- #1

Konhbri

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- 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.

$$\hat{\phi}=-sin(\phi)\hat{x}+cos(\phi)\hat{y} \therefore \hat{\phi} =-sin(tan^{-1}(x/y))\hat{x}+cos(tan^{-1}(x/y))\hat{y}$$

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