Electricity and Magnetism Magnetic Field from Moving Charges

Transcription

1 Electricity and Magnetism Magnetic Field from Moving Charges Lana Sheridan De Anza College Nov 17, 2015

2 Last time force on a wire with a current in a B-field torque on a wire loop in a B-field motors relating a current loop to a magnet magnetic dipole moment torque and potential energy of magnetic dipole magnetism of matter

3 Overview magnetic fields from moving charges magnetic fields around current-carrying wires forces between parallel wires Gauss s law

4 Magnetic fields from moving charges and currents We are now moving into chapter 29. Anything with a magnet moment creates a magnetic field. This is a logical consequence of Newton s third law.

5 Magnetic fields from moving charges A moving charge will interact with other magnetic poles. This is because it has a magnetic field of its own. The field for a moving charge is given by the Biot-Savart Law: B = µ 0 q v ˆr 4π r 2

6 Magnetic fields from moving charges B = µ 0 q v ˆr 4π r 2 1 Figure from rakeshkapoor.us.

7 Magnetic fields from currents B = µ 0 q v ˆr 4π r 2 We can deduce from this what the magnetic field do to the current in a small piece of wire is. Current is made up of moving charges! q v = q s t = q s = I s t We can replace q v in the equation above.

8 Magnetic fields from currents This element of current creates a magnetic field at P, into the page. current-length element a differential magnetic nt P. The green (the ) at the dot for point P db : is directed into the i ids θ ds ˆ r r P Current distribution db(into page) This is another version of the Biot-Savart Law: B seg = µ 0 I s ˆr 4π r 2 where B seg is the magnetic field from a small segment of wire, of length s.

9 mmation Magnetic fields from currents The magnetic field vector ecause of at any point is tangent to s a scalar, Magnetic field a circle. around a wire segment, viewed end-on: being the Wire with current into the page a current- (29-1) hat points constant, B B (29-2) the cross Fig The magnetic field lines produced by a current in a long straight wire

10 Magnetic fields from currents NETIC How FIELDS to determine DUE TO the CURRENTS direction of the field lines (right-hand rule): s the dia curig. 29-2, ld B : at erpend din of the ) If the o the ed rahe page, (a) B i (b) Here is a simple right-hand rule for finding the direction of the mag set up by a current-length element, such as a section of a long wire: B i The thumb is current's dire The fingers re the field vecto direction, whi tangent to a c Right-hand rule: Grasp the element in your right hand with your extended th

11 Magnetic field from a long straight wire The Biot-Savart Law, B seg = µ 0 I s ˆr 4π r 2 implies what the magnetic field is at a perpendicular distance R from an infinitely long straight wire: B = µ 0I 2πR (The proof requires some calculus.)

12 Force between 2 wires With two current carrying wires, each creates its own magnetic field: B = µ 0I 2πR The result is that the wires interact, much like two bar magnets producing magnetic fields would.

13 Force between 2 wires Currents in opposite directions repel, currents in the same direction attract. 1 Figure from salisbury.edu.

14 Force between 2 wires It is a bit more intuitive to think about the force per unit length on the wires (since longer wires will experience larger forces). The force per unit length on a wire due to another parallel wire at a distance d: F B l = µ 0I 1 I 2 2πd

15 Force between 2 wires It is a bit more intuitive to think about the force per unit length on the wires (since longer wires will experience larger forces). The force per unit length on a wire due to another parallel wire at a distance d: F B l = µ 0I 1 I 2 2πd Where does this come from? The force on a current carrying wire is: F = IL B

16 Force between 2 wires Suppose that wire a produces a field: B a = µ 0I a 770 CHAPTER 29 MAGNETIC 2πd FIELDS DUE TO CURRENTS The force on wire b is: The field due to a at the position of b creates a force on b. i a i b F ba Fig Two parallel wires carrying currents in the same direction attract each other. B : a is the magnetic field at wire b produced by the current in wire a. F : ba is the resulting force acting on wire b because it carries current in B : a. a L L d b B a (due to i a ) F = I b L µ 0I a 2πd sin(90 ) F B L = µ 0I a I b 2πd 29-3 Force Between Tw Two long parallel wires carrying shows two such wires, separated Let us analyze the forces on thes We seek first the force on w current produces a magnetic fie causes the force we seek. To fi direction of the field B : a at the sit wire b is, from Eq. 29-4, The curled straight right-hand down, as Fig shows. Now that we have the field Equation tells us that the nal magnetic field is L : B : a where is the length vector of t dicular to each other, and so with

17 Force between 2 wires 1 Figure from Stonebrook Physics ic.sunysb.edu.

18 10 6 g,and then launches it with a speed of 10km/s,all within 1ms.S rail Question guns may be used to launch materials into space from mining opera e Moon or an asteroid. The figure here shows three long, straight, parallel, equally spaced HECKPOINT 1 wires with identical currents either into or out of the page. Rank e figure thehere wires shows according three to long, thestraight, magnitude parallel, of theequally force on spaced each wires due towith iden rrents the either currents into or in out the of other the two page.rank wires, greatest the wires first. according to the magnitud e force on each due to the currents in the other two wires, greatest first. a b c A a, b, c 4 Ampere s Law B b, c, a an find C c, the b, net a electric field due to any distribution of charges by first w ifferential electric field de : due to a charge element and then summin ributions of de : from all the elements. However, if the distribution is co d, we may have to use a computer. Recall, however, that if the distrib 1 Halliday, Resnick, Walker, pg 771.

