μo ∫ J ⋅ dA= μo I Continuous Charge Distribution Note: For spheres, cylinders, infinite planes, rods, etc with high symmetry, the electric field is found using Gauss’s Law which we’ll do next chapter! dq = dE k ∫ 2 rˆ E ∫= r Q λ≡ linear charge density L Q σ≡ surface charge density A Q ρ≡ volume charge density V dq = λ dx dq = σ dA dq = ρ dV Gauss’s Law The net electric flux through ANY closed surface is due to the net charge contained inside that surface (divided by ε0)! qin Φ E= ∫ E ⋅ dA = ε0 Biot-Savart Law • Empirical observations are summarized in the mathematical equation called the Biot-Savart law: µo I ds × rˆ B= 4π ∫ r 2 • The magnetic field described by the law is the field due to the current-carrying conductor – Don’t confuse this field with a field external to the conductor μ I ds × ˆr dB = o 4π r 2 Ampere’s Law Ampere’s law states that 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= μo ∫ J ⋅ dA= μo I s I B ⋅ d = μ o ∫ Gauss’ Law in Magnetism • • • • The magnetic flux associated with a magnetic field is defined in a way similar to electric flux The unit of magnetic flux is T.m2 = Wb is a weber Magnetic fields do not begin or end at any point – The number of lines entering a surface equals the number of lines leaving the surface Gauss’ law in magnetism says the magnetic flux through any closed surface is always zero: Φ B = ∫ B ⋅ dA Φ B= ∫ B ⋅ dA = 0 Maxwell’s Equations Gauss's law ( electric ) : q E A d ⋅ = S∫ εo Gauss's law in (magnetic): 0 ∫ B ⋅ dA = S Faraday's law: d ΦB − ∫ E ⋅ ds = dt Ampere-Maxwell law ∫ B ⋅ ds = μo I + μo d ΦE dt Currents Produce Magnetic Fields Smallest Magnet: Electron How to Find the B Field? If there is high symmetry: Use Ampere’s Law μ I ∫ B ⋅ ds = o Otherwise use the Biot-Savart Law: µo I ds × rˆ B= 2 ∫ r 4π Ampere’s Law Ampere’s law states that 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. ⋅ = I B d s μ o ∫ ∫ B ⋅ ds = μo I = μo ∫ J ⋅ dA Field Due to a Long Straight Wire The current is uniformly distributed through the cross section of the wire • Outside of the wire, r > R ⋅ ds ∫ B = B= B( 2πr = ) μo I μo I 2πr • Inside the wire, r < R s ∫ B ⋅ d= B( 2πr= ) μo I ' → μo I B= r 2 2πR r2 I= ' I 2 R Magnetic Fields due to Currents Magnitude of the Field: µ0 I B= 2π r µ0 4π x10 T ⋅ m / A −7 “mew not” : The permeability of free space. Permeability is a measure of a materials ability to be permeated by magnetic fields – ferromagnetic materials have a higher permeability. Currents Produce Magnetic Fields Smallest Magnet: Electron B for a Long, Straight Conductor • The magnetic field lines are circles concentric with the wire • The field lines lie in planes perpendicular to to wire • The magnitude of B is constant on any circle of radius a • The right-hand rule for determining the direction of B is shown Direction for a loop Right Hand Rules What is the direction of the current? Electromagnets What’s the direction of the B field? A cross-sectional view of a coaxial cable is shown. The center conductor is surrounded by a rubber layer, which is surrounded by an outer conductor, which is surrounded by another rubber layer. In a particular application, the current in the inner conductor is 1.00 A out of the page and the current in the outer conductor is 3.00 A into the page. Determine the magnitude and direction of the magnetic field at points a and b. Adding B Fields The wires carry a current of 8.0A. Find the magnitude of the net B field at A and B. First, define positive directions. Up out of plane is positive. µ 0 I1 µ 0 I 2 µ 0 I 1 1 −= B= − netA 2π r1 2π r2 2π r1 r2 4π x10−7 Tm / A(8 A) 1 1 ( ) − 2π .03m .15m −5 BnetA = 4.27 x10 T Is there any place along the line where the fields