Ferromagnetism. So that once magnetized the material will stay that way even in the absence of external current it is a permanent magnet.

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Tracy Ross
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1 Ferroagnetis We now turn to the case where is not proportional to. We distinguish two cases: soft and hard ferroagnets. In a soft ferroagnet a graph of vs looks like If is now reduced, will retrace the sae curve. Since is often produced by an external current this akes possible electroagnets when the current is turned off, the agnetization returns to zero. In a hard ferroagnet the graph looks like So that once agnetized the aterial will stay that way even in the absence of external current it is a peranent agnet. We will first consider peranent agnets with no free currents. Then the equations of agnetostatics because H B But B H H These are just the equation of electrostatics with a charge density, nˆ

2 except for the factor /ε. Hence we have converted agnetostatics to electrostatics, at least for peranent agnets. All we need do is replace ρ by Exaple Consider a bar agnet of length and radius having a unifor agnetization ˆ z lying along the axis. Then everywhere except at the ends. There we get a charge density at the top and at the botto. We can then find by integration over this known charge density. We use cylindrical coordinates with center at the center of the agnet and along the axis. Then

3 ˆ ˆ ˆ x cos 'cos ' y sin 'sin ' z z 'd 'd ' H 3/ ' ' z 'cos ' ' xˆ cos 'cos ' yˆ sin 'sin ' zˆ z d ' 3/ ' z 'cos ' 'd ' ' ' 3/ ' z 'cos ' ˆx cos 'cos ' d ' 'd ' ' ' ' z 'cos ' ' ' ' ' ' ' ŷ sin 'sin ' " 'dp'd ' ẑ z 'dp'd ' ' z 'cos ' z 'dp'd ' ' z 'cos ' 3/ 3/

4 Because of the cylindrical syetry we know that this ust have only and coponents. Hence we pick = and the ρ coponent to be H x. Then 3/ ' 3/ ' ' z 'cos ' H 'dp' 'cos ' d ' ' z 'cos ' 3/ ' ' z 3/ ' ' 'd 'd ' z ' z 'cos ' H 'd 'd ' z ' z 'cos ' Once we have we find fro B H Where = inside and outside the agnet. As a check we evaluate these at z =. Then H ρ = as expected fro syetry.

5 H 'dp'd ' ' 'cos ' z 3/ ' ' / d d ' 3/ cos ' ' We can now evaluate this for any a values of /ρ and /ρ. Take /ρ = ½, /ρ = 5. Then Hz d d ' / 5 cos ' We can also check known liits. At large distance the agnet should just be a dipole z ˆ H, r H H zˆ ˆ z

6 3/ z ' z 'cos ' H 'd ' 'cos ' ' 3/ z ' z 'cos ' ' 'd ' 'cos ' z ' z 3/ ' ' z z 'cos ' z z 'cos ' z 3 z 'cos ' z 'd ' 3/ 'cos ' z, z ' 3 z 'cos ' z 3 3 z z 'd ' 'cos ' d ' z z z 3/ 3/ 3/ 3/

7 H 'd 'd ' z 3/ ' ' z 'cos ' ' z z z 'd 'd ' z 3/ ' z 'cos ' ' ' z 3/ 3z 'cos ' z 'd 'd ' ' z ' 3/ 3/ z, z 3z 'cos ' z 'd 'd ' ' z ' z 3/ 3 z 3 z z z z z 3/ 3/ 3z 3z z z z z 3z Bz Hz 3 r r B 3z 3z Hp 3 3 r r r r The dipole field fro a dipole is

8 3zr ˆ r Br zˆ 3zˆ 3 3 zcos ˆ ˆsinzcos ˆ ˆsinzˆ r r r r ẑ 3cos 3cossin 3 ˆ z cos, sin r r z z B r 3 zˆ 3 3 ˆ r r r Hence the field at long distance is indeed that due to a dipole zˆ as expected. Exaple Consider two identical bar agnets of radius and length, each having agnetization as shown. Find the force between the. This tie we have charge sheets

9 The force between charges is given by k da k q q da F r r r Then we need the forces between (-3), (-), (-3), (-). They differ only in sign and z distance 3 total 3 3 F F F F F F repulsive Clearly we need only worry about the coponent of force. et s be the distance between the sheets. Then Fs 'dd 'dd 's 3/ ' ' cos 'cos ' sin 'sin ' s 'dd 'dd ' s 3/ ' s 'cos ' 'd 'dd s ' ' s 'cos 3/ s a b adabdbd s a b abcos 3/

10 We can now evaluate the integral for any s/. For F 3 we have s h For F we have +h/. For F 3 we get h/. In our case take =.5, =.5, h=. The for F we have s For F we have abdadbd 3 3/ F 5.3 a b 3 ab cos s For F 3 we have F s..5 Hence 3 F F rep 6 Hence the agnets attract with a force of We can estiate as follows.

#atos dipole vol ato qv qv ato IA #atos vol Aw 9 6 qv 8.6 3 5 6 ap/ 7 Aw.6 57 which is clearly about right F 8n lbs Exaple 3 Consider a bar agnet of radius and length with agnetization as shown.

11 #atos dipole vol ato qv qv ato IA #atos vol Aw 9 6 qv ap/ 7 Aw.6 57 which is clearly about right F 8n lbs Exaple 3 Consider a bar agnet of radius and length with agnetization as shown. It is placed on a flat sheet of iron (a soft ferroagnetic aterial). What force is required to reove the agnet? For a soft ferroagnet the effective μ is very high tiny H akes large. Since B H H Since B n is continuous, B in the iron B in the agnet, which is finite. Hence H in the iron is very sall. In the liit μ >> (it is typically thousands in the portion of the curve where is changing) we ust have H in the iron =. Further, since H tan is continuous (no free current on the surface) and inside the iron it ust be zero just outside the surface. Hence the iron is behaving for as a conductor does for. We can therefore solve this by the ethod of iages. We have

12 We therefore need the forces between (-3), (,3), (,), (,). Clearly (-3) and (-) are the repulsive. (-3) and (-) are attractive. We find (-3) and (-) just as in exaple. (-3) is easier. We know fro the electrostatic case that the force will be given by k k attractive For (,3) we have s/ = /. For (,) s/ = / Fk s abdadbd 3/ a b s As an exaple take =.5, =.5 as before. Then a b abcos abdadbd 3 3/ F k 3 a b 3 ab cos 5 k k.7 abdadbd 3/ F k 6 a b 6 ab cos 6 k k.36 F k.7.36 k n 6

13 Note that the contribution fro (-3) is by far the largest. With this in ind we can estiate the radius of an electroagnet needed to lift a 5 lb car in a junk yard / / w 5.5 w k.53 7 k Not uch! Electroagnets are very effective.

Electromagnetic scattering. Graduate Course Electrical Engineering (Communications) 1 st Semester, Sharif University of Technology

Electromagnetic scattering. Graduate Course Electrical Engineering (Communications) 1 st Semester, Sharif University of Technology Electroagnetic scattering Graduate Course Electrical Engineering (Counications) 1 st Seester, 1388-1389 Sharif University of Technology Contents of lecture 5 Contents of lecture 5: Scattering fro a conductive