Calculation of torque and force on magnetic dipole near infinite wire

  • #1
zenterix
524
74
Homework Statement
Consider the figure below, which shows an infinite wire carrying a current ##I_1## and a semicircular conducting loop with current ##I_2## placed near the wire.
Relevant Equations
What happens to to the semicircular loop?
1709968982554.png

Please see the image further below for the definition of the coordinate system.

The magnetic dipole moment of the loop is

$$\vec{\mu}=\mu\hat{k}\tag{1}$$

The magnetic field for an infinite wire depends only on the distance from the wire.

For the specific configuration above, we have

$$\vec{B}(y)=B(y)\hat{k}=\frac{\mu_0I}{2\pi y}\hat{k}\tag{2}$$

Note that I calculated this field considering an infinite wire and a point ##P## above it. That was a 2d problem (more on why I am saying this further below in my question).

Then

$$\vec{\tau}=\vec{\mu}\times \vec{B}=0$$

That is, there is no torque on the loop.

$$\vec{F}=\nabla(\vec{\mu}\cdot \vec{B})$$

$$=\nabla(\mu B(y))$$

$$=\mu\frac{\partial B}{\partial y}\hat{j}$$

Since the partial derivative above is negative, the loop translates towards the wire.

The problem I just solved online entailed reaching the two conclusions above, namely that the loop experiences no torque and translates towards the wire.

However, I am a little bit confused about some things.

I simply applied the formula I just learned: ##\vec{F}=\nabla(\vec{\mu}\cdot \vec{B})##, which comes from the force being the negative of the gradient of potential energy (the latter being ##-\vec{\mu}\cdot \vec{B}##).

However,

1709969404070.png


The way I solved the problem, I considered ##\vec{B}## to be a function of ##y## which is position in the ##\hat{j}## direction.

But doesn't ##\vec{B}## have a non-zero derivative relative to ##z## as well, ie in the ##\hat{k}## direction?

After all, ##y## is a function of ##z##.

There is something wrong with my interpretation of the equation ##\vec{F}=\nabla(\vec{\mu}\cdot \vec{B})## I believe.
 
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  • #2
Your equation for the force in terms of the gradient of the potential energy requires a negative sign up front. Otherwise it is correct.

If the semi circular loop is in the ##xy## plane, the magnetic field in that plane has only a ##z##-component and ##y## does not depend on ##z.##

You realize that to find the force you need to do an integral. Look up the vector calculus identity for the gradient of the dot product.
 
  • #3
kuruman said:
Your equation for the force in terms of the gradient of the potential energy requires a negative sign up front. Otherwise it is correct.

If the semi circular loop is in the ##xy## plane, the magnetic field in that plane has only a ##z##-component and ##y## does not depend on ##z.##

You realize that to find the force you need to do an integral. Look up the vector calculus identity for the gradient of the dot product.
But potential energy already has a minus sign (##-\mu\cdot\vec{B}##). Thus, ##\vec{F}=-\nabla(-\mu\cdot\vec{B})## so the negative signs cancel, no?

With regard to integration for finding the force, you can also find a force by differentiation of the appropriate function, right? Namely, the potential energy function. No integration involved in this case.
 
  • #4
zenterix said:
But potential energy already has a minus sign (##-\mu\cdot\vec{B}##). Thus, ##\vec{F}=-\nabla(-\mu\cdot\vec{B})## so the negative signs cancel, no?
Yes. I was thinking about something else.
zenterix said:
With regard to integration for finding the force, you can also find a force by differentiation of the appropriate function, right? Namely, the potential energy function. No integration involved in this case.
That will work in the approximation that the semi circular loop is a point dipole, i.e. its distance from the wire is much larger than the radius of the semicircle. In that approximation it doesn't matter whether the loop is a semi circle, a circle or a rectangle or an irregular closed loop.
 

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