2. Chapter 30: Sources of Magnetic fields Reading assignment: Chapter 30 Homework 30.1 (due Thursday, Nov. 13): OQ1, OQ4, OQ5, OQ6, OQ8, OQ9, QQ1, 2, 3, 4, 21, 23, 40, 41 Homework 30.2 (due Monday, Nov. 17): OQ13, OQ15, 47, 48, 51 Electric currents produce magnetic fields Field of a straight current-carrying wire, field of a wire loop, field of a solenoid Force between two parallel conductors/wires Magnetic flux & Gauss’s law in magnetism Magnetism in matter (ferromagnetism, paramagnetism, diamagnetism) Extra material: Tools to calculate/evaluate magnetic fields. Biot-Savart law Ampere’s law Key concept: Electric currents produce magnetic fields

3. Observation: Electric currents produce magnetic fields Right hand rule: Grasp the wire with your right hand so that the thumb points in the direction of the conventional current (positive); then your fingers will encircle the wire in the direction of the magnetic field.

4. Magnetic field of a long, straight wire is circular around the wire (right hand rule). Proportional to I and to 1/r It’s strength is given by: I – current in the wire r – distance from wire  0 – permeability of free space  0 = 4  ·10 -7 T·m/A Magnetic field due to an infinitely long, straight wire Magnetic field lines of a long, straight, current-carrying wire

5. i-clicker A battery establishes a steady current around the circuit below. A compass needle is placed successively at points P, Q, and R (slightly above the wire) to measure the magnetic field. Rank the strength of the magnetic field from strongest to weakest. A)P, Q, R. B)Q, R, P. C)R, Q, P. D)P, R, Q. E)Q, P, R

6. Magnetic field lines produced from a wire loop Again, we an use right hand rule to find direction of field Looks like field of a tiny magnetic moment (bar magnet)

7. Magnetic field of a solenoid Strong, nearly homogeneous field inside Fields from all loops add up Field inside solenoid: I … current in the wire n = N/L … number of coils per length  0 … permeability of free space Note: Outside of an ideal (long, tightly wound) solenoid the field is (close to) zero.

8. White Board problems In 1962 measurements of the magnetic field of a large tornado were made at the Geophysical Observatory in Tulsa, Oklahoma. If the magnitude of the tornado's field was B = 15.00 nT pointing north when the tornado was 9.00 km east of the observatory, what current was carried up or down the funnel of the tornado? Model the vortex as a long, straight wire carrying a current. http://info.umkc.edu A certain superconducting magnet in the form of a solenoid of length 0.500 m can generate a magnetic field of 9.00 T in its core when its coils carry a current of 75.0 A. Find the number of turns in the solenoid.

9. Magnetic force between two parallel conductors (wires) Consider two wires carrying currents I 1 and I 2, separated by a distance a. Each current produces a magnetic field that is felt by the other. For example I 2 “feels” the field from I 1: The force (per unit length) on conductor 2 is: Thus, the force on I 2 due to the field of I 1 : Attractive if currents are in same direction; repulsive if currents are in opposite direction.

10. White board example and i-clicker Force between two current carrying wires. The two wires are 2.0 m long and are 3.0 mm apart and carry a current of 8.0 A. 1)Calculate the force between the wires. 2)Looking at the figure, is the force attractive of repulsive? aA)Attractive B)Repulsive C)No force

11. Magnetic Flux; Gauss’s law in Magnetism Magnetic flux is defined exactly the same way for magnetism as it was for electricity: Unit of magnetic flux: Tesla·meter 2 ; also called a Weber (Wb)

12. A cylindrical solenoid of radius 10 cm has length 50 cm and has 1000 turns of wire going around it. A current of 2.00 A is passing through the wire? a)What is the magnetic field inside the solenoid? b)What is the magnetic flux through the inside of the solenoid? White board example

13. i-clicker A regular tetrahedron (four sides, all the same) has a cylindrical magnet placed in the middle of the bottom face. There is a total magnetic flux of 0.012 T  m 2 entering the bottom face. What is the total flux from one of the three top faces? A) 0.006 T  m 2 B) 0.004 T  m 2 C) 0.003 T  m 2 D) 0.012 T  m 2 E) None of the above Gauss’s Law for Magnetism Magnetic field lines always go in circles – there are no magnetic monopole sources. For closed surfaces, any flux in must go out somewhere else.

