1February 24, 2005 Physics for Scientists&Engineers 2 1Physics for Scientists &Physics for Scientists &EngineersEngineers 22Spring Semester 2005Lecture 23February 24, 2005 Physics for Scientists&Engineers 2 2ReviewReviewµ0 is the magnetic permeability of free space whose value is The magnitude of the magnetic field at a distance r from along, straight wire carrying currrent i is given by The magnitude of the magnetic field at the center of a loopwith radius R carrying current i is given byµ0= 4!"10#7 TmAB(r ) =µ0i2!rB =µ0i2RFebruary 24, 2005 Physics for Scientists&Engineers 2 3Review (2)Review (2) Ampere’s Law is where the integral is carried out around an Amperian loopand ienc is the current enclosed by the loop The magnitude of the magnetic field inside a long wire withradius R carrying a current i at a radius r⊥ is given by !B ! d!s""=µ0iencB(r!) =µ0i2"R2#$%&'(r!February 24, 2005 Physics for Scientists&Engineers 2 4Force on a Current CarryingForce on a Current Carrying WireWire Consider a long, straight wirecarrying carrying a current iin a constant magnetic field B The magnetic field will exerta force on the moving chargesin the wire The charge q flowing in the wirein a given time t in a length L of wire is given by where v is the drift velocity of the electronsq = ti =Lvi2February 24, 2005 Physics for Scientists&Engineers 2 5Force on a Current CarryingForce on a Current Carrying Wire (2)Wire (2) The magnitude of the magnetic force is thenθ is the angle between the current and the magnetic field The direction of the force is perpendicular to both thecurrent and the magnetic field and is given by the righthand rule This equation can be expressed as a vector cross productiL represents the current in a length L of wireF = qvB sin!=Lvi"#$%&'vB = iLBsin! !F = i!L !!BFebruary 24, 2005 Physics for Scientists&Engineers 2 6Parallel Current Carrying WiresParallel Current Carrying Wires Consider the case in which two parallel wires are carryingcurrent The two wires will exert a magnetic force on each otherbecause the magnetic field of one wire will exert a force onthe moving charges in the second wire The magnitude of the magnetic field created by a currentcarrying wire is given by This magnetic field is always perpendicular to the wire witha direction given by the right hand rule.B(r ) =µ0i2!rFebruary 24, 2005 Physics for Scientists&Engineers 2 7Parallel Current Carrying Wires (2)Parallel Current Carrying Wires (2) Let’s start with wire one carryinga current i1 to the right The magnitude of the magneticfield a distance d from wire one is Now consider wire two carryinga current i2 in the same directionas i1 placed a distance d fromwire one The magnetic field due to wire onewill exert a magnetic force on themoving charges in the currentflowing in wire twoB1=µ0i12!dFebruary 24, 2005 Physics for Scientists&Engineers 2 8Parallel Current Carrying Wires (3)Parallel Current Carrying Wires (3) The charge q2 flowing inwire two in a given time tin a length L of wire isgiven by where v is the drift speed of the charge carriers The magnetic force is then Putting in our expression for B1 we getq2= ti2=Lvi2F = qvB =Lvi2!"#$%&vB1= i2LB1F12= i2Lµ0i12!d"#$%&'=µ0i1i2L2!d3February 24, 2005 Physics for Scientists&Engineers 2 9Torque on a Current-Carrying LoopTorque on a Current-Carrying Loop Electric motors rely on the magnetic force exerted on a current carrying wire This force is used to create a torque that turns a shaft A simple electric motor is depicted below consisting of a single loop carryingcurrent i in a constant magnetic field B The two magnetic forces, F and -F, shown in the figure are of equal magnitudeand opposite direction These forces create a torque that tends to rotate the loop around its axisFebruary 24, 2005 Physics for Scientists&Engineers 2 10Torque on a Current-Carrying Loop (2)Torque on a Current-Carrying Loop (2) As the coil turns in the field, the forces on the sides of the loopperpendicular to the magnetic field will change The forces on the square loop with sides are illustrated below where θis the angle between a normal vector, n, and the magnetic field B The normal vector is perpendicular to the plane of the wire loop andpoints in a direction given by the right hand rule based on the currentflowing in the loopFebruary 24, 2005 Physics for Scientists&Engineers 2 11Torque on a Current-Carrying Loop (3)Torque on a Current-Carrying Loop (3) Here the current is flowing upward in the topsegment and downward in the lower segmentas illustrated by the arrow feathers andarrowhead The force each of the vertical segments is The force on the other two sides is parallel or anti-parallel to the axisof rotation and cannot cause a torque The sum of the torque on the upper side plus the torque on the lowerside gives the torque exerted on the coil about the center of the loop where A = a2F = iaB!1= iaB( )a2"#$%&'sin(+ iaB( )a2"#$%&'sin(= ia2B sin(= iAB sin(February 24, 2005 Physics for Scientists&Engineers 2 12Magnetic Dipole MomentMagnetic Dipole Moment If we replace this loop with N loops wound close togetherwe can write Although we derived this expression for a square loop, thisexpress applies to circular loops as well as long as themagnetic field is uniform We can describe this coil with one parameter consisting ofinformation about the coil only, combined with informationabout the magnetic field We define the magnitude of the magnetic dipole momentof the coil above to be!= N!1= NiAB sin"µ= NiA4February 24, 2005 Physics for Scientists&Engineers 2 13Magnetic Dipole Moment (2)Magnetic Dipole Moment (2) The direction of the magnetic dipolemoment, µ, is given by the right handrule and points in the direction of thesurface normalvector n We can rewrite our expression for the torque as which we can generalize to The torque will always be perpendicular themagnetic field magnetic dipole moment and themagnetic field!= NiA( )B sin"=µB sin" !!=!µ"!B !µ !niFebruary 24, 2005 Physics for Scientists&Engineers 2 14PotentialPotential Energy of aEnergy of a Magnetic DipoleMagnetic Dipole A magnetic dipole has a potential energy in an externalmagnetic field• If the magnetic dipole is aligned with the magnetic field, it is in itsminimum energy condition• If the magnetic dipole oriented in a direction opposite to theexternal
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