Finding energy from dipole moment - Helmholtz pair?

AI Thread Summary
The discussion revolves around calculating the energy associated with a dipole moment in a magnetic field, specifically addressing the formula U = -m · B. There is confusion regarding the sign of the energy and the angle θ used in the calculation, with one participant suggesting that θ might be 90 degrees, which would yield zero energy. The correct approach involves determining the directions of the dipole moment and the magnetic field to find the appropriate angle for θ. Clarification is sought on whether to double the energy value to avoid a negative result, indicating a misunderstanding of the underlying physics. The conversation emphasizes the importance of correctly identifying the vector directions to resolve the energy calculation.
smileandbehappy
Messages
66
Reaction score
0
Firstly apologies for not typing this out - but I need the diagram. And I have no idea where to start. I 'think' most of it is correct. BUT - I have no idea what to do with the last part of c. I thought I could just double the energy. But I'm going to get a negative energy for the system... Clearly that isn't right. I have also considered that the energies just cancel each other out... But that doesn't seem to make sense either. Any ideas?

17631688445_e0915d965e_b.jpg


17444099010_438e7b2a08_k.jpg


17631761911_0b93ee8e7b_b.jpg
 
Physics news on Phys.org
Your work looks very good to me. But what should be the sign of the dot product of m and B?
 
TSny said:
Your work looks very good to me. But what should be the sign of the dot product of m and B?

The sign shoud be: U = -B.m

So I still get a minus sign... Or am I missing something obvious? Should I just double the U and make it positive to get the answer?
 
Sorry, I wasn't very clear.

##U = -\vec{m} \cdot \vec{B}= -m \; B \cos \theta##. What is the value of ##\theta##?
 
TSny said:
Sorry, I wasn't very clear.

##U = -\vec{m} \cdot \vec{B}= -m \; B \cos \theta##. What is the value of ##\theta##?

I though ti was 90 degrees. But I am guessing I am wrong...
 
Note that ##\cos 90^o = 0##, so that's not what you would like to have.

Consider the loop on the left. What is the direction of ##\vec{m}## for that loop? What is the direction of the magnetic field ##\vec{B}## produced by the loop on the right at the location of the loop on the left?
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top