14. i-clicker A sphere of radius R is placed near a long, straight wire that carries a steady current I. The magnetic field generated by the current is B. The total magnetic flux passing through the sphere is  0 I  0 I/(4  R 2 ) C)4  R 2 /   I D)Zero E)Need more information

15. A cube of edge length = 2.50 cm is positioned as shown below. A uniform magnetic field given by B = (5.00 i + 4.00 j + 3.00 k) T exists throughout the region. (A)Calculate the flux through the shaded region. (B)What is the total flux through the six faces? White board example

16. Extra material Tools to calculate/evaluate magnetic fields. Biot-Savart law Ampere’s law

17. The Biot-Savart Law (Used to calculate magnetic fields of currents, e.g., a current-carrying wire) Magnetic fields go around the wire – they are perpendicular to the direction of current Magnetic fields are perpendicular to the separation between the wire and the point where you measure it - Sounds like a cross product! Permeability of free space

18. Magnetic Field from a Finite Wire Magnetic field from a finite straight wire: Let a be the distance from the wire to point P. Let x be the horizontal separation r I dsds P a x-x 1 x2x2 22 11 O for long straight wire 11 22

19. Ampere’s Law Suppose we have a wire coming out of the plane Let’s integrate the magnetic field around a closed path There’s a new symbol for such an integral Circle means “over a closed loop” The magnetic field is parallel to direction of integration What if we pick a different path? I dsds ds cos  r dd We have demonstrated this is true no matter what path you take Wire doesn’t even need to be a straight infinite wire All that matters is that current passes through the closed Ampere loop

20. Understanding Ampere’s Law If multiple currents flow through, add up all currents that are inside the loop Use right-hand rule to determine if they count as + or – Curl fingers in direction of Ampere loop If thumb points in direction of current, plus, otherwise minus The wire can be bent, the loop can be any shape, even non-planar 6 A 2 A 1 A 4 A 7 A i-clicker There are currents going in and out of the screen as sketched at right. What is the integral of the magnetic field around the path sketched in purple? A)  0 (4 A) B)  0 (-4 A) C)  0 (12 A) D)  0 (-12 A) E) None of the above Right hand rule causes thumb to point down Downward currents count as +, upwards as –

21. Using Ampere’s Law Ampere’s Law can be used – rarely – to calculate magnetic fields Need lots of symmetry – usually cylindrical A wire of radius a has total current I distributed uniformly across its cross-sectional area. Find the magnetic field everywhere. II End-on view Draw an Ampere loop outside the wire – it contains all the current Magnetic field is parallel to the direction of this loop, and constant around it Use Ampere’s Law: But we used a loop outside the wire, so we only have B for r > a

22. Using Ampere’s Law (2) Now do it inside the wire Ampere loop inside the wire does not contain all the current The fraction is proportional to the area End-on view a

23. Solenoids Consider a planar loop of wire (any shape) with a current I going around it. Now, stack many such loops Treat spacing as very closely spaced Assume stack is tall compared to size of loop Can show using symmetry that magnetic field is only in vertical direction Can use Ampere’s Law to show that it is constant inside or outside the solenoid But magnetic field at infinity must be zero

24. Field Inside a Solenoid It remains only to calculate the magnetic field inside We use Ampere’s law Recall, no significant B-field outside Only the inside segment contributes There may be many (N) current loops within this Ampere loop Let n = N/L be loops per unit length L Works for any shape solenoid, not just cylindrical For finite length solenoids, there are “end effects” Real solenoids have each loop connected to the next, like a helix, so it’s just one long wire