Magnetic moment and Minimum energy

[Arman777]

In exam we have a question like this ,

I said its false cause $U=-\vec μ⋅\vec B$ , $U=-μBcosθ$ when $cosθ=1$ it is in the most stable point.But when $cosθ=-1$ its not, since.In both cases $\vec μ$ and $\vec B$ are parallel ?

 

[rude man]

In exam we have a question like this , View attachment 199309
I said its false cause $U=-\vec μ⋅\vec B$ , $U=-μBcosθ$ when $cosθ=-1$ it is in the most stable point.But when $cosθ=1$ its not, since.In both cases $\vec μ$ and $\vec B$ are parallel ?

edit: OK , didn't see the last part of your post.

The statement is correct but misleading.
When the B and μ vectors point in the same direction (θ = 0) the potential energy (p.e.) is negative and minimum. When the B and μ vectors are at θ = +/-π/2 the p.e. is zero. When θ = -1 you have what is called metastability, but the p.e. is maximum: a slight motion away from θ = +/-π will move the magnet towards θ = 0. As it passes +/-90 deg it already has developed kinetic energy = p.e..

 

[Daniel Gallimore]

Poorly phrased question. All I can say is that minimum potential energy (p.e.) is when B and μ are aligned, at which point the p.e. is usually defined to be zero. When B and μ are at 90 deg. it is usually defined as negative.

It is perfectly legitimate to define a negative potential energy when the moment is parallel to the field and positive potential energy when the moment is antiparallel. What's important is that the most stable orientation has the lowest potential energy. Recall the definition of $U$ given by @Arman777. When the dot product is positive (i.e., the moment is parallel to the field), $U$ is negative due to the negative sign tacked onto $\vec{\mu}$. Similarly, when the dot product is negative (i.e., the moment is antiparallel to the field), $U$ is positive. Thus, the lowest potential energy is achieved when the moment is parallel to the field.

 

[Arman777]

If we say $\vec u =-\vec r$ and u and r parallel is it wrong ?

 

[rude man]

If we say $\vec u =-\vec r$ and u and r parallel is it wrong ?

It's a semantic issue. Some peole say parallel means pointing in the same direction, so if u = -v then the answer is u and v are not parallel but antiparallel.

However, I think the more common meaning of parallel is as defined by (for example) Wolfram: " Two vectors

and

are parallel if their cross product is zero, i.e.,

." In which case u and -v are parallel

 

[Arman777]

It's a semantic issue. Some peole say parallel means pointing in the same direction, so if u = -v then the answer is u and v are not parallel but antiparallel.

However, I think the more common meaning of parallel is as defined by (for example) Wolfram: " Two vectors View attachment 199359 and View attachment 199360 are parallel if their cross product is zero, i.e., View attachment 199361." In which case u and -v are parallel

Yes we need to define what's "parallel" means.This is the critical point of this question.We need general mathematical rule.If its defined as your claimed then.The answer should be false but as you said again its sementic issue...

Here what I found In Online Pauli notes ;
So, let’s suppose that a and b are parallel vectors. If they are parallel then there must be a number c so that,
a=cb
So, two vectors are parallel if one is a scalar multiple of the other.

 

[rude man]

Here what I found In Online Pauli notes ;
So, let’s suppose that a and b are parallel vectors. If they are parallel then there must be a number c so that,
a=cb
So, two vectors are parallel if one is a scalar multiple of the other.

In which case, with c negative, two antiparallel vectors are parallel!

 

[Arman777]

In which case, with c negative, two antiparallel vectors are parallel!

there's no two anti-parallel vectors

 

[Arman777]

I see ok problem solved.Thanks rude man