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As shown in the diagram , the torus of radius of \$~ R ~\$ exists .

The wire is wrapped around the torus .

$$ N:=\text{total number of the turns of the wire around the torus } $$

The current is flown inside the wire .

$$ I:=\text{current } $$

$$ r:=\text{radius of cross section of the torus } $$

$$ H:=\text{magnetic field at } ~ \left( r,\theta_{} \right) $$

Currently I have the dought of the below equation . I've been unable to derive the below rightmost formula .

$$ NI = \int_{ }^{ } \boldsymbol{H}\cdot\boldsymbol{ds} = H\cdot 2\pi \left( R+r \cos^{}\left(\theta_{} \right) \right) $$

I confidentially guess that the above equation is gained by the law of Ampere .

We assume the circle of radius \$~ \left( R+ r \cos^{}\left(\theta_{} \right) \right) ~\$ which circumference is inside the flesh of the torus .

We assume the closed path . The path is made by the circumference of the circle which we defined previously .

\$~ NI ~\$ makes sense of penetrations of each current(\$~ I ~\$ ) to the closed surface .

I assume \$~ ds= \left( R+ r \cos^{}\left(\theta_{} \right) \right) d\theta ~\$

$$ NI = \int_{ }^{ } \boldsymbol{H}\cdot\boldsymbol{ds} = H\cdot 2\pi \left( R+r \cos^{}\left(\theta_{} \right) \right) $$

$$ = \int_{0 }^{2\pi } H \cdot \left\{ \left( R+ r \cos^{}\left(\theta_{} \right) \right)d\theta \right\} $$

$$ = \int_{0 }^{2\pi } H \cdot \left( R+ r \cos^{}\left(\theta_{} \right) \right)d\theta $$

I've been stucked from here since \$~ H ~\$ includes the information of \$~ \theta_{} ~\$ and, the form of formula of \$~ H ~\$ is unknown so far .

What I should focus for next?

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  • \$\begingroup\$ \$\alpha\$ is the internal cross sectional radius of the toroid not \$r\$. Dimension \$r\$ is the distance of point \$P\$ from the centre of the toroid's cross section. \$\endgroup\$
    – Andy aka
    Sep 26, 2021 at 12:24

1 Answer 1

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I assume $$ds= \left( R+ r \cos^{}\left(\theta_{} \right) \right) d\theta$$

No. ds refers to a small displacement around the torus, (i.e. around the torus's axis of symmetry. ) By symmetry, B or H is the same at all points in a circle around the axis of symmetry, so for such a path, B or H is constant and can be taken outside of the integral. This modified integral is then just the circumference of that circle.

$$\oint ds= 2\pi \left( R+ r \cos^{}\left(\theta_{} \right) \right)$$

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  • \$\begingroup\$ About "around the torus's axis of symmetry" , needless to say that this "around" includes inside the flesh of the torus? \$\endgroup\$ Sep 27, 2021 at 4:13
  • \$\begingroup\$ Using your $$\oint ds= 2\pi \left( R+ r \cos^{}\left(\theta_{} \right) \right)$$ , I wrote the below equations. $$ NI = \int_{ }^{ } \boldsymbol{H}\cdot\boldsymbol{ds} = H \oint ds = H \cdot \left\{ 2\pi \left( R+ r \cos^{}\left(\theta_{} \right) \right) \right\} $$ \$\endgroup\$ Sep 27, 2021 at 4:17
  • \$\begingroup\$ "needless to say that this "around" includes inside the flesh of the torus? " Yes, the B or H is constant along any circle centered around the axis of symmetry of the torus, whether inside the "flesh" or outside (where it is 0). \$\endgroup\$ Sep 27, 2021 at 4:22
  • \$\begingroup\$ It is quite weird for me that H can be taken out from the integrator . \$\endgroup\$ Sep 27, 2021 at 4:23
  • \$\begingroup\$ Since H includes the information of \$\theta ~~,~~ r \$ \$\endgroup\$ Sep 27, 2021 at 4:24

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