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Quiz 1 100 90 80 70 60 Borderline Series1 50 Trouble 40 30 Deep Trouble 20 10 0 10 20 30 40 From Last time Temperature effects on resistance Permanent magnets and gaussmeter. Electromagnets - BigBite for Research Force on a wire in a magnetic field - Demo Force on a coil in a magnetic field - Demo Crossed E/B fields. Discovery of the electron Lecture 9 starts here Torques on current loops Electric motors operate by connecting a coil in a magnetic field to a current supply, which produces a torque on the coil causing it to rotate. F i . P a i F B B b Above is a rectangular loop of wire of sides a and b carrying current i. B is in the plane of the loop. Equal and opposite forces F = iaB are exerted on the sides a. No forces exerted on b since i B Since net force is zero, we can evaluate t (torque) about any point. Evaluate it about P (cm). For N loops we mult by N t = Fbsinq F = iaB t = iabBsinq t = NiabBsin q = NiABsin q A=ab=area of loop Torque on a current loop t = NiABsin q n̂ q = NiA = magnetic dipole moment B t = Bsinq t =B Torque tends to rotate loop until plane is^ to B n̂ ( n̂ parallel to B). n̂ q Must reverse current using commutator to keep loop turning. Loop will rotate and stop when the plane is perpendicular to B Magnetic dipole moment is in the same direction as the normal. http: //hyperphysics .phy-astr .gsu.edu/hbase/magnetic/motdc .html#c4 Split ring commutator + - Show Galvanometer Demo Electron moving with speed v in a crossed electric and magnetic field in a cathode ray tube. F = qE qvxB e y Electric field bends particle upwards Magnetic field bends it downwards Discovery of the electron by J.J. Thompson in 1897 1. E=0, B=0 Observe spot on screen 2. Set E to some value and measure y the deflection qEL2 y= 2mv 2 3. Now turn on B until spot returns to the original position qE = qvB E v= B 4 Solve for m B2L2 = q 2yE Show demo of CRT This ratio was first measured by Thompson to be lighter than hydrogen by 1000 How do you calculate the magnetic field from a moving charge or from a current flowing in a wire? What do the flux lines look like? Lecture 9 Magnetic Fields due to Currents Ch. 30 • • • • Cartoon - Shows magnetic field around a long current carrying wire and a loop of wire Topics – Magnetic field produced by a moving charge – Magnetic fields produced by currents. Big Bite as an example. – Using Biot-Savart Law to calculate magnetic fields produced by currents. – Examples: Field at center of loop of wire, at center of circular arc of wire, at center of segment of wire. – Amperes’ Law : Analogous to Gauss’ L:aw in electrostatics, Useful in symmetric cases. – Infinitely long straight wire of radius a. Find B outside and inside wire. – Solenoid and Toroid Find B field. – Forces between current carrying wires or parallel moving charges Demos – Torque on a current loop(galvanometer) – Iron filings showing B fields around wires with currents. – Compass needle near current carrying wire – Big Bite as an example of using a magnet as a research tool. – Force between parallel wires carrying identical currents. Polling Magnetic Fields due to Currents • • • • • So far we have used permanent magnets as our source of magnetic field. Historically this is how it started. In early decades of the last century, it was learned that moving charges and electric currents produced magnetic fields. How do you find the Magnetic field due to a moving point charge? How do you find the Magnetic field due to a current? – Biot-Savart Law – direct integration – Ampere’s Law – uses symmetry Examples and Demos Topic: Moving charges produce magnetic fields q q r̂ v r 0 v rˆ B= q 2 4 r Determined from Experiment 1. Magnitude of B is proportional to q, v, and 1/r2. 2. B is zero along the line of motion and proportional to sin qat other points. 