Ben Gurion University of the Negev

1. Ben Gurion University of the Negev www.bgu.ac.il/atomchip Physics 2B for Materials and Structural Engineering Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shay Inbar Week 7. Sources of the magnetic field – Biot-Savart law • parallel currents • Ampère’s law • Gauss’s law for B• towards Maxwell’s equations Source: Halliday, Resnick and Krane, 5th Edition, Chap. 33.

2. Biot-Savart law The last lecture was all about how moving charges respond to magnetic fields. This lecture is all about how moving charges create magnetic fields! In 1820, the Danish scientist Hans Christian Oersted gave a lecture-demonstration about how an electric current in a wire generates heat. For another part of the demonstration, he had a compass nearby. While lecturing, he noticed to his surprise that every time the electric current was switched on, the compass needle moved. He said nothing but, after a few months of experiments, published his discovery (in Latin) with no explanation.

3. dB r ds Biot-Savart law Within months, Jean-Baptiste Biot and Félix Savart had made careful quantitative experiments on the force exerted by an electric current on a nearby magnet. They summarized their results in the law named after them.

4. dB r ds Biot-Savart law Each current element ds contributes to the magnetic field B(r) a contribution where μ0 = 4π × 10–7 T·m/A . “the permeability(פרמאביליות, חלחלות) of free space”

5. dB dB r ds Biot-Savart law Each current element ds contributes to the magnetic field B(r) a contribution where μ0 = 4π × 10–7 T·m/A . • The vector dB is perpendicular to ds and to r. • The vector dB is proportional to I, to 1/r2 and to |ds|. • When ds and r are parallel, dB vanishes. • When ds and r are not parallel, dB is perpendicular to both of them, according to the right-hand rule, and proportional to the sine of the ds → r angle.

6. Right-hand rule (for the Biot-Savart law): ds r dB

7. y dB r a θ1 θ θ2 x ds = dx Biot-Savart law Example 1: A point makes angles θ1 and θ2 with the ends of a straight wire carrying current I. The distance of the point from the wire is a. What is B? Answer: Introduce x- and y-axes. Note that r = (0,a) – (x,0) = (–x, a), that a = r sin θ and thus r = a/sin θ. Note also that x = –r cos θ= –a cot θ, so dx = a dθ /sin2θ.

8. y dB r a θ1 θ θ2 x ds = dx Biot-Savart law Example 1: A point makes angles θ1and θ2 with the ends of a straight wire carrying current I. The distance of the point from the wire is a. What is B? Answer: The magnetic field is the integral but dx/r2 = dθ/a, so

9. y dB r a θ1 θ θ2 x ds = dx Biot-Savart law Example 1: A point makes angles θ1 and θ2 with the ends of a straight wire carrying current I. The distance of the point from the wire is a. What is B? Answer: The magnetic field is the integral

10. y dB r a θ1 θ θ2 x ds = dx Biot-Savart law Example 2: What is B if the wire is straight and infinitely long? Answer: Then θ1 = 0 and θ2 = π so

11. y dB r a θ1 θ θ2 x ds = dx Biot-Savart law In this example, a different version of the right-hand rule is useful: If your thumb shows the direction of the current, your fingers curl in the direction of the magnetic field.

12. y r dB R θ Biot-Savart law Example 3: What is B a distance x along the symmetry axis of a circular loop of radius R carrying current I? Answer: The magnitude of dB is dB = μ0 I ds/4πr2 , where r2 = x2 + R2, but only the x-component of dB survives. We multiply dB by cos θ = R/r and replace ds by 2πR to get θ x

13. Parallel currents Now let’s look at magnetic fields and moving charges as a complete system. Our first question is simple but important: Two straight, infinitely long parallel wires, of separation a, carry currents I1 and I2 . What is their mutual force per unit length? The field due to one current (say, I1) is μ0 I1/2πa. It exerts a magnetic force (μ0 I1/2πa) I2L. on any section of length L in the other wire. Therefore the magnetic force per unit length FB /L is FB /L= μ0 I1I2/2πa. This force per unit length is invariant if we interchange I1 and I2. If we check the direction of each force, we find that the mutual forces are equal and opposite, as Newton’s third law requires.

