4 The Biot–Savart LawBiot and Savart conducted experiments on the force exerted by an electric current on a nearby magnetThey arrived at a mathematical expression that gives the magnetic field at some point in space due to a currentThe magnetic field is at some point PThe length element isThe wire is carrying a steady current of I
5 Calculation of B near a wire Experiments show that the magnetic field at a point near a long straight wire isThis relationship is valid as long as r, the perpendicular distance to the wire, is much less than the distance to the ends of the wireWhere o is called the permeability of free space.Permeability is a measure of the effect of a material on the magnetic field by the material.
6 dB the magnetic field produced by a small section of wire ds a vector the length of the small section of wire in the direction of the currentr the positional vector from the section of wire to where the magnetic field is measuredI the current in the wire angle between ds & r
7 Biot-Savart Law – Observations The vector is perpendicular to both and to the unit vector directed from toward PThe magnitude of is inversely proportional to r2, where r is the distance from to PThe magnitude of is proportional to the current and to the magnitude ds of the length elementThe magnitude of is proportional to sin q, where q is the angle between the vectors and
8 Permeability of Free Space The observations are summarized in the mathematical equation called the Biot-Savart law:The magnetic field described by the law is the field due to the current-carrying conductorDon’t confuse this field with a field external to the conductorPermeability of Free SpaceThe constant mo is called the permeability of free space mo = 4p x 10-7 T. m / A
9 Total Magnetic Fieldis the field created by the current in the length segment dsTo find the total field, sum up the contributions from all the current elements IThe integral is over the entirecurrent distribution
10 Biot-Savart Law – Final Notes The law is also valid for a current consisting of charges flowing through spacerepresents the length of a small segment of space in which the charges flowFor example, this could apply to the electron beam in a TV set
11 Compared to Distance Direction The magnitude of the magnetic field varies as the inverse square of the distance from the sourceThe electric field due to a point charge also varies as the inverse square of the distance from the chargeDirectionThe electric field created by a point charge is radial in directionThe magnetic field created by a current element is perpendicular to both the length element and the unit vector
12 Compared toSourceAn electric field is established by an isolated electric chargeThe current element that produces a magnetic field must be part of an extended current distributionTherefore you must integrate over the entire current distribution
13 for a Long, Straight Conductor, Special Case If the conductor is an infinitely long, straight wire, q1 = p/2 andq2 = -p/2The field becomesr
14 for a Long, Straight Conductor, Direction The magnetic field lines are circles concentric with the wireThe field lines lie in planes perpendicular to to wireThe magnitude of the field is constant on any circle of radius rThe right-hand rule for determining the direction of the field is shown
15 Magnetic Force Between Two Parallel Conductors, final The result is often expressed as the magnetic force between the two wires, FBThis can also be given as the force per unit length:
16 Definition of the Ampere The force between two parallel wires can be used to define the ampereWhen 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 ADefinition of the CoulombThe SI unit of charge, the coulomb, is defined in terms of the ampereWhen a conductor carries a steady current of 1 A, the quantity of charge that flows through a cross section of the conductor in 1 s is 1 C
17 Magnetic Field of a Wire A compass can be used to detect the magnetic fieldWhen there is no current in the wire, there is no field due to the currentThe compass needles all point toward the Earth’s north poleDue to the Earth’s magnetic fieldHere the wire carries a strong currentThe compass needles deflect in a direction tangent to the circleThis shows the direction of the magnetic field produced by the wireUse the active figure to vary the current
18 Ampere’s LawThe product of can be evaluated for small length elements on the circular path defined by the compass needles for the long straight wireAmpere’s law states that the line integral ofaround any closed path equals moI where I is the total steady current passing through any surface bounded by the closed path:
19 Ampere’s law describes the creation of magnetic fields by all continuous current configurations Most useful for this course if the current configuration has a high degree of symmetryPut the thumb of your right hand in the direction of the current through the amperian loop and your fingers curl in the direction you should integrate around the loop
20 Magnetic Field of a Wire, 3 The circular magnetic field around the wire is shown by the iron filings
21 dB perpendicular to dsdB perpendicular to r|dB| inversely proportional to |r|2|dB| proportional to current I|dB| proportional to |ds||dB| proportional to sin q
22 We already used the Biot-Savart Law to show that, for this case, . Let’s show it again, using Ampere’s Law:First, we are free to draw an Amperian loop of any shape, but since we know that the magnetic field goes in circles around a wire, let’s choose a circular loop (of radius r).Then B and ds are parallel, and B is constant on the loop, soAmpere’s Law
23 Magnetic Field Produced by Long Straight Current-Carrying Wire Any time an electric current flows in a wire, it produces a magnetic field.In a long, straight wire, the direction of the magnetic field can be determined using the right hand rule.Point the thumb of your right hand in the direction of the current in the wire and curl your fingers. Your fingers now point in the direction of the magnetic field caused by the current in the wire.The strength of the field is directly proportional to the current in the wire and inversely proportional to the distance from the wire. The equation is:B = μ0I/2πrwhere μ0 is a proportionality constant called the permeability of free space. Its value is 4π x 10-7 Tm/A.
24 Direction of the Field of a Long Straight Wire Draw some magnetic field lines (loops in this case) along the wire.Using x’s and dots to represent vectors out of and into the page, show the magnetic field for the same wire. Note B diminishes with distance from the wire.B out of pageB into pageDirection of the Field of a Long Straight WireRight Hand RuleGrasp the wire in your right handPoint your thumb in the direction of the currentYour fingers will curl in the direction of the field
25 F = qvBsinθ F = (Δq/Δt)(vΔt)sinθ F = ILBsinθ The force on an electric current in a wire can be calculated much the same way since a current consists of charges moving through a wire.F = qvBsinθF = (Δq/Δt)(vΔt)sinθSince Δq/Δt = I and vΔt = L(length of the wire), the equation becomes:F = ILBsinθ
26 Magnetic forces on moving charges When a charge moves through a B-field, it feels a force.It has to be movingIt has to be moving across the B-fieldDirection of force is determined by another right-hand rule
27 30.3 Ampère’s Law A current-carrying wire produces a magnetic field The compass needle deflects in directions tangent to the circleThe compass needle points in the direction of the magnetic field produced by the current
28 Magnitude of the Field of a Long Straight Wire Magnetic fieldIFor a long straight wire Ampere’s lawThe magnitude of the field at a distance r from a wire carrying a current of I is given by the formula above, where µo = 4 x 10-7 T m / Aµo is called the permeability of free space
29 Ampère’s LawAndré-Marie Ampère found a procedure for deriving the relationship between the current in a arbitrarily shaped wire and the magnetic field produced by the wireAmpère’s Circuital LawB|| Δℓ =µo I Integral (=sum) over the closed path
30 Choose an arbitrary closed path around the current Sum all the products of B|| ds around the closed path; B|| is the component of B parallel to ds .Amperian LoopWire surfaceDirection of Integration
31 Rules for finding direction of B field from a current flowing in a wire
32 B I into page, B clockwise I Permanent magnets aren’t the only things that produce magnetic fields. Moving charges themselves produce magnetic fields. We just saw that a current carrying wire feels a force when inside an external magnetic field. It also produces its own mag-netic field. A long straight wire produces circular field lines centered on the wire. To find the direction of the field, we use another right hand rule: point your thumb in the direction of the current; the way your fingers of your right hand wrap is the direction of the magnetic field. B diminishes with distance from the wire. The pics at the right show cross sections of a current carrying wire.I out of page, B counterclockwiseI into page, B clockwiseIB
33 The positive charge is moving straight out of the page The positive charge is moving straight out of the page. What is the direction of the magnetic field at the position of the dot?1. Right2. Left3. Up4. Down
34 The magnetic field of a straight, current-carrying wire is 1. parallel to the wire.2. inside the wire.3. perpendicular to the wire.4. around the wire.5. zero.