cancel? I1 I2 Current Affair An electric current is flowing through two parallel wires in the same direction. Do the wires tend to a) repel each other b) Attract each other c) Exert no force on each other d) Twist at right angles to each other e) Spin F Current Affair F F Magnetic Force between two Current Carrying Wires ON Each Other: Opposite CURRENTS repel! Like CURRENTS attract! Magnetic Force between two Current Carrying Wires ON Each Other: F = IL × B F = ILB sin θ Fon 2 = I 2 LB1at 2 sin θ µ 0 I1 = I 2 L( ) sin 90 2π r Fon 2 µ 0 I1 I 2 L = = − Fon1 2π r Problem The wires carry a current of 8.0A. What is the force per meter on each wire? What direction? Fon 2 µ 0 I1 I 2 L = = − Fon1 2π r N T= A⋅ m 4π x10−7 Tm / A(8 A) 2 Fon 2 µ 0 I1 I 2 = = 2π (.12m) L 2π r = 1.07 x10−4 N / m I1 They repel each other. I2 Questions Is the net B field greater, less or the same at A and B? Do the wires exert forces on each other? Do the wires exert torques on each other? If so, what directions? Question Is the wire attracted or repelled by the coil? Compare E & B Electric Force and Field q1q2 F12 = k 2 rˆ r dq = dE k ∫ 2 rˆ E ∫= r 1 Φ E= ∫ E ⋅ dA = ε0 ∫ ρ dV= qin ε0 Magnetic Force and Field F = qv × B Φ B= ∫ B ⋅ dA = 0 µo I ds × rˆ B= 2 ∫ r 4π ∫ B ⋅ ds= μo ∫ J ⋅ dA= μo I Compare E & B Symmetry Use Gauss’s Law qin E dA ⋅ = ∫ ε0 λl E ( 2πrl ) = ε0 λ E= 2πε0 r Use Ampere’s Law μ I ∫ B ⋅ ds = o B( 2πr ) = μo I μo I B= 2πr Compare E & B: Infinite Current Sheet Use Ampere’s Law J B = µ0 2 Infinite Charged Plane Use Gauss’s Law σ E= 2ε 0 Circular Charge Distribution Ering = (x kxQ 2 +a ) 2 3/ 2 Bx = μo I R 2 2(x + R 2 2 ) Far from ring: Dipole Field Ering kxQ = 3 a μo I R 2 Bx = 2x 3 3 2 Compare Finite Rods λ E= 2πε 0 y μo I B= 2πa Compare Gauss’s Law Flux for Dipoles Φ B= B ⋅ d A = 0 ∫ qin Φ E= ∫ E ⋅ dA = ε0 Compare Dipoles p ≡ 2aq, τ = p × E, U = -p ⋅ E μ × E, U = µ ≡ IA, τ = -μ ⋅ B Compare Dipoles E0 & B0: External Field Enet = E0 − Eind Reduces External Field Bnet= B0 + Bind Increases External Field (ferromagnetic) How to Find the B Field? If there is high symmetry: Use Ampere’s Law μ I ∫ B ⋅ ds = o Otherwise use the Biot-Savart Law: µo I ds × rˆ B= 2 ∫ r 4π Ampere’s Law Ampere’s law states that 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= μo ∫ J ⋅ dA= μo I s I B ⋅ d = μ o ∫ A long cylindrical conductor of radius R carries a current I as shown. The current density J, however, is not uniform over the cross section of the conductor but is a function of the radius according to J = br, where b is a constant. Find an expression for the magnetic field B (a) at a distance r1 < R and (b) at a distance r2 > R, measured from the axis. Biot-Savart Law • Empirical observations are summarized in the mathematical equation called the Biot-Savart law: µo I ds × rˆ B= 4π ∫ r 2 • The magnetic field described by the law is the field due to the current-carrying conductor – Don’t confuse this field with a field external to the conductor μ I ds × ˆr dB = o 4π r 2 B for a Long, Straight Conductor μ I ds × ˆr dB = o 4π r 2 • The thin, straight wire is carrying a constant current ds × ˆr= dxˆi × (cos θˆi + sin θˆj )= ( dx sin θ ) kˆ a r= , x = r cos θ = a cot θ ,dx = −a csc 2 θdθ sin θ 2 sin ds × ˆr θ 1 2 ˆ = − sin csc = sin θdθkˆ θk a θdθ 2 2 r a a μo I θ2 B= sin θ dθ ∫ θ 4πa 1 μo I ( cos θ1 − cos θ2 ) 4πa • If the conductor is an infinitely long, straight wire, θ1 = 0 and θ2 = π μo I B= 2πa Same result with Ampere’s Law which is way easier • Outside of the wire, r > R ⋅ ds ∫ B = B= B( 2πr = ) μo I μo I 2πr • Inside the wire, r < R s ∫ B ⋅ d= B( 2πr= ) μo I ' → μo I B= r 2 2πR r2 I= ' I 2 R The segment of wire carries a current of I = 5.00 A, where the radius of the