What Surface Does ds Represent in Calculating Magnetic Flux Through a Toroid?

[Abdulwahab Hajar]

Homework Statement

The general method to find the total magnetic flux through an object is found by:
Φ =∫B*ds (dot product)
what is the ds over which we take the integral on??
what surface is it?, is it the surface over which the magnetic flux enters?

Homework Equations

Φ =∫B*ds (dot product)

The Attempt at a Solution

well the problem is that a magnetic field can extend so far right??
but in a toroid we assume that almost B= 0 anywhere other than in between the inner and outer radius.
Therefore the surface should be one in between the inner and outer radii of the toroid, in a toroid like the in the figure attached the magnetic field is µ₀NI/2πr
therefore the total flux Φ =∫(µ₀NI/2πr)*ds
In the book ds is defined as h*dr (in the direction of phi)
why can't ds simply be h*(b-a) namely the height of the toroid multiplied by the outer radius - the inner radius
why is the term dr necessary??
furthermore what is ds usually in general?

Thank you

 

[NascentOxygen]

H is not constant, it varies with r. So each of the small B.dA terms that you need to sum are going to have different values depending on r, hence the need for integral calculus.

 

[Abdulwahab Hajar]

Thank you sir, now I get it