19 10 6 g,and then launches it with a speed of 10km/s,all within 1ms.S rail Question guns may be used to launch materials into space from mining opera e Moon or an asteroid. The figure here shows three long, straight, parallel, equally spaced HECKPOINT 1 wires with identical currents either into or out of the page. Rank e figure thehere wires shows according three to long, thestraight, magnitude parallel, of theequally force on spaced each wires due towith iden rrents the either currents into or in out the of other the two page.rank wires, greatest the wires first. according to the magnitud e force on each due to the currents in the other two wires, greatest first. a b c A a, b, c 4 Ampere s Law B b, c, a an find C c, the b, net a electric field due to any distribution of charges by first w ifferential electric field de : due to a charge element and then summin ributions of de : from all the elements. However, if the distribution is co d, we may have to use a computer. Recall, however, that if the distrib 1 Halliday, Resnick, Walker, pg 771.

20 Magnetic Permeability A constant we will need is: µ 0 = 4π 10 7 Tm/A µ 0 is called the magnetic permeability of free space. It arises when we look at magnetic fields because of our choice of SI units. Whenever we use µ 0 we assume we are considering the magnetic field to be in a vacuum or air. µ 0 is not the magnetic dipole moment µ! Another notation coincidence.

21 Definition of the Ampère (Amp) This relation: F B L = µ 0I 1 I 2 2πd gives us the formal definition of the Ampère. Ampère Unit Two long parallel wires separated by 1 m are said to each carry 1 A of current when the force per unit length on each wire is N/m.

22 Gauss s Law for Magnetic Fields There is more we can say about magnetic fields. When studying electric fields we used Gauss s law to understand the how the electric field looked around a point charge. There is also Gauss s law for magnetic fields, but it tells us something different about magnetic fields. Reminder about Gauss s law for electric fields...

23 ics in attaining this ctric Gauss s field E : of Law the for Electric Fields basic idea, we considered the Then we simplified dicular Gauss s components law relates the electric field across a closed surface (eg. a sphere) to the amount of net charge enclosed by the surface. save far more work matician and physing the fields de : of siders a hypothetical is Gaussian surface, s our calculations of distribution. For exse the sphere with a en, as we discuss in t that? Spherical Gaussian surface ian surface to the E field on a Gaussian s a limited example, lly outward from the ediately tells us that Fig A spherical Gaussian surface. If the electric field vectors are of uniform magnitude and point radially outward at all surface points,

24 Flux 726 Chapter 24 Gauss s Law A The number of field lines that go through the area A is the same as the number that go through area A. w u A w, Normal u S E From th coulomb lines pe If the through the norm that the cross th lar to th A 5,w. by w 5 areas ar

25 Reminder: Gauss s Law for Electric Fields Gauss s Law for Electric fields: Φ E = E da = q enc ɛ 0 The electric flux through a closed surface is equal to the charge enclosed by the surface, divided by ɛ 0. There is a similar expression for magnetic flux! First we must define magnetic flux, Φ B.

26 Magnetic(30.18) Flux Definition of magnetic flux field B S that makes an case is (30.19), then u and the he plane as in Figure aximum value). weber (Wb); 1 Wb 5 Magnetic flux plane is a agnetic to the plane. B S u S d A Figure The magnetic flux through an area element da is S B? d S A 5 B da cos u, where d S A is a vector perpendicular to the surface. The magnetic flux of a magnetic field through a surface A is Φ B = B A da S Units: Tm 2 If the surface is a flat plane and B is uniform, that just reduces to: Φ B = B A

27 Gauss s Law for Magnetic Fields Gauss s Law for magnetic fields.: B da = 0 Where the integral is taken over a closed surface A. (This is like a sum over the flux through many small areas.) We can interpret it as an assertion that magnetic monopoles do not exist. The magnetic field has no sources or sinks.

28 Gauss s Law for Magnetic Fields 32-3 INDUCED MAGNETIC FIELDS B da = PART 3 re complicated than does not enclose the Fig encloses no x through it is zero. nly the north pole of l S. However, a south ce because magnetic like one piece of the encloses a magnetic Surface II N B Surface I S tom faces and curved B of the uniform and and B are arbitrary of the magnetic flux

29 B-Field around a wire revisited Gauss s law will not help us find the strength of the B-field around a wire, or wires carrying current: the flux through any closed surface will be zero. Another law can: Ampère s Law.

30 Summary field from a moving charge field from a current force between two parallel wires Guass s law Homework Halliday, Resnick, Walker: PREVIOUS: Ch 28, Problems: 54, 55, 57, 61, 65 NEW: Ch 29, onward from page 783. Questions: 3; Problems: 1, 11, 21, 23