1. 27. 7. 20031 III–3 Magnetic Dipoles

2. 27. 7. 20032 Main Topics Magnetic Dipoles The Fields they Produce Their Behavior in External Magnetic Fields Calculation of Some Magnetic Fields Solenoid Toroid Thick Wire with Current

3. 27. 7. 20033 Magnetic Dipoles I In electrostatics we defined electric dipoles. We can imagine them as solid rods which hold one positive and one negative charge of the same absolute values some distance apart. Although their total charge is zero they are sources of fields with special symmetry which decrease faster than fields of point sources. External electric field is generally trying to orient and shift them.

4. 27. 7. 20034 Magnetic Dipoles II Their analogues in magnetism are either thin flat permanent magnets or loops of current. These also are sources of fields with a special symmetry which decrease faster than fields from straight currents In external magnetic fields they are affected similarly as electric dipoles. Later we shall describe magnetic behavior of matter using the properties of magnetic dipoles.

5. 27. 7. 20035 Magnetic Dipoles III Let us have a circular conductive loop of the radius a and a current I flowing in it. Let us describe the magnetic field at some distance b on the axis of the loop. We can “cut” the loop into little pieces dl = ad  and vector add their contribution to the magnetic induction using the Biot- Savart law.

6. 27. 7. 20036 Magnetic Dipoles IV For symmetry reasons the direction of is the same as the direction of the z-axis perpendicular to the loop and integration in this case means only to add the projections dB z = dB sin . And from the geometry: sin  = a/r  1/r 2 = sin 2  /a 2 r 2 = a 2 + b 2 Let us perform the integration.integration

7. 27. 7. 20037 Magnetic Dipoles V Since magnetic dipoles are sources of magnetic fields they must also be affected by them. In uniform magnetic field they will experience a torque directing them in the direction of the field. We shall illustrate it using a special case of rectangular loop a x b carrying current I.

8. 27. 7. 20038 A Quiz: What is the total force and torque on this rectangular loop if it lies perpendicularly to the field lines of the uniform magnetic field? What would be the difference, if the loop is circular?

9. 27. 7. 20039 The Answer Force on each side is perpendicular to it and lies in the plane of the rectangle. Its particular direction depends on the directions of the magnetic field and the current. The forces acting on opposite sides will cancel, however, since the current direction in them is the opposite. In a circular loop a force acting on its piece dl will cancel with the force acting on the piece dl’ across the diameter.

10. 27. 7. 200310 Magnetic Dipoles VI Form the drawing we see that forces on the sides a are trying to stretch the loop but if it is stiff enough they cancel. Forces on the sides b are horizontal and the upper acts into the blackboard and the lower from the blackboard. Clearly they are trying to stretch but also rotate the loop.

11. 27. 7. 200311 Magnetic Dipoles VII To find the contribution of each of the b sides to the torque we have to find the projection of the force perpendicularly to the loop: T/2 = F b sin  a/2 Since both forces act in the same sense: T = BIabsin  We can generalize this using the magnetic dipole moment :

12. 27. 7. 200312 A Galvanometer A loop with current in a uniform magnetic field whose torque would be compensated by a spring is a possible principle of measuring the current. The scale of such device would not be linear! So special radial but constant (radial uniform) field is used to keep the two torque forces always perpendicular to the loop.

13. 27. 7. 200313 Magnetic Field of a Solenoid I Solenoid is a long coil of wire consisting of many loops. In the case of finite solenoid the magnetic field must be calculated as a superposition of magnetic inductions generated by all loops. In the case of almost infinite we can neglect effects close to the ends and use the Ampere’s law in a very elegant way.

14. 27. 7. 200314 Magnetic Field of a Solenoid II As a closed path we choose a rectangle whose two sides are parallel to the axis of the solenoid. From symmetry we can expect that the field lines will be also parallel to the axis direction. Since the closed field lines return through the whole Universe outside the solenoid we can expect they are infinitely “diluted”.

15. 27. 7. 200315 Magnetic Field of a Solenoid III Only the part of the path along the side inside the solenoid will make non-zero contribution to the loop integral. If the rectangle encircles N loops with current I and its length is l then: Bl =  0 NI And if we introduce the density of loops n = N/l  B =  0 nI

16. 27. 7. 200316 Magnetic Field of a Solenoid IV For symmetry reason we didn’t make any assumptions about how deep is our rectangle immersed in the solenoid. So the magnetic field in the long solenoid can be expected to be uniform or homogeneous. Many physical methods rely on uniform magnetic field, e.g. NMR and mass spectrography. A reasonably uniform magnetic field can be obtained if we shorten thick solenoid and cut it into halves - Helmholtz coils.

17. 27. 7. 200317 Magnetic Field of a Toroid I We can think of the toroid as of a solenoid bent into a circle. Since the field lines cant escape we do not have to make any assumptions about the size. If the toroid has a radius R to its central field line and N loops of current I, we can simply show that all the field is inside and what is the magnitude on a particular field line.

18. 27. 7. 200318 Magnetic Field of a Toroid II Let’s us choose the central filed line as our path then the integration simplifies and: B(r) 2  r =  0 NI  B(r) =  0 NI/2  r His is also valid for any r within the toroid. The field: is non uniform since it depends on r. is zero outside the loops of the toroid

19. 27. 7. 200319 Magnetic Field of a Thick Wire I Let’s have a straight wire of a diameter R in which current I flows and let us suppose that the current density j is constant. We use Ampere’s law. We use circular paths one outside and one inside the wire. Outside the field is the same as if the wire was infinitely thin. Inside we get linear dependence on r.

20. 27. 7. 200320 Magnetic Field of a Thick Wire II If we take a circular path of the radius r inside the wire we get: B 2  r =  0 I enc The encircled current I enc depends on the area surrounded by the path I enc = j  r 2 = I  r 2 /  R 2  B =  0 Ir/2  R 2

21. 27. 7. 200321 Conceptual Material In a region where are no currents magnetic field with parallel field lines must also be uniform. Field from right-angle wire of current. A conductive rod lying on two rails in uniform magnetic field when contacted to a power source.

22. 27. 7. 200322 Homework Please add 28 – 37, 38 to your homework assignment due Monday!

23. 27. 7. 200323 Things to read This Lecture Covers Chapter 27 – 5 and 28 – 4, 5 Advance Reading Chapter 27 – 5 and 28 – 4, 5

24. Circular Loop of Current I

25. Circular Loop of Current II A =  a 2 is the area of the loop and its normal has the z direction. We can define a magnetic dipole moment and suppose that we are far away so b>>a. Then: Magnetic dipole is a source of a special magnetic field which decreases with the third power of the distance. ^

26. The vector or cross product I Let Definition (components) The magnitude |c| Is the surface of a parallelepiped made by.

27. The vector or cross product II The vector is perpendicular to the plane made by the vectors and and they have to form a right-turning system.  ijk = {1 (even permutation), -1 (odd), 0 (eq.)} ^