1. CHAPTER OUTLINE 30.1 The Biot–Savart Law 30.2 The Magnetic Force Between Two Parallel Conductors 30.3 Ampère’s Law 30.4 The Magnetic Field of a Solenoid 30.5 Magnetic Flux 30.6 Gauss’s Law in Magnetism 30.7 Displacement Current and the General Form of Ampère’s Law

2. 30.1 The Biot–Savart Law Biot and Savart arrived at a mathematical expression that gives the magnetic field at some point in space in terms of the current that produces the field. That expression is based on the following experimental observations for the magnetic field dB at a point P associated with a length element ds of a wire carrying a steady current I :

3. The total magnetic field B created at some point by a current of finite size

4. Magnetic Field on the Axis of a Circular Current Loop

5. 30.2 The Magnetic Force Between Two Parallel Conductors parallel conductors carrying currents in the same direction attract each other, and parallel conductors carrying currents in opposite directions repel each other. The force per unit length:

6. Example 4 Interaction between current-carrying wires 2 parallel wires A & B, carry the same size of current. (a)Draw a few magnetic field lines due to the current through wire A. AB

7. Example 4 Interaction between current-carrying wires 2 parallel wires A & B, carry the same size of current. (b)Mark the direction of this magnetic field at wire B with an arrow. AB field at B due to current through A

8. Example 4 Interaction between current-carrying wires 2 parallel wires A & B, carry the same size of current. (c)Mark the direction of magnetic force acting on B due to magnetic field of the current through A. AB field at B due to current through A magnetic force

9. Example 4 Interaction between current-carrying wires 2 parallel wires A & B, carry the same size of current. (d)Deduce the direction of magnetic force acting on A due to magnetic field of the current through B. AB field at B due to current through A magnetic force field at A due to current through B

10. Example 4 Interaction between current-carrying wires 2 parallel wires A & B, carry same size of current. (e)Do A & B attract or repel each other? AB field at B due to current through A magnetic force field at A due to current through B They attract each other.

11. The force between two parallel wires is used to define the ampere as follows: The value 2 X 10-7 N/m is obtained from Equation 30.12 with I 1 = I 2 = 1 A and a = 1 m.

13. 30.3 Ampère’s Law

15. Example 30.4 The Magnetic Field Created by a Long Current-Carrying Wire

16. We need to find I’

17. A superconducting wire carries a current of 10 4 A. Find the magnetic field at a distance of 1.0 m from the wire. ( µ 0 = 4∏ x 10 -7 A-m/T) 1. 1. 2.0 x 10 -3 T 2. 2. 8.0 x 10 -3 T 3. 3. 1.6 x 10 -2 T 4. 4. 3.2 x 10 -2 T

18. A 2.0 m wire segment carrying a current of 0.60 A oriented parallel to a uniform magnetic field of 0.50 T experiences a force of what magnitude? 1. A. 6.7 N 2. B. 0.30 N 3. C. 0.15 N 4. D. zero

19. A proton moving with a speed of 3 x 10 5 m/s perpendicular to a uniform magnetic field of 0.20 T will follow which of the paths described below? (q e = 1.6 x 10 -19 C and m p = 1.67 x 10 -27 kg) 1. 1. a straight line path 2. 2. a circular path of 1.5 cm radius 3. 3. a circular path of 3.0 cm radius 4. 4. a circular path of 0.75 cm radius

20. A proton which moves perpendicular to a magnetic field of 1.2 T in a circular path of 0.08 m radius, has what speed? (q p = 1.6 x 10 -19 C and m p = 1.67 x 10 -27 kg) 1. A) 3.4 x 10 6 m/s 2. B) 4.6 x 10 6 m/s 3. C) 9.6 x 10 6 m/s 4. D) 9.2 x 10 6 m/s

21. A current carrying wire of length 50 cm is positioned perpendicular to a uniform magnetic field. If the current is 10.0 A and it is determined that there is a resultant force of 3.0 N on the wire due to the interaction of the current and field, what is the magnetic field strength? a) 0.6 T b) 1.50 T c) 1.85 x 10 3 T d) 6.7 x 10 3 T

