Calculating crossproduct integral, Parametrization

In summary, i) an approximate solenoid can be described by a cylinder with height L and radius R. The surface occupied by the cylinder can be described by all vectors ##\vec x =(x,y,z)## so that ##x^2+y^2=R^2## , 0 ≤ z ≤ L. One parametrization to this could be ##x(t) = R\cdot cos(t), y(t)=R\cdot sin(t)## , 0 ≤ z ≤ L , 0≤t ≤ ##2\pi## , t = 0 at the x-axis and positive rotation is counterclockvise around the z axis.
  • #1
Karl Karlsson
104
12
Homework Statement
Consider a circular solenoid with length L and radius R. The solenoid is wound with n revolutions per unit length and is traversed by a current ##I_0##. If the solenoid is tightly wound (ie, n is large), the current density can be approximated by a surface current of the form ##d\vec I = I_0n\vec e_3\times d\vec S##.

The magnetic force ##d\vec F## from an external magnetic field ##\vec B## on a small surface element is given by ##d\vec F = d\vec I\times \vec B##

i) Motivate and write down a parameterization of the surface occupied by the solenoid. (Start by placing the solenoid in an appropriately selected way in a Cartesian coordinate system.) Specify in particular how the Cartesian coordinates depend on your parameters and which parameter values are allowed.

ii) Calculate the surface element ##d\vec S## in your parameterization.

iii) The solenoid is placed in an arbitrary constant magnetic field ##\vec B##. Write down the expression for dF and simplify this as far as possible.

iiii) The total moment of force on the solenoid relative to a position ##\vec x_0## is obtained by integrating ##\vec T = \int_S (\vec x - \vec x_0 )\times d\vec F## over the entire solenoid surface S. Calculate this integral for the situation described in iii) when ##\vec x_0## is the center of the solenoid.
Relevant Equations
##d\vec I = I_0n\vec e_3\times d\vec S##
##d\vec F = d\vec I\times \vec B##
##\vec T = \int_S (\vec x - \vec x_0 )\times d\vec F##
i) I approximate the solenoid as a cylinder with height L and radius R. I am not sure how I am supposed to place the solenoid in the coordinate system but I think it must be like this, right?
IMG_0703.jpg


The surface occupied by the cylinder can be described by all vectors ##\vec x =(x,y,z)## so that ##x^2+y^2=R^2## , 0 ≤ z ≤ L. One parametrization to this could be ##x(t) = R\cdot cos(t), y(t)=R\cdot sin(t)## , 0 ≤ z ≤ L , 0≤t ≤ ##2\pi## , t = 0 at the x-axis and positive rotation is counterclockvise around the z axis. ##\vec x = (R\cdot cos(t), R\cdot sin(t), z)##

ii) I get ##d\vec S = \frac {d\vec x} {dt}\times \frac {d\vec x} {dz} \cdot dzdt = (R\cdot cos(t), R\cdot sin(t), 0) \cdot dzdt##. Do you get the same and is this correct?

iii) From here the expressions get so long that I will post a picture with what I wrote by hand and I wrote ρ instead of t while doing the calculations below so t = ρ
IMG_0704.jpg

Is there any faster way to get to the result above?

iiii) Here the calculations will be even longer. If I insert the expression i got in iii) above in the integral in iiii) as the picture below shows, then I get a long expression
IMG_0705.jpg

So as you can see I only started calculating the vectorproduct within the integral above until I had a feeling this is not the fastest way of solving this problem and maybe iii) as well? But how else are you supposed to do it?

Thanks in advance!
 
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  • #2
Cylindrical polar coordinates [itex](r, \phi, z)[/itex] are the obvious choice. With an eye to part (iv) I would take [itex]-\frac12L \leq z \leq \frac12L[/itex] rather than [itex]0 \leq z \leq L[/itex] so that [itex]\mathbf{x}_0 = 0[/itex].

The surface element of the cylinder with outward normal is then [tex]d\mathbf{S} = R\mathbf{e}_r\,d\phi\,dz[/tex] where [tex]\begin{align*}\mathbf{e}_r &= \cos \phi\,\mathbf{e}_x + \sin \phi\,\mathbf{e}_y \qquad\mbox{and} \\ \mathbf{e}_\phi &= -\sin \phi\,\mathbf{e}_x + \cos \phi\,\mathbf{e}_y \end{align*}[/tex] are the unit vectors in the directions of increasing [itex]r[/itex] and [itex]\phi[/itex] respectively.

