of 5

Physics Courseware Electromagnetism

0 views5 pages

Download

All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.
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
We Need Your Support
Thank you for visiting our website and your interest in our free products and services. We are nonprofit website to share and download documents. To the running of this website, we need your help to support us.

Thanks to everyone for your continued support.

No, Thanks