Ch 30 - Sources of Magnetic Field

Transcription

1 Ch 30 - Sources of Magnetic Field

2 Currents produce Magnetism? 1820, Hans Christian Oersted: moving charges produce a magnetic field. The direction of the field is determined using a RHR.

3 Oersted (1820) f the wire is grasped with the right hand, with the thumb in the direction of current flow, the fingers curl around the wire in the direction of the magnetic field. The magnitude of B is the same everywhere on a circular path perpendicular to the wire and centered on it. Experiments reveal that B is proportional to, and inversely proportional to the distance from the wire.

4 Example 3 Predict the orientation of the compass needles.

5 The Biot-Savart Law magnetic field at a point P perpendicular to segment of wire ds & perpendicular to unit vector r from ds to P. d B = k m d s ˆ r steady current in direction ds r 2 db out P element of wire determines direction of field k m = 10-7 T m/a k m = µ o /4!, where µ o = permeability of free space µ o =4! x10-7 ˆr r θ ds

6 Example 1 Determine the magnitude and direction of the magnetic field at the point O in the diagram. (Current flows from top to bottom, radius of curvature = r.) For straight segments, ds r = 0. For curved segment, ds r = ds, because s and r are. B = µ o 4π B = µ o 4πr 2 B = µ o θ 4πr ds r 2 rdθ θ ds

7 Example 2 Determine the direction and total magnitude of the magnetic field at the point P shown here, near a long, thin, wire. r ds a P The magnetic field for an infinitely long straight wire is B = µ o 2πa

8 Magnetic Force between wires F = B = Bsinθ B (due to wire) = µ o 2πa % F = 1 µ ( o 2 ' *, or & 2πa ) F = µ o 1 2 2πa

9 Ampère s Law B d s f we evaluate the dot-product " in a circle around this wire, and sum this product over the entire path of the circle, we get B d $ s = B ds = µ o ' & ) 2πr % 2πr( ( ) = µ o ds B Line integral, taken around a closed path that surrounds the current. This result is valid for any closed path that encloses a wire conducting a steady current, and is known as Ampère s Law: B d s = µ o

10 Gauss & Ampère E da = q in ε o B d s = µ o

11 Example 3 Use Ampère s Law to calculate the magnetic field at a distance r away from a current-carrying wire. B B ds = µ o ds = µ o B(2πr) = µ o B = µ o 2πr (for r > R) r < o r o = πr2 πr 2, so r = r2 R 2 o r r R B = µ o r 2 R 2 o 2πr = µ or o (for r < R) 2 2πR

12 Example 5 Find the magnetic field B for a thin, infinite sheet of current, carrying current of linear density J in the z direction. B B ds = µ o ds = µ o (J) L B(2) = µ o (J) B = µ o J 2 w

13 Example 4 Use Ampère s Law to calculate to calculate the magnetic field inside a toroid of N loops, at a distance r from the center. To determine magnetic field inside torus, draw Amperian line inside coil. The wire passes around that Amperian line N times. B B ds = µ o ds = µ o B(2πr) = µ o N B = µ o N 2πr Assumptions: r of torus is large compared to cross-sectional radius, so B is ~ uniform within torus. Also, for an ideal torus (closely spaced turns of wire), the external magnetic field is close to 0.

14 Electromagnets & Solenoids A solenoid is a long wire wound in the form of a helix. f the turns are closely space and the solenoid is finite in length, the field lines from a solenoid look very similar to those of a bar magnet.

15 B Field for a Solenoid nside = strong B field Outside = weak B field Between coils = 0 field

16 Example 6 Use Ampère s Law to calculate the magnetic field of a solenoid B B ds = µ o ds = µ o Bl = µ o N B = µ on l = µ o n

17 Electric Flux (Review) Brief review of Electric Flux: Φ = E A Α θ Ε θ Φ = EAcosθ The net electric flux through any closed surface is equal to the net charge inside the surface divided by o. Φ c = E da = q in ε o

18 Magnetic Flux Φ B = B A Α θ Ε Φ B = BAcosθ θ

19 Example 7 Find the total magnetic flux through the loop shown here. Φ B = B da Φ B = Φ B = Φ B = µ o b 2π µ o 2πr da µ o 2πr b dr a +c c 1 r dr Φ B = µ ob 2π ln % a + c ( ' * & c ) c a b

20 Gauss s Laws Electric field intensity depends only on the net internal charge. Magnetic field lines are continuous, and form closed loops. Magnetic field lines created by currents don t begin or end at any point.therefore, the net magnetic flux through any closed surface is always zero! Φ B = Φ c = E da B da = 0 = q in ε o N S

21 Difficulty w/ampère s Law B d s = µ o Two problems: What if current is changing? What if Amperian path doesn t enclose current?

22 Ampère s Law (redux) dφ ε E d o dt B d s = µ ( + ) o d B d s = µ o ( + ε o dφ E dt )

23 Ampère-Maxwell Law B d s = µ o ( + ε o dφ E dt )