Lecture 17: THU 18 MAR 10Lecture 17: THU 18 MAR 10AmpereAmpere’’s laws lawPhysics 2102Jonathan DowlingAndré Marie Ampère (1775 – 1836)Given an arbitrary closed surface, the electric flux through it isproportional to the charge enclosed by the surface.qFlux!=!0!q!="#$Surface0%qAdE!!Remember Gauss Law for E-Fields?Remember Gauss Law for E-Fields?GaussGauss’’s Law for B-Fields!s Law for B-Fields!No isolated magnetic poles! The magnetic flux throughany closed “Gaussian surface” will be ZERO. This is oneof the four “Maxwell’s equations”.!=• 0AdBThere are no SINKS or SOURCES of B-Fields!What Goes IN Must Come OUT!The circulation of B(the integral of Bscalar ds) along animaginary closed loop isproportional to the netamount of currenttraversing the loop.i1i2i3dsi4)(3210loopiiisdB !+="#µ!!Thumb rule for sign; ignore i4If you have a lot of symmetry, knowing the circulation of Ballows you to know B.AmpereAmpere’’s law: Closed Loopss law: Closed Loops !B ! d!sLOOP""=µ0ienclosedThe circulation of B(the integral of Bscalar ds) along animaginary closed loop isproportional to the netamount of currenttraversing the loop. ! B !d! s loop"=µ0(i1# i2)Thumb rule for sign; ignore i3If you have a lot of symmetry, knowing the circulation of Ballows you to know B.AmpereAmpere’’s law: Closed Loopss law: Closed Loops !B ! d!sLOOP""=µ0ienclosedenc Calculation of . We curl the fingersof the right hand in the direction in which the Amperian loop was traversed. We notethei direction of the thumb. All currents inside the loop to the thumb are counted as .All currents inside the loop to the thumb are counted as .All currents outside the loop are not countpaantiparalrallel positivelel negativeenc 1 2ed. In this example : .i i i= !Sample ProblemSample Problem• Two square conducting loops carry currentsof 5.0 and 3.0 A as shown in Fig. 30-60.What’s the value of ∫B·ds through each ofthe paths shown?Path 1: ∫B·ds = µ0•(–5.0A+3.0A)Path 2: ∫B·ds = µ0•(–5.0A–5.0A–3.0A)AmpereAmpere’’s Law: Example 1s Law: Example 1• Infinitely long straightwire with current i.• Symmetry: magnetic fieldconsists of circular loopscentered around wire.• So: choose a circular loopC so B is tangential to theloop everywhere!• Angle between B and ds is0. (Go around loop in samedirection as B field lines!)!="CisdB0µ!!!==CiRBBds0)2(µ"RiB!µ20=RMuch Easier Way to Get B-FieldAround A Wire: No Calculus.AmpereAmpere’’s Law: Example 2s Law: Example 2• Infinitely longcylindrical wire of finiteradius R carries a totalcurrent i with uniformcurrent density• Compute the magneticfield at a distance rfrom cylinder axis for:r < a (inside the wire)r > a (outside the wire)iCurrent outof page,circular fieldlines!="CisdB0µ!!AmpereAmpere’’s Law: Example 2 (cont)s Law: Example 2 (cont)!="CisdB0µ!!Current out ofpage, fieldtangent to theclosedamperian loopenclosedirB0)2(µ!=22222)(RrirRirJienclosed===!!!riBenclosed!µ20=202 RirB!µ=For r < RFor r>R, ienc=i, soB=µ0i/2πR = LONG WIRE!Need Current Density J!R02iRµ!rBOAmpereAmpere’’s Law: Example 2 (cont)s Law: Example 2 (cont)r < RB ! rr > RB ! 1 / rr < RB =µ0ir2!R2r > RB =µ0i2!rOutside Long!Wire!SolenoidsSolenoidsinBinhBhisdBinhhLNiiNihBsdBisdBenchencenc0000)/(000µµµµ=!=!=•===+++=•=•"""!!!!!!The n = N/L is turns per unit length.Magnetic Field in a Toroid “Doughnut” Solenoid Toriod Fusion Reactor: Power NYC For a Day on a Glass of H2O B =µ0Ni2!rMagnetic Field of a Magnetic DipoleMagnetic Field of a Magnetic Dipole302/32202)(2)(zzRzBµ!µµ!µ!!!"+=All loops in the figure have radius r or 2r. Whichof these arrangements produce the largestmagnetic field at the point indicated?A circular loop or a coil currying electrical current is amagnetic dipole, with magnetic dipole moment ofmagnitude µ=NiA. Since the coil curries a current, itproduces a magnetic field, that can be calculated usingBiot-Savart’s law:Magnetic Dipole in a B-Field !F = 0!!=!µ"!BU = #!µi!BForce is zero in homogenous fieldTorque is maximum when θ!=!90°Torque is minimum when θ!=!0° or 180°Energy is maximum when θ!=!180°Energy is minimium when θ!=!0°Neutron Star a Large Magnetic
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