Magnetic Dipole Moment: Solving Homework Questions

  • Thread starter Thread starter shivam01anand
  • Start date Start date
  • Tags Tags
    Magnetism
AI Thread Summary
The discussion centers on the equation m/l = q/2M, which relates the magnetic dipole moment to charge and mass. It is clarified that this equation is applicable only to classical charged particles and does not account for quantum mechanics or non-uniform charge distributions. The validity of using this equation in various homework scenarios is questioned, particularly in cases with complex charge distributions. Participants emphasize the importance of understanding the limitations of classical formulas in the context of magnetism. The conversation concludes with a focus on using these results for exam preparation, despite their potential inaccuracies.
shivam01anand
Messages
37
Reaction score
0

Homework Statement



in the topic related to magnetic dipole moment.

The result is derived that m/l = q/2M

m= Magnetic Dipole moment of the substance[of mass M whose angular momenta=L] with a charge q



Homework Equations




Now my question to you is whether this "result" holds true for all the questions related to this topic.

Even in cases of non uniform charge distribution?..Or not in the cases wherein q= k*x/l or something like that when its given

The Attempt at a Solution


Homework Statement





Homework Equations



The Attempt at a Solution

 
Physics news on Phys.org
The equation you have given, is the dipole moment of a classical charged particle - for instance an electron. The equation itself is incorrect, because such particles must be described by quantum mechanics. So it doesn't really "hold" for anything. Be careful about accepting the classical formulas :-)

However, I don't understand the rest of your question? The equation you gave is only for a single particle. So the equation has nothing to do with the total charge distribution in a material, since it only describes single charges.

You shouldn't try to understand too much of magnetism at the early stages. It is a major topic in solid state physics, where you get to understand better, how different charge distributions behave, and how many different type of magnetic properties, you can encounter.
 
I'd be lying if i said i got what you said completely.

Also, I completely agree with what you said.

I'm just planning on using these so called results in some questions directly.

Just for the purpose of acing the exam that's all.

It's just like getting the job done. Thanks btw :)
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top