circular arc is R = 3.00 cm. Determine the magnitude and direction of the magnetic field at the origin. B for a Circular Current Loop • The loop has a radius of R and carries a steady current of I • Find B at point P Bx = μo I R 2 2(x + R 2 2 ) 3 2 • At x = 0: Bx μo I R 2 μo I = 3 2 2 2 ( x + R ) 2 2R μ I ds × ˆr dB = o 4π r 2 Magnetic Field of a Coil (Not a Solenoid) Magnetic Field at the center of the loop: B= µ0 I 2R N Magnetic Field of a Solenoid • A solenoid is a long wire wound in the form of a helix • The field distribution is similar to that of a bar magnet • As the length of the solenoid increases – the interior field becomes more uniform – the exterior field becomes weaker The Field Outside is nearly zero because the fields from different loops cancel each others Ideal Solenoid The field inside is uniform the field outside is ~zero. ∫ B ⋅ ds = ∫ B ⋅ ds = B ∫ ds = B path1 path1 ∫ B ⋅ ds = B = µ oNI N = B μ= I μo n I o What current is required in the windings of a long solenoid that has 1 000 turns uniformly distributed over a length of 0.400 m, to produce at the center of the solenoid a magnetic field of magnitude 1.00 × 10–4 T? N = B μ= I μo n I o Magnetic Field of a Toroid • Find the field at a point at distance r from the center of the toroid • The toroid has N turns of wire ) μo N I ds B( 2πr = ∫ B ⋅= μo N I B= 2πr A magnetic field of 1.30 T is to be set up in an iron-core toroid. The toroid has a mean radius of 10.0 cm, and magnetic permeability of 5 000 μ0. What current is required if the winding has 470 turns of wire? The thickness of the iron ring is small compared to 10 cm, so the field in the material is nearly uniform. μo N I B= 2πr Coil, Solenoid, Toroid Coil: B= µ0 I 2R N N B μ= I μo n I Solenoid:= o Toroid: μo N I B= 2πr Magnetic Dipoles The magnetic dipole moment of a current loop enclosing an area A is defined as The SI units of the magnetic dipole moment are A m2. The on-axis field of a magnetic dipole is Smallest Magnet: Electron • The electrons move in circular orbits • The orbiting electron constitutes a tiny current loop • The magnetic moment of the electron is associated with this orbital motion • L is the angular momentum • µ is magnetic moment 1 = μ I= A evr 2 Classical Atom q e e = I = = ∆t T 2πr / v In Bohr’s 1913 model of the hydrogen atom, the electron is in a circular orbit of radius 5.29 × 10–11 m and its speed is 2.19 × 106 m/s. (a) What is the magnitude of the magnetic moment due to the electron’s motion? (b) If the electron moves in a horizontal circle, counterclockwise as seen from above, what is the direction of this magnetic moment vector? Domains • All ferromagnetic materials are made up of microscopic regions called domains – The domain is an area within which all magnetic moments are aligned • The boundaries between various domains having different orientations are called domain walls Domains, Unmagnetized Material • The magnetic moments in the domains are randomly aligned • The net magnetic moment is zero Domains, External Field Applied • A sample is placed in an external magnetic field • The size of the domains with magnetic moments aligned with the field grows • The sample is magnetized Domains, External Field Applied • The material is placed in a stronger field • The domains not aligned with the field become very small • When the external field is removed, the material may retain a net magnetization in the direction of the original field Ferromagnetism • Some substances exhibit strong magnetic effects called ferromagnetism • Some examples of ferromagnetic materials are: – – – – – iron cobalt nickel gadolinium dysprosium • They contain permanent atomic magnetic moments that tend