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Physics Curseware Electrmagnetism Magnets Prblem 1.- A permanent magnet has a square shape f side L and thickness w. It is magnetized unifrmly thrugh its thickness with magnetizatin M. Calculate the intensity

Physics Curseware Electrmagnetism Magnets Prblem 1.- A permanent magnet has a square shape f side L and thickness w. It is magnetized unifrmly thrugh its thickness with magnetizatin M. Calculate the intensity f the magnetic field B just abve the center f the square magnet. Cnsider that w l. Slutin: We can replace the permanent magnet with a distributin f electric current using the equatins: r r J b M, which is a vlume current density inside the magnet. It has units f amps per meter squared, s yu need t multiply it by the area that the current is ging thrugh t get the value f I. In the case f this prblem it is zer because the magnetizatin is unifrm. K r b M r nˆ, which is als a current density, but a surface current density this time, and it is at the bundary f the magnet. K b has units f amps per meter and yu need t multiply it by the length that the current is ging thrugh t get the value f I. We ntice that the tp surface and bttm surface will give zer K b because the magnetizatin and the unit vectr pint alng the same directin (r ppsite). On the ther hand, the sides f the magnet cntribute a value f K b M with the directin f the vectr as shwn in the figure belw: A surface current density current is I wm. K b M flwing n the surface f with w means that the T find the magnetic field at the center f the magnet we can use the equatin that we fund fr a plygn: I π N sin N, where R is the distance frm the center t the side f the plygn and N is the number f sides. In ur case N4, RL/ and I wm, s: wm π 4sin 4 wm πl Prblem.- Tw permanent magnets in the shape f thin disks f radius R and thickness w are separated by a distance equal t the radius R. calculate the magnetic field at the midpint between the center f the disks. Take w r. Slutin: In a similar way as the previus prblem we replace the permanent magnets with tw circular lps with current IMw. We knw hw t find the magnetic field abve a circular lp with current I. It is the equatin: IR 3 / ( z + R ) In this case we substitute IMw and zr/. We als ntice that we have tw cils, s we need t multiply the magnetic field by tw. 3 / MwR Mw B 4 3 / ( R / 4 + R ) 5 R Prblem 3.- Tw disks f thickness d and radius R, where R d, are magnetized thrugh their thickness with unifrm magnetizatin M. The disks are arranged symmetrically as shwn in figure 1. Calculate the magnetic field prduced at pint P. Slutin: We substitute the disks with circular lps that carry a current IMd. Each disk prduces a field at pint P equal t: B IR 3 / ( z + R ) In this case we substitute IMd and zl. We als ntice that we have tw vectrs, but nly the vertical cmpnent needs t be added (the hrizntal cmpnents cancel each ther) s the ttal field at pint P is: MdR MdR 3 cs30 3 / 3 / ( L R ) + ( L + R ) Prblem 4.- A very lng rd is magnetized unifrmly with magnetizatin M. The rd is cut and a gap w is pen as shwn in figure. Calculate the magnetic field in the middle f the gap if w r l Slutin: The lng rd with the gap can be substituted by an infinite slenid and a circular lp with current in the ppsite directin. The surface density f current is K b M. T find the effect f the infinite slenid n the center we use Ampere s law n a simple rectangular circuit as shwn in the figure. r r Bdl I Ba K a Ma M enclsed b The circular lp that replaces the gap has current in the ppsite directin, with a value f IMW. And the magnetic field prduced is: I MW R R S, the ttal magnetic field is: MW W B M M 1 R R Prblem 5.- A permanent magnet has an hexagnal shape f side L and thickness W. It is magnetized unifrmly thrugh its thickness with magnetizatin M. Calculate the intensity f the magnetic field B just abve its center. Cnsider that W L. Slutin: We replace the effect f the hexagnal magnet with a hexagnal circuit: The distance between the center f the hexagn and ne side is The magnetic field is: R I MW MW B π π N 3 sin 6sin N 3 6 πl π L 3 L

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