3. The direction is given by the RHR rotating 4. Magnetic permeability 0 = 4 10 7 v into N A2 Demo Sprinkle iron filings around a wire carrying current r̂ Example: A point charge q = 1 mC (1x10-3C) moves in the x direction with v = 108 m/s. It misses a mosquito by 1 mm. What is the B field experienced by the mosquito? r̂ 90o 108 m/s 0 v B= q 2 4 r B = 10 7 N A2 B = 10 4 T 3 10 C 10 8 m s 1 10 6 m 2 Topic: A current produces a magnetic field Recall the E field of a charge distribution “Coulombs Law” kdq dE = 2 r̂ r To find the B field of a current distribution use: r̂ Biot-Savart Law 0 ids r̂ dB = 4 r 2 This Law is found from experiment Find B field at center of loop of wire lying in a plane with radius R and total current i flowing in it. i 0 i dl r̂ B= 4 r 2 d l r̂ dl q P R r̂ is a vector coming out of the paper The angle between dl and r is constant and equal to 90 degrees. dl r̂ = dl sin q k̂ sin q = sin 90 = 1 dl r̂ = dl k̂ B = dB = 0 idl 0 i 0 i k̂ 2 = k̂ 2 dl = k̂ 2 2 R 4 R 4 R 4 R Magnitude of B field at center of loop. Direction is out of paper. i R k̂ 0i B= k̂ 2R Magnitude is Direction is k̂ 0i 2R Example Loop of wire of radius R = 5 cm and current i = 10 A. What is B at the center? Magnitude and direction i B= 0i 2R B = 4 107 10 A 2(.05m) B = 1.2 10 6 10 2 T N A2 B = 1.2 104 T = 1.2 Gauss Direction is out of the page. What is the B field at the center of a segment or circular arc of wire? v 0 i dl r̂ B= 4 r 2 i B = dB = dl r̂ = 0 i k̂ dl 2 4 R 0 i k̂ S 2 4 R Total length of arc is S. dl r̂ dl r̂ q0 R P dl r = rdl sinq k̂ sin q = sin 0 = 0 where S is the arc length S = Rq0 q0 is in radians (not degrees) B= 0 i q 0 k̂ 4 R q 0 = 100 deg=100 deg 6.28 rad = 100 0.0174 = 1.74 rad 360 deg Why is the contribution to the B field at P equal to zero from the straight section of wire? Suppose you had the following loop. Find magnetic field at center of arc length What is the magnitude and direction of B at the origin? B= 0 i S 4 R 2 0 i q 4 R 0 i i i B = 0 ( q0 q0 ) = 0 q0 4 R R/2 2 R B= Next topic: Ampere’s Law Allows us to solve certain highly symmetric current problems for the magnetic field as Gauss’ Law did in electrostatics. Ampere’s Law is Examples B dl = 0 Ic Current enclosed by the path This dl is along the Amperian loop not along the wire. dl is the same as ds below 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). dl r i By symmetry B i dl B dl = Bdl = B dl = B2r = 0i 0 i B= 2 r Suppose i = 10 A R = 10 cm 0 = 4 10 7 N A2 2 10 7 10 B= 10 1 B = 2 10 5 T Show Fe fillings around a straight wire with current, current loop, and solenoid. Rules for finding direction of B field from a current flowing in a wire 0i B= 2r Force between two current carrying wires Find the force due to the current element of the first wire and the magnetic field of the second wire. Integrate over the length of both wires. This will give the force between the two wires. Fba ia ib 0 = L 2 d 0i a Ba = 2d (Given by Ampere’s Law) Fba = i bL x Ba 0i a = i bL 2d The force is directed towards the wire (wires are attracted). Show Demo Suppose one of the currents is in the opposite direction? What direction is Fba ? Example: What is the magnetic field inside the wire? Find the magnetic field inside a long, thick wire of radius a 2r B dl = Bdl = B dl = B2r B2r = u0Ic 0 IC = current enclosed by the circle whose radius is r r 2 r2 IC = 2 i = 2 i a a r2 1 B = 0 2 i a 2r 0ir B= 2a 2 Example: Find field inside a solenoid. See next slide. Solenoid B = 0ni n is the number of turns per meter dl d c dl dl B a dl b First evaluate the right side of the integral , it’s easy B dl = 0 IC IC is the total current enclosed by the path Ic = nhi n is the the number of loops of current per meter and h is the length of one side. Right side = 0 nhi d dl dl c dl a B dl b Evaluate left side of the integral : c d a b B dl = B dl B dl B dl B dl a b c d = Bh 0 0 0 Bh = 0 nhi B = 0 ni n = the number of loops or turns per meter B dl = 0 IC Toroid Bdl = 0 Ni B 2r = 0 Ni 0 Ni B= 2r N is the total number of turns a<r<b a b dl B dl = Bdl cos 0 o = 1 Tokamak Toroid at Princeton I = 73,000 Amps for 3 secs B = 5.2 T Magnetic dipole inverse cube law z P . 0 B= 2 z 3 Magnetic field lines for a bar magnet and a current loop are similar Chapter 29 Problem 5 In Figure 29-36, two circular arcs have radii a = 13.1 cm and b = 10.6 cm, subtend angle θ = 74.0°, carry current i = 0.388 A, and share the same center of curvature P. What is the magnitude and direction of the net magnetic field at P? (Take out of the page to be positive.) Figure 29-36 Chapter 29 Problem 41 A 180 turn solenoid having a length of 22 cm and a diameter of 10 cm carries a current of 0.39 A. Calculate the magnitude of the magnetic field B inside the solenoid. Chapter 29 Problem 49 A student makes a short electromagnet by winding 150 turns of wire around a wooden cylinder of diameter d = 7.0 cm. The coil is connected to a battery producing a current of 4.0 A in the wire. (a) What is the magnetic moment of this device? (b) At what axial distance z d will the magnetic field of this dipole have the magnitude 5.0 µT (approximately onetenth that of Earth's magnetic field)?