to align parallel to each other even in a weak external magnetic field Paramagnetism • Paramagnetic substances have small but positive magnetism • It results from the presence of atoms that have permanent magnetic moments – These moments interact weakly with each other • When placed in an external magnetic field, its atomic moments tend to line up with the field – The alignment process competes with thermal motion which randomizes the moment orientations Diamagnetism • When an external magnetic field is applied to a diamagnetic substance, a weak magnetic moment is induced in the direction opposite the applied field • Diamagnetic substances are weakly repelled by a magnet – Weak, so only present when ferromagnetism or paramagnetism do not exist Curie Temperature • • The Curie temperature is the critical temperature above which a ferromagnetic material loses its residual magnetism – The material will become paramagnetic Above the Curie temperature, the thermal agitation is great enough to cause a random orientation of the moments Meissner Effect • Certain types of superconductors also exhibit perfect diamagnetism in the superconducting state – This is called the Meissner effect • If a permanent magnet is brought near a superconductor, the two objects repel each other Earth’s Magnetic Field • The Earth’s magnetic field resembles that achieved by burying a huge bar magnet deep in the Earth’s interior • The Earth’s south magnetic pole is located near the north geographic pole • The Earth’s north magnetic pole is located near the south geographic pole Earth’s interior consists of a rocky mantle and an iron rich core The Earth’s magnetic field is caused by dynamo movements in Earth’s core Earth’s magnetic field flips poles. Consecutive reversals are spaced 5 thousand years to 50 million years apart. The last reversal happened 740,000 years ago. Some researchers think our planet is overdue for another one, but nobody knows exactly when the next reversal might occur. The Sun's north magnetic pole is pointing through the Sun's southern hemisphere, until the year 2012 when they will reverse. This transition happens, as far as we know, at the peak of every 11year sunspot cycle -- like clockwork. Babcock’s magnetic dynamo is one possible explanation of the sunspot cycle where magnetic field lines become complexly entangled after many solar rotations The Sun’s magnetic fields create sunspots Zeeman effect - spectral lines split in regions of high magnetic fields Magnetic field lines connect sunspots on the Sun’s photosphere Solar magnetic fields also create other atmospheric phenomena • plages • filaments Solar magnetic fields also create other atmospheric phenomena • plages • filaments • prominences Solar magnetic fields also create other atmospheric phenomena • • • • plages filaments prominences solar flares Solar magnetic fields also create other atmospheric phenomena • • • • • plages filaments prominences solar flares coronal holes Solar magnetic fields also create other atmospheric phenomena • • • • plages filaments prominences solar flares • coronal holes • coronal mass ejections (CMEs) Reversals of the Earth’s Magnetic Field • The direction of the Earth’s magnetic field reverses every few million years – Evidence of these reversals are found in basalts resulting from volcanic activity – The rocks provide a timeline for the periodic reversals of the field • The rocks are dated by other means to determine the timeline This Week’s Lab Using a Tangent Galvanometer to Measure the Earth’s Magnetic Field Magnetic Field at the center of the loop: B= µ0 I 2r N Horizontal Earth Field Total Field

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