Square wire in a cylindrical magnetic field

In summary, the conversation discusses the conversion of a main integral to cartesian coordinates, involving the use of the axis of symmetry being oriented along the y-axis. The problem mentioned in the conversation may be difficult and it is unclear what the variable "s" stands for. It is suggested to simplify the work by avoiding unnecessary expressions and using trigonometry to find values for sine and tangent.
  • #1
Konhbri
2
1
Homework Statement
A cylindrical region contains a magnetic field $$\vec{B}=k\hat{\phi}/s$$. A square loop of wire of side d is centered at the origin in the xz plane, with two sides parallel to the z axis. It carries a current I. In the segment of wire at x=d/2, the current is in the $$\hat{z}$$ direction. Find the force on the wire loop.
Relevant Equations
$$\vec{F_m}=\vec{I} \times \vec{B}$$

$$\vec{F_m}=(\int_{-d/2}^{d/2} I\hat{z} \times \frac{k}{s} \hat{\phi} dz)+(\int_{-d/2}^{d/2} I\hat{x} \times \frac{k}{s} \hat{\phi} dx)+(\int_{-d/2}^{d/2} I\hat{-z} \times \frac{k}{s} \hat{\phi} dz)+(\int_{-d/2}^{d/2} I\hat{-x} \times \frac{k}{s} \hat{\phi} dz)$$

$$x=rcos(\phi); y=rsin(\phi); z=z$$

$$r=\sqrt{x^2+y^2}; \phi=tan^-1(\frac{x}{y}); z=z$$
For if the axis of symmetry is oriented along the y-axis I have gotten as far as converting the main integral entirely to cartesian coordinates.

$$\hat{\phi}=-sin(\phi)\hat{x}+cos(\phi)\hat{y} \therefore \hat{\phi} =-sin(tan^{-1}(x/y))\hat{x}+cos(tan^{-1}(x/y))\hat{y}$$
$$\vec{F_m}=(\int_{-d/2}^{d/2} I\hat{z} \times \frac{k}{s} \hat{\phi} dz)+(\int_{-d/2}^{d/2} I\hat{x} \times \frac{k}{s} \hat{\phi} dx)+(\int_{-d/2}^{d/2} I\hat{-z} \times \frac{k}{s} \hat{\phi} dz)+(\int_{-d/2}^{d/2} I\hat{-x} \times \frac{k}{s} \hat{\phi}dz)$$is equal to
$$(\int_{-d/2}^{d/2} I\hat{z} \times \frac{k}{s}*(-sin(tan^{-1}(x/z))\hat{x}+cos(tan^{-1}(x/z))\hat{z}) dz)+(\int_{-d/2}^{d/2} I\hat{x} \times \frac{k}{s}*(-sin(tan^{-1}(x/z))\hat{x}+cos(tan^{-1}(x/z))\hat{z}) dx)+(\int_{-d/2}^{d/2} I\hat{-z} \times \frac{k}{s}*(-sin(tan^{-1}(x/z))\hat{x}+cos(tan^{-1}(x/z))\hat{z} dz)+(\int_{-d/2}^{d/2} I\hat{-x} \times \frac{k}{s}*(-sin(tan^{-1}(x/z))\hat{x}+cos(tan^{-1}(x/z))\hat{z}) dz)$$

I have no idea how to do any of these integrals, and being that the this is the first homework of the quarter I am beginning to think that it wasn't meant to be this hard.

Did I miss something in the description of the problem? is the field oriented so that the axis of symmetry is z? because that would evaluate the magnetic force to be 0 on the top and bottom wires and left on the wires positioned vertically at x=d/2 and x=-d/2.
 
Physics news on Phys.org
  • #2
Konhbri said:
is the field oriented so that the axis of symmetry is z
Yes, to be precise, the axis of rotational symmetry. And:

In the x-z plane, isn't ##\phi=0## for ##x>0\ \Rightarrow \hat\phi = \hat y\ \ ## and ##\phi=\pi## for ##x<0\ \ \Rightarrow \hat\phi = -\hat y\ \ ## ?

In our old template, the first item was 'problem statement and known/unknown variables' ; in you case it isn't clear to me what ##s## stands for ?
 
  • #3
Your work would be simpler if you avoided inane expressions like ##\sin[\tan^{-1}(\frac{x}{z})]##. You are looking for ##\sin\theta## when you know ##\tan\theta=\frac{x}{z}##. Well, you know that ##\tan\theta=\frac{opposite}{adjacent}## which means that you can identify "opposite" with ##x## and "adjacent" with ##z##. You also know that ##\sin\theta=\frac{opposite}{hypotenuse}##, therefore ##\sin\theta=\frac{x}{\sqrt{x^2+z^2}}.~## It's magical, no?
 

FAQ: Square wire in a cylindrical magnetic field

1. What is square wire in a cylindrical magnetic field?

Square wire in a cylindrical magnetic field refers to a wire with a square cross-section that is placed inside a cylindrical magnetic field. This setup is often used in experiments to study the behavior of charged particles in a magnetic field.

2. How does the shape of the wire affect its interaction with the magnetic field?

The shape of the wire affects its interaction with the magnetic field because it determines the amount of surface area that is exposed to the field. In the case of a square wire, all four sides are exposed to the field, allowing for a stronger interaction compared to a circular wire with only one side exposed.

3. What is the significance of using a cylindrical magnetic field instead of a uniform magnetic field?

A cylindrical magnetic field is non-uniform, meaning the strength of the field varies at different points. This allows for a more complex and realistic study of the behavior of charged particles, as they experience different forces at different points in the field.

4. How is the motion of charged particles affected by the presence of a square wire in a cylindrical magnetic field?

The motion of charged particles in a cylindrical magnetic field is affected by the presence of a square wire because the wire creates a non-uniform magnetic field. This causes the particles to experience a force that changes in magnitude and direction as they move through the field, resulting in a curved path.

5. What are some potential applications of studying square wire in a cylindrical magnetic field?

Studying square wire in a cylindrical magnetic field can have various applications, such as understanding the behavior of charged particles in magnetic confinement devices, developing new technologies for particle accelerators, and investigating the properties of materials in strong magnetic fields.

Back
Top