14. L I1 B2 a F2 I2 a Parallel currents Do parallel currents attract or repel each other? What about anti-parallel currents?

15. Parallel currents A large current begins to flow through a spring. Does the spring expand or contract?

16. Parallel currents We calculated the magnetic force per unit length to be Now let’s try some numbers. Assume I1 = I2 = I, a = 1 m, and FB/L = 2 × 10–7 N/m. Since μ0 = 4π × 10–7 T·m/A, we have Since 1 T = N /A·m (recall FB= qv × B), we obtain I2 = 1 A2 and an operational definition of the ampere: When the force per unit length between two long, parallel wires separated by 1 m and carrying identical currents I is FB= 2 × 10–7 N/m, the current in each wire is 1 A.

17. Parallel currents We likewise obtain an operational definition of the coulomb: When a wire carries a steady current of 1 A, the amount of charge that flows through the wire in 1 s is 1 C.

18. I B dr I = 0 Ampère’s law Oersted’s discovery that an electric current deflects a compass needle suggests the following experiment: B(r) depends only on r. From the Biot-Savart law, B = μ0 I/2πr and so in this case, at least,

19. I B dr I = 0 Ampère’s law Oersted’s discovery that an electric current deflects a compass needle suggests the following experiment: Ampère discovered that this law holds in general, and the equation is called Ampère’s law.

20. d 5 A 1 A c b a 2 A Ampère’s law Example 1: Rank for the four closed curves a, b, c, d using Ampère’s law.

21. B I Ampère’s law Example 2: What is B inside and outside an infinitely long conducting wire of radius R, carrying uniform current I? Answer: By symmetry, B must circulate around the axis of the wire. Outside the wire, we choose a concentric circle with radius r > R. Then Ampère’s law tells us that 2πrB= μ0 I so B = μ0 I/2πr.

22. I Ampère’s law Example 2: What is B inside and outside an infinitely long conducting wire of radius R, carrying uniform current I? Answer: By symmetry, B must circulate around the axis of the wire. Inside the wire, we choose a concentric circle with radius r < R. Then Ampère’s law tells us that 2πrB = μ0 I(πr2/πR2)so B = μ0 Ir/2πR2. B

23. Js B B Ampère’s law Example 3: What is B for an infinite current sheet carrying surface current density Js? Answer: We use the rectangular loop shown to evaluate The top and bottom sides don’t contribute, while the right and left sides each contribute BL, where L is the height of the rectangle. We have 2BL =μ0JsL, so B =μ0Js/2.

24. W 2 3 L 1 4 B Ampère’s law Example 4: Ampère’s law, like Gauss’s law, is especially useful in cases of high symmetry. Let’s apply Ampère’s law to find the field of a long solenoid, which would be awkward using the Biot-Savart law. Experiment shows that B outsidesuch a solenoid is very weak. What is the field B inside the solenoid?

25. W 2 3 L 1 4 B Ampère’s law Example 4: Ampère’s law, like Gauss’s law, is especially useful in cases of high symmetry. Let’s apply Ampère’s law to find the field of a long solenoid, which would be awkward using the Biot-Savart law. Consider the integral of B·dr around a rectangular path as shown, in the direction shown. On Side 3 , B is negligible. On Sides 2 and 4 the integrals of B·ds sum to 0. So B·dr on Side 1 is what must equal μ0 times the current through the rectangle.

26. W 2 3 L 1 4 B Ampère’s law Example 4: Ampère’s law, like Gauss’s law, is especially useful in cases of high symmetry. Let’s apply Ampère’s law to find the field of a long solenoid, which would be awkward using the Biot-Savart law. We have BL = μ0NI, where N windings, each carrying current I, pass through the rectangle. Letting n = N/L denote the number of windings per unit length, we conclude that B = μ0nI. This is the magnetic field anywhere inside the solenoid; it is uniform.

27. Gauss’s law for B We have seen that the field lines of E and B are similar. One striking difference is that there are no magnetic monopoles. This similarity of the field lines is not just a coincidence. We will see that the fundamental equations for E and B are very similar. First, there is a version of Gauss’s law for the magnetic field. It states that total magnetic flux out of a closed surface equals 0: This law can be derived from the law of Biot and Savart.