35 Example: Use Ampere’s Law to find B near a very long, straight wire Example: Use Ampere’s Law to find B near a very long, straight wire. B is independent of position along the wire and only depends on the distance from the wire (symmetry).riBy symmetrySupposeI = 10 AR = 10 cm
36 Ampere’s Law - a line integral blue - into figure ( -)red - out of figure (+)
37 Two Parallel WiresTwo long parallel wires suspended next to each other will either attract or repel depending on the direction of the current in each wire.B-field produced by each wire interacts with current in the other wireProduces magnetic deflecting force on other wire.Wires exert equal and opposite forces on each other.
38 Parallel wires with current flowing in the same direction, attract each other. Parallel wires with current flowing in the opposite direction, repel each other.Note that the force exerted on I2 by I1 is equal but opposite to the force exerted on I1 by I2.
39 Currents in Same Direction… Wires AttractCurrents in Opposite Directions…Wires RepelB-field produced by wire 2 at the position of wire 1 is given bywhere a is the distance between the wires.
40 B-field produced by wire 2 at the position of wire 1 is given by where a is the distance between the wires.
41 Example 1Determine the magnetic field at point A.A
42 Example 2Determine the magnetic field at point A.A
43 Example 3Two parallel conductors carry current in opposite directions. One conductor carries a current of 10.0 A. Point A is at the midpoint between the wires, and point C is a distance d/2 to the right of the 10.0-A current. If d = 18.0 cm and I is adjusted so that the magnetic field at C is zero, find (a) the value of the current I and (b) the value of the magnetic field at A.d/2
44 Example 3Two parallel conductors carry current in opposite directions. One conductor carries a current of 10.0 A. Point A is at the midpoint between the wires, and point C is a distance d/2 to the right of the 10.0-A current. If d = 18.0 cm and I is adjusted so that the magnetic field at C is zero, find (a) the value of the current I and (b) the value of the magnetic field at A.d/2
45 Field due to a long Straight Wire Need to calculate the magnetic field at a distance r from the center of a wire carrying a steady current IThe current is uniformly distributed through the cross section of the wire
46 Field due to a long Straight Wire The magnitude of magnetic field depends only on distance r from the center of a wire.Outside of the wire, r > R
47 Field due to a long Straight Wire The magnitude of magnetic field depends only on distance r from the center of a wire.Inside the wire, we need I’, the current inside the amperian circle
48 Field due to a long Straight Wire The field is proportional to r inside the wireThe field varies as 1/r outside the wireBoth equations are equal at r = R
49 Magnetic forces on moving charges FB = qvBIf a current I in a wire of length L passes through the field, the force is:FB = BIL
50 Magnetic force on current F = BILWire can be levitated by magnetic force when a current passes through it
51 Force on a current-carrying conductor The direction of magnetic force always perpendicular to the direction of the magnetic field and the direction of current passing through the conductor.
52 Magnetic Force Between Two Parallel Conductors F1=B2I1ℓB2=m0I2/(2d)F1=m0I1I2 ℓ /(2d)The field B2 at wire 1 due to the current I2 in wire 2 causes the force F1 on wire 1.
53 Two Parallel Wires a = separation l = length of wire However, it is often more appropriate to determine the force per unit length on the wire.Since the wires are parallel, the force exerted by wire 2 on wire 1 is given bya = separationl = length of wire
54 Force Between Two Conductors, cont Parallel conductors carrying currents in the same direction attract each otherParallel conductors carrying currents in the opposite directions repel each other
55 Defining Ampere and Coulomb The force between parallel conductors can be used to define the Ampere (A)If two long, parallel wires 1 m apart carry the same current, and the magnitude of the magnetic force per unit length is 2 x 10-7 N/m, then the current is defined to be 1 AThe SI unit of charge, the Coulomb (C), can be defined in terms of the Ampere (A)If a conductor carries a steady current of 1 A, then the quantity of charge that flows through any cross section in 1 second is 1 C
56 If I1 = 2 A and I2 = 6 A in the figure below, which of the following is true: (a) F1 = 3F2, (b) F1 = F2, or (c) F1 = F2/3?(b). The two forces are an action-reaction pair. They act on different wires, and have equal magnitudes but opposite directions.
