Hi! I have a question regarding calculating the diamagnetic force:
You guys said "to calculate the diamagnetic force, use the magnetic susceptibility of bismuth to calculate the induced counter- magnetic field in the bismuth... the magnetization of bismuth will cause it to have a magnetic dipole that is opposite to that of the induced magnet." So, by my understanding you guys are saying that the magnetization of bismuth (which I can calculate by multiplying the magnetic susceptibility of bismuth by the strenght of the applied magnetic field) will cause the induced counter-magnetic field in bismuth? In other words: THE MAGNETIZATION OF BISMUTH WILL CAUSE THE INDUCED COUNTER-MAGNETIC FIELD IN BISMUTH? I CAN CALCULATE THE INDUCED COUNTER-MAGNETIC FIELD IN BISMUTH BY CALCULATING THE MAGNETIZATION OF BISMUTH? THE MAGNETIZATION OF BISMUTH WILL GIVE YOU THE INDUCED COUNTER- MAGNETIC FIELD IN BISMUTH, WHICH WILL GIVE YOU THE DIAMAGNETIC FORCE? Again, you guys said "to calculate the diamagnetic force, use the magnetic susceptibility of bismuth to calculate the induced counter-magnetic field in the bismuth" DO I USE THE FORMULA TO CALCULATE THE MAGNETIZATION OF BISMUTH TO CALCULATE THIS TOO? Again, to calculate the magnetization of bismuth:I can calculate by multiplying the magnetic susceptibility of bismuth by the strenght of the applied magnetic field
|
Answer 1:
If you use the website you mentioned in your above question, and your piece of bismuth is small enough, then yes, you can use the "magnetic susceptibility" of bismuth to calculate the electromagnetic force this would generate. Because the magnetic susceptibility of bismuth is negative, bismuth is diamagnetic: it will be repelled away from your magnet, rather than pulled towards it, as a piece of iron or other metal with a positive magnetic susceptibility would be.
Click Here to return to the search form.
|
