# Finding out the magnetic field inside the cross section of the torus which is wrapped by conductive wire and the current flows it

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?

• $\alpha$ 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. Sep 26, 2021 at 12:24

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.

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

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