57 Magnetic Field in Wires Moving current carries a magnetic fieldWires in your house generate a magnetic fieldCurrent in same direction Force is attractive
58 Current in opposite directions Force is repulsive
59 ExampleTwo wires in a 2.0 m long cord are 3.0 mm apart. If they carry a dc current of 8.0 A, calculate the force between the wires. F = (4p X 10-7 T m/A)(8.0A)(8.0A)(2.0m) (2p)(3.0 X 10-3 m) F = 8.5 X 10-3 N
60 ExampleThe top wire carries a current of 80 A. How much current must the lower wire carry in order to leviate if it is 20 cm below the first and has a mass of 0.12 g/m?Solve for I2I2 = 15 A
61 Magnetic Field of a Solenoid Solenoids are one of the most common electromagnetsSolenoids consist of a tightly wrapped coil of wire, sometimes around an iron core. The multiple loops and the iron magnify the effect of the single loop electromagnet.A solenoid behaves as just like a simple bar magnet but only when current is flowing.The greater the current and the more turns per unit length, the greater the field inside.An ideal solenoid has a perfectly uniform magnetic field inside and zero field outside.
62 The cross section of a solenoid is shown The cross section of a solenoid is shown. At point P inside the solenoid, the B field is a vector sum of the fields due to each section of wire. Note from the table that each section of wire produce a field vector with a component to the right, resulting in a strong field inside. In the ideal case the magnetic field would be uniform inside and zero outside.B = 0I out of the pageBPxI into the page
63 A cross-sectional view of a tightly wound solenoid Length LA cross-sectional view of a tightly wound solenoidIf the solenoid is long compared to its radius, we assume the field inside is uniform and outside is zeroApply Ampère’s Law to the red dashed rectangle
64 Magnetic Field of a Solenoid An ideal solenoid is approached when:the turns are closely spacedthe length is much greater than the radius of the turnsConsider a rectangle with side ℓ parallel to the interior field and side w perpendicular to the fieldThe side of length ℓ inside the solenoid contributes to the fieldThis is path 1 in the diagram
65 Magnetic Field of a Solenoid Applying Ampere’s Law givesThe total current through the rectangular path equals the current through each turn multiplied by the number of turnsSolving Ampere’s law for the magnetic field isn = N / ℓ is the number of turns per unit lengthThis is valid only at points near the center of a very long solenoid
66 the magnetic field inside a solenoid is andSo we havewherenumbers of loops / length of coilor. The magnetic field inside a solenoid is constant. If the length of the solenoid is much greater than its diameter the magnetic field outside the solenoid is nearly zero.The magnetic field is strongest at the centre of the solenoid and becomes weaker outside.
67 Magnetic field at the center of a current-carrying solenoid (N is the number of turns): B=m0NI/L, where L is the length of the solenoid and with n=N/L (number of turns per unit lengths) we get:B=m0nI ( Ampère’s law)The longer the solenoid, the more uniform is the magnetic field across the cross-sectional area with in the coil.The exterior field is nonuniform, much weaker, and in the opposite direction to the field inside the solenoid
68 Magnetic Field in a Solenoid, final The field lines of the solenoid resemble those of a bar magnet
69 Solenoids and Bar Magnets A solenoid produces a magnetic field just like a simple bar magnet. Since it consists of many current loops, the resemblance to a bar magnet’s field is much better than that of a single current loop.
