1. Fig 30-CO, p.927

2. Ch 30 Sources of the Magnetic Field 30.1 The Biot-Savart Law dB =  0 Ids x r /4r 2 (cross or vector product)  

3. Fig 30-1, p.927 dB =  0 Ids x r /4r 2  

4.    ds = dx r r x axis I x = 0 a Long Thin Wire (a << L) x

5. B =  0 I/2a

6. CT1: The magnitude of the magnetic fields are the same at the outer loop in each diagram below and the drawing is to scale. The ratio I 1 /I 2 is A. four B. two C. one D. one-half E. one-quarter I1I1 I2I2

7. P30.5 (p.859)

8. Ch 30 Sources of the Magnetic Field 30.2 The Magnetic Force between Two Parallel Current-Carrying Conductors F/ l =  0 I 1 I 2 /2a opposite direction currents repel same direction currents attract

9. FTFT FRFR FBFB FLFL B1B1 X non- uniform P30.17 (p.860)

10. Ch 30 Sources of the Magnetic Field 30.3 Ampère’s Law B·ds =  0 I net 

11. I dd B ds r rd Arbitrary Current Carrying Loop (yellow) B·ds = Brd

12. Ch 30 Sources of the Magnetic Field 30.3 Ampère’s Law Method for Ampère’s Law: Use symmetry Choose an Amperian loop such that on a portion of the loop : i. B || ds and B is constant. ii. B. ds is zero because B = 0. iii. B. ds is zero because B  ds. iv. Any combination of i-iii.

13. Amperian Loops P30.31 (p.862)

14. Coaxial Cable End View CT2: Use Ampere’s Law to determine the magnetic field. The field between the wire and shielding is A.  0 I/2r clockwise B.  0 I/2r counterclockwise C.  0 I/2r clockwise D.  0 Ir counterclockwise E. zero I into plane I out of plane

15. Coaxial Cable End View CT3: Use Ampere’s Law to determine the magnetic field. The field outside the shielding is A.  0 I/2r clockwise B.  0 I/2r counterclockwise C.  0 I/2r clockwise D.  0 Ir counterclockwise E. zero I into plane I out of plane

16. Ch 30 Sources of the Magnetic Field 30.4 The Magnetic Field of a Solenoid

17. B for an ideal solenoid: B = 0 outside and uniform inside. B =  0 nI n = turns/length

18. Toroid: B =  0 IN/2r N = number of turns

19. P30.37 (p.863)

20. Ch 30 Sources of the Magnetic Field 30.5 Magnetic Flux and Gauss’s Law in Magnetism  B = B·dA = 0 Units: Tm 2 = Weber (Wb) Recall  E = E·dA = q in / 0 P30.39 (p.863) P30.42 (p.863)  

21. A. remains the same. B. reverses. C. changes in magnitude, but not in direction. D. changes to some other direction. E. other CT4: On a computer chip, two conducting strips carry charge from P to Q and from R to S. If the current direction is reversed in both wires, the net magnetic force of strip 1 on strip 2

22. Wrong Answer/Right Reason: Magnets attract other magnets and some metals. Right Answer/Wrong Reason: Magnets only interact with other magnets. Wrong Answer/Wrong Reason: Magnets are objects with permanent electric charge on their ends. 

23. CT5: Cosmic rays (atomic nuclei stripped bare of their electrons) would continuously bombard Earth’s surface if most of them were not deflected by Earth’s magnetic field. Given that Earth is, to an excellent approximation, a magnetic dipole, the intensity of cosmic rays bombarding its surface is greatest at the A. poles. B. mid-latitudes. C. equator.

24. CT6: A rectangular loop is placed in a uniform magnetic field with the plane of the loop perpendicular to the direction of the field. If a current is made to flow through the loop in the sense shown by the arrows, the field exerts on the loop: A. a net force. B. a net torque. C. a net force and a net torque. D. neither a net force nor a net torque.

25. CT7: A rectangular loop is placed in a uniform magnetic field with the plane of the loop parallel to the direction of the field. If a current is made to flow through the loop in the sense shown by the arrows, the field exerts on the loop: A. a net force. B. a net torque. C. a net force and a net torque. D. neither a net force nor a net torque.