22. A horizontal wire of length 3.0 m carries a current of 6.0 A and is oriented so that the current direction is 50° S of W. The horizontal component of the earth's magnetic field is due north at this point and has a strength of 0.14 x 10 -4 T. What is the size of the force on the wire? a)0.28 x 10 -4 N b)2.5 x 10 -4 N c)1.9 x 10 -4 N d)1.6 x 10 -4 N

23. 30.4 The Magnetic Field of a Solenoid A solenoid is a long wire wound in the form of a helix A reasonably uniform magnetic field can be produced in the space surrounded by the turns of wire—which we shall call the interior of the solenoid—when the solenoid carries a current. When the turns are closely spaced, each can be approximated as a circular loop, and the net magnetic field is the vector sum of the fields resulting from all the turns.

24. The contributions from sides 2 and 4 are both zero, again because B is perpendicular to ds along these paths An ideal solenoid is approached when the turns are closely spaced and the length is much greater than the radius of the turns. We can use Ampère’s law to obtain a quantitative expression for the interior magnetic field in an ideal solenoid. The contribution along side 3 is zero because the magnetic field lines are perpendicular to the path in this region. Side 1 gives a contribution to the integral because along this path B is uniform and parallel to ds

25. If N is the number of turns in the length !, the total current through the rectangle is NI.Therefore, Ampère’s law applied to this path gives

26. A solenoid with 500 turns, 0.10 m long, carrying a current of 4.0 A and with a radius of 10 -2 m will have what strength magnetic field at its center? (magnetic permeability in empty space µ 0 = 4 ∏ x 10 -7 T-m/A) 1. A) 31 x 10 -4 T 2. B) 62 x 10 -4 T 3. C) 125 x 10 -4 T 4. D) 250 x 10 -4 T

27. 30.5 Magnetic Flux Consider an element of area dA on an arbitrarily shaped surface, as shown in Figure. If the magnetic field at this element is B, the magnetic flux through the element is B. dA, where dA is a vector that is perpendicular to the surface and has a magnitude equal to the area dA. Therefore, the total magnetic flux Φ B through the surface is: Φ B =0

29. Magnetic Flux Graphical Interpretation of Magnetic Flux The magnetic flux is proportional to the number of magnetic flux lines passing through the area.

30. Magnetic Flux

31. A General Expression for Magnetic Flux The SI unit of magnetic flux is the weber (Wb), named after the German Physicist W.E. Weber (1804-1891). 1 Wb = 1 T.m 2.

32. EXAMPLE : Magnetic Flux A rectangular coil of wire is situated in a constant magnetic field whose magnitude is 0.50 T. The coil has an area of 2.0 m 2. Determine the magnetic flux for the three orientations,  = 0°, 60.0°, and 90.0°, shown below.

33. Example 30.8 Magnetic Flux Through a Rectangular Loop

34. So

35. 30.6 Gauss’s Law in Magnetism The magnetic field lines of a bar magnet form closed loops. Note that the net magnetic flux through a closed surface surrounding one of the poles (or any other closed surface) is zero. (The dashed line represents the intersection of the surface with the page.)

36. The electric field lines surrounding an electric dipole begin on the positive charge and terminate on the negative charge. The electric flux through a closed surface surrounding one of the charges is not zero.

37. 30.7 Displacement Current and the General Form of Ampère’s Law When a conduction current is present, the charge on the positive plate changes but no conduction current exists in the gap between the plates. For surface S 1 For surface S 2 because no conduction current passes through S 2

38. Thus, we have a contradictory situation that arises from the discontinuity of the current! Maxwell solved this problem by postulating an additional term on the right side of Equation 30.13, which includes a factor called the displacement current I d, defined as we can express the general form of Ampère’s law (sometimes called the Ampère–Maxwell law) as

39. The electric flux through S 2 is simply Hence, the displacement current through S 2 is

40. Example 30.9 Displacement Current in a Capacitor