70 The field lines in the interior are Magnetic Field of a SolenoidThe field lines in the interior areapproximately parallel to each otheruniformly distributedclose togetherThis indicates the field is strong and almost uniform
71 Magnetic Field of a Solenoid The field distribution is similar to that of a bar magnetAs the length of the solenoid increasesthe interior field becomes more uniformthe exterior field becomes weaker
72 When two current carrying wires are near each other, each one exerts a force on the other. This can be a repulsive force or an attractive force depending on the current direction.Consider wire 1 to be fixed. In diagram (a) wire 2 experiences a force due to the magnetic field generated by wire 1 and the current flowing through wire 2. Right hand rule # 2, when applied to wire 1, tells us that the magnetic field at wire 2 acts upward.Right hand rule # 1 when applied to wire 2 tells us that the force acting on wire 2 is to the right, perpendicular to both the current direction and the magnetic field direction.The magnitude of the force between two current carrying wires can be calculated using the equations:F2 = I2LBsinθ
73 Magnetic Flux ФB = B · A = B A cos Magnetic flux,, is a measure of the amount of magnetic field lines going through an area. If the field is uniform, flux is given by:ФB = B · A = B A cosThe area vector in the dot product is a vector that points perpendicular to the surface and has a magnitude equal to the area of the surface.Imagine you’re trying to orient a window so as to allow the maximum amount of light to pass through it. To do this you would, of course, align A with the light rays. With = 0, cos = 1, and the number of light rays passing through the window (the flux) is a max. Note: with the window oriented parallel to the rays, = 90° and ФB = 0 (no light enters the window).The SI unit for magnetic flux is the tesla-square meter: T m2. This is also know as a weber (Wb).
74 Magnetic FluxThe first step to understanding the complex nature of electromagnetic induction is to understand the idea of magnetic flux.BFlux is a general term associated with a FIELD that is bound by a certain AREA. So MAGNETIC FLUX is any AREA that has a MAGNETIC FIELD passing through it.AWe generally define an AREA vector as one that is perpendicular to the surface of the material. Therefore, you can see in the figure that the AREA vector and the Magnetic Field vector are PARALLEL. This then produces a DOT PRODUCT between the 2 variables that then define flux.
75 Magnetic Flux How could we CHANGE the flux over a period of time? We could move the magnet away or towards (or the wire).We could increase or decrease the area.We could ROTATE the wire along an axis that is PERPENDICULAR to the field thus changing the angle between the area and magnetic field vectors.
76 Magnetic flux is used in the calculation of the induced emf.
79 The unit of flux density is the Weber per square meter. Magnetic flux lines are continuous and closed.Direction is that of the B vector at any point.Flux lines are NOT in direction of force but ^.When area A is perpendicular to flux:The unit of flux density is the Weber per square meter.
80 Calculating Flux Density When Area is Not Perpendicular The flux penetrating the area A when the normal vector n makes an angle of q with the B-field is:The angle q is the complement of the angle a that the plane of the area makes with the B field. (Cos q = Sin a)
83 You can change the magnetic flux through a given surface by ExampleYou can change the magnetic flux through a given surface bychanging the magnetic field.changing the surface area over which the magnetic field is distributed.changing the angle between the magnetic field and surface in question.any combination of a through c.none of these strategies.
84 ExampleThe figure shows a flat surface with area 3cm2 in a uniform B-field. If the magnetic flux through this area is 0.9mWb, Find the magnitude of B-field.
85 ExampleSuppose you double the magnetic field in a given region and quadruple the area through which this magnetic field exists. The effect on the flux through this area would be toleave it unchanged.double it.quadruple it.increase it by a factor of six.increase it by a factor of eight.
86 ExampleWhat is the flux density in a rectangular core that is 8 mm by 10 mm if the flux is 4 mWb?
87 Summary Calculate the B field due to a current using Biot-Savart Law Permiability constant:B due to long straight wire: circular arc: complete loop:Force between two parallel currentsAnother way to calculate B is using Ampere’s Law (integrate B around closed Amperian loops):m0 = permeability constantexactly