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Chapter 28

Chapter 28. Gravitational and Electric Fields. We expect that B will have a 1/r 2 dependence Directed along the radius Recall F = i L x B and F =q E and F =m g Based on the pattern . Broken Symmetry. But those other fields diverge and B does not

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Chapter 28

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  1. Chapter 28

  2. Gravitational and Electric Fields • We expect that B will • have a 1/r2 dependence • Directed along the radius • Recall F= iLxB and F=qE and F=mg • Based on the pattern

  3. Broken Symmetry • But those other fields diverge and B does not • Based on our experience with bar magnets, B must return to the magnet

  4. Biot-Savart Law B into page X X X q i ds Point P X X X • B goes into the page • i ds and r are both perpendicular to B • B proportional to i ds x r • But needs to be 1/r2 so

  5. A new constant, m0 • Called the permeability of free space • Value = 4p x 10-7 T*m/A • And yes, m0/ 4p= 10-7 T*m/A

  6. Modified Biot-Savart • If a charged particle has a constant velocity, v, then I can modify Biot-Savart:

  7. Solving Biot-Savart Problems • Biot-Savart problems are typically one of geometry • You must integrate about the limits of the current loop while evaluating the cross-product • Sometimes, ds and r are parallel, which eliminates the contribution. You should examine the problem to see where this is true. • Sometimes, ds and r are perpendicular, which forces the cross-product to its maximum value. • Mostly, the ds and r are related only through sin q. This means that you may have to create an integrand over the angle, not necessarily the length of the current loop. In conclusion, Biot-Savart problems are strongly dependent on creating the appropriate geometry over which to integrate.

  8. Magnetic Field Due to a Long, Straight Wire ds q Direction of current s r X X X R Direction of B X Point P X i

  9. Integrating

  10. Magnetic Field due to Current of a Circular Arc of Wire r always perpendicular to ds so cross-product is ds R ds r B out of page

  11. A Complete Loop • We can take the previous result and evaluate it at f=2p • The result is the same as your text on pg. 1077 Eq. 28-17

  12. Force Between Two Long Wires with Parallel Currents Force on b caused by the magnetic field of a Fa on b =ibL x Ba d X X X X ia L ib By RH rule, Fa on b is towards a

  13. By Symmetry • Fb on a is equal and opposite to Fa on b • So the two wires with parallel currents are attracted to one another • If I reverse the current on b (anti-parallel), then the forces generated by the wires will repel one another. • Parallel currents attract; anti-parallel currents repel.

  14. Derivation of Ampere’s Law Consider a distance R from a wire carrying current I. Now consider all of the points which are a distance R from the wire. They form a circle of circumference 2pR. Now we evaluate the closed loop integral at this point. So the circle that we made is called an “Amperian Loop”.

  15. Ampere’s Law • Ampere’s Law says that the magnitude of B is proportional to the net current enclosed within the Amperian Loop. • Amperian loops do not have to be circles. They could be rectangles but a circle is usually more convenient to integrate. The only rule is that they have to be closed!

  16. What is the direction of the magnetic field? • We have an agreed convention: • Curl your fingers around the Amperian loop with your fingers curling in the direction of the integration. If the current is in the same direction as your thumb, then current is assigned a positive value. If your thumb is opposite the direction of the current, then the current is assigned a negative value.

  17. Magnetic Field of Long, Straight Wire Revisited Integrate from 0 to 2p R In this case, the enclosed current is defined to be in the negative direction. I

  18. Differential Version of Ampere’s Law • Recall A current density creates a steady magnetic field which circulates around the current density

  19. Solenoid • Solenoid is a number of coils packed tightly together. • It solves an incredibly tough problem. • Consider: It is easy to make a uniform E-field. All you need are two parallel plates. But how do you make a uniform field out of a bunch of circles (i.e. a B-field).

  20. Finding the magnetic field of a solenoid The solenoid is characterized by the number of turns per unit length, N The magnetic field outside of the solenoid is so weak that it is consider equal to zero

  21. Toroid– A circular solenoid Consider a circular solenoid of radius, R Current into page Current out of page Amperian Loop

  22. Bohr Magneton • In the last chapter, we discussed how atoms could be thought of as current loops • I=e/T where T=2pr/v • v=velocity of the electron • r= radius of the electron’s orbit • If m=IA then m=((ev)/2pr)*pr2 • m=evr/2 • Recall for an orbiting point, the angular momentum, L, is equal to mvr • So m=eL/(2m)

  23. Quantization of angular momentum • Each orbital is an integer value of Planck’s constant, h divided by 2p • =h/ 2p • L=n  • m=e /2m • Called the “Bohr Magneton” • Represents smallest amount of dipole moment possible

  24. Magnetic Materials This represents any material. The arrows indicate the direction of the individual magnetic moments of the atoms. As you can see, their orientation is random and the vector sum of them is zero. This represents a permanent magnet. The magnetic moments or dipoles have a net vector sum point to the left.

  25. Magnetic Materials in an External Magnetic Field Before Let’s turn on an external B which points to the left Possibility 2: The dipoles align to oppose the external field Possibility 1: The dipoles align with the external field Bexternal

  26. The degree to which the magnet moments align or oppose the magnetic field determines their classification • Ferromagnetic—Strongly aligns with the magnetic field • Paramagnetic—Weakly aligns with an external magnetic field • Diamagnetic—Weakly opposes the external magnetic field • For any of these cases, we define the total magnetic field as B=Bexternal + Bm

  27. Bm represents the magnetic field created by the alignment/opposition of the dipoles • Bm= cm*Bexternal • cmis called the magnetic “susceptibility” of the material • Paramagnetic materials and their susceptibilities • Al 2.2 x 10-5 • Cu 1.4 x 10-5 • Air 3.6 x 10-7 • Diamagnetic materials and their susceptibilities • Bi -1.7 x 10-5 • Ca -2.2 x 10-5 • H2O -9.1 x 10-6 • Ferromagnetic materials have susceptibilities from 103 to 106

  28. m--The Permeability of a Material • B=Bexternal + cm*Bexternal • B=Bexternal (1+ cm) • Relative permeability, Km is defined as • Km=1+ cm • Actually, we could replace m0with a new “m” called the permeability of the material. • m= Km m0

  29. Hysteresis • Let’s say you had a ferromagnet and you measured its magnetic field • You applied an external B-field and the dipoles are aligned with the field • Now you remove the field but some of the dipoles get “stuck” in their new position. • Now you measure the magnetic field of the ferromagnet and find that it is different. • You repeat the process and yet, you never get back to your original value as more or less of the dipoles stick or unstick in their new positions. • This behavior is called hysteresis.

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