28. Towards Maxwell’s equations The fundamental equations for E and B are very similar. First, there is a version of Gauss’s law for the magnetic field. It states that total magnetic flux out of a closed surface equals 0: Second, we have already seen a statement about E that is very similar to Ampère’s law:

29. Remember Faraday? • Now we can add a new basic rule about field lines: • They never start or stop in empty space – they stop or start on a charge or extend to infinity. • They never cross – if they did, a small charge placed at the crossing would show the true direction of the field there. • The density of field lines in one direction is proportional to the strength of the field in the perpendicular direction. • There are no closed loops! E E E E

30. (Gauss’s law) (Ampère’s law) Towards Maxwell’s equations The set of four fundamental equations for E and B, together with the Lorentz force law FEM = q (E + v × B), sum up everything we have learned so far about electromagnetism!

31. (Gauss’s law) (Ampère’s law) Towards Maxwell’s equations The set of four fundamental equations for E and B, are correct for all static charges and steady-state currents, i.e. for electrostatics and magnetostatics. But if E and B change in time, two of the equations need correction. Guess which two?

32. Halliday, Resnick and Krane, 5th Edition, Chap. 33, Prob. 9: A long solenoid, centered along the z-axis, has 100 turns per centimeter. An electron moves within the solenoid in a circle of radius 2.30 cm in the xy-plane. The speed of the electron is 0.0460c, where c = 3.00 × 108 m/s is the speed of light. What is the current I in the wire of the solenoid?

33. Halliday, Resnick and Krane, 5th Edition, Chap. 33, Prob. 9: A long solenoid, centered along the z-axis, has 100 turns per centimeter. An electron moves within the solenoid in a circle of radius 2.30 cm in the xy-plane. The speed of the electron is 0.0460c, where c = 3.00 × 108 m/s is the speed of light. What is the current I in the wire of the solenoid? Answer: The cyclotron frequency ω of the electron is ω = v/r = 0.0460c/0.0230 m = 6.00 × 108 Hz and we derived ω = eB/m where e is the electron charge and m is its mass. Also B = μ0nI for the field in a solenoid, hence I = mω/neμ0 . Putting in the numbers (m = 9.11× 10–31 kg and e = 1.60 × 10–19 C) we obtain I = 0.272 A.

34. –q ∂S B q r Surface S2 I Surface S1 The problem with Ampère’s law How do we apply Ampère’s law, to this electrical circuit? Current flows through Surface S1. No current flows through Surface S2. Hence we have a μ0I/2πr = B = 0: CONTRADICTION!

35. –q q I ∂S The problem with Ampère’s law You may say, “This is silly. Ampère’s law is not like Faraday’s law, no matter what Ampère thought. Forget about S1 and S2! Just consider how much current flows through the closed loop ∂S.” Yes, but suppose the closed loop straddles the capacitor: How much current is flowing through the loop ∂S now?

36. (Gauss’s law) (Faraday’s law) (Ampère’s law as modified by Maxwell) Maxwell’s fix This set of four fundamental equations for E and B, is called Maxwell’s equations!

37. –q ∂S B q r Surface S2 I Surface S1 Maxwell’s fix Let’s check that Maxwell’s modification of Ampère’s law, fixes the problem, for the case of a parallel- plate capacitor. If we consider Surface S1, we have (2πr)B = μ0I .

38. –q ∂S B q r Surface S2 I Surface S1 Maxwell’s fix Let’s check that Maxwell’s modification of Ampère’s law, fixes the problem, for the case of a parallel- plate capacitor. If we consider Surface S2, we have (2πr)B = μ0ε0 dΦE/dt. For a parallel-plate capacitor, it still equals μ0I!

39. Maxwell’s fix Let’s check that Maxwell’s modification of Ampère’s law, fixes the problem, for the case of a parallel-plate capacitor. If we consider Surface S1, we have (2πr)B = μ0I . If we consider Surface S2, we have (2πr)B = μ0ε0 dΦE/dt. For a parallel-plate capacitor of area A and plate separation d, ΦE = AE = AV/d = Aq/Cd, and C = ε0 A/d, so ΦE = qA/Cd = q/ε0. We get dΦE/dt = I/ε0 so (2πr)B = μ0ε0 dΦE/dt = μ0I! Note I = ε0 dΦE/dt is often called the displacement current.