Now set [itex]\mathbf{B} = B_r \mathbf{e}_r + B_\phi \mathbf{e}_\phi + B_z \mathbf{e}_z[/itex] and calculate the cross products using the identities [tex]
\mathbf{e}_r \times \mathbf{e}_\phi = \mathbf{e}_z, \qquad
\mathbf{e}_\phi \times \mathbf{e}_z = \mathbf{e}_r, \qquad
\mathbf{e}_z \times \mathbf{e}_r = \mathbf{e}_\phi.[/tex]
 
  • #3
pasmith said:
Cylindrical polar coordinates [itex](r, \phi, z)[/itex] are the obvious choice. With an eye to part (iv) I would take [itex]-\frac12L \leq z \leq \frac12L[/itex] rather than [itex]0 \leq z \leq L[/itex] so that [itex]\mathbf{x}_0 = 0[/itex].

The surface element of the cylinder with outward normal is then [tex]d\mathbf{S} = R\mathbf{e}_r\,d\phi\,dz[/tex] where [tex]\begin{align*}\mathbf{e}_r &= \cos \phi\,\mathbf{e}_x + \sin \phi\,\mathbf{e}_y \qquad\mbox{and} \\ \mathbf{e}_\phi &= -\sin \phi\,\mathbf{e}_x + \cos \phi\,\mathbf{e}_y \end{align*}[/tex] are the unit vectors in the directions of increasing [itex]r[/itex] and [itex]\phi[/itex] respectively.

Now set [itex]\mathbf{B} = B_r \mathbf{e}_r + B_\phi \mathbf{e}_\phi + B_z \mathbf{e}_z[/itex] and calculate the cross products using the identities [tex]
\mathbf{e}_r \times \mathbf{e}_\phi = \mathbf{e}_z, \qquad
\mathbf{e}_\phi \times \mathbf{e}_z = \mathbf{e}_r, \qquad
\mathbf{e}_z \times \mathbf{e}_r = \mathbf{e}_\phi.[/tex]
Hi! Apperently the magnetic field is supposed to be constant. So I had to write B with cartesian coordinates for it to be constant. Is my attempt on iiii) below correct and is this the shortest way of solving iiii)? I assumed I had done iii) correctly. Is iii) correct?
IMG_0706.jpg

IMG_0707.jpg

Thanks in advance!
 
  • #4
I agree with you as far as [tex]
\mathbf{T} = nI_0 R \int_{-L/2}^{L/2} \int_0^{2\pi} (RB_r +zB_z) \mathbf{e}_\phi\,d\phi\,dz.[/tex]

Unfortunately at this point you have substituted the wrong value for [itex]\mathbf{e}_\phi = -\sin\phi\,\mathbf{e}_x + \cos\phi\,\mathbf{e}_y[/itex].

Also, because the solenoid is axially symmetric, you can choose the direction of the [itex]x[/itex] axis such that [itex]\mathbf{B} = B_x\mathbf{e}_x + B_z\mathbf{e}_z[/itex]. This should simplify the calculations.
 
Last edited:
  • #5
pasmith said:
I agree with you as far as [tex]
\mathbf{T} = nI_0 R \int_{-L/2}^{L/2} \int_0^{2\pi} (RB_r +zB_z) \mathbf{e}_\phi\,d\phi\,dz.[/tex]

Unfortunately at this point you have substituted the wrong value for [itex]\mathbf{e}_\phi = -\sin\phi\,\mathbf{e}_x + \cos\phi\,\mathbf{e}_y[/itex].
Thanks for spotting that misstake!
pasmith said:
Also, because the solenoid is axially symmetric, you can choose the direction of the [itex]x[/itex] axis such that [itex]\mathbf{B} = B_x\mathbf{e}_x + B_z\mathbf{e}_z[/itex]. This should simplify the calculations.
I did not think of that. That makes the expressions shorter, thanks!
 
  • #6
Could somebody please check if my solutions on i - iiii are correct in the pdf attached to this post?

Thanks in advance!
 

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Related to Calculating crossproduct integral, Parametrization

1. What is a crossproduct integral?

A crossproduct integral is a type of integral that is used to calculate the area of a surface in three-dimensional space. It involves taking the cross product of two vectors and integrating them over a given region.

2. How is a crossproduct integral calculated?

To calculate a crossproduct integral, you first need to parametrize the surface by defining two vector functions that describe the surface. Then, you take the cross product of these two vectors and integrate it over the region of the surface. This will give you the area of the surface.

3. What is the purpose of parametrization in a crossproduct integral?

Parametrization is used to define a surface in terms of two variables, usually u and v, which can then be used to describe the surface in three-dimensional space. This allows for the calculation of the cross product and subsequent integration to find the area of the surface.

4. What are some common applications of crossproduct integrals?

Crossproduct integrals are commonly used in physics, engineering, and other scientific fields to calculate the area of surfaces such as curved objects, fluid flows, and electromagnetic fields.

5. Are there any limitations to using crossproduct integrals?

One limitation of crossproduct integrals is that they can only be used for surfaces that can be parametrized by two variables. Additionally, they may be difficult to calculate for complex surfaces or in cases where the parametrization is not well-defined.

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