# Flux linkage inside of a conductor

Can someone explain to me why the flux linkage inside of a conductor is dependent on the cross sectional area of the conductor?

My book says that d$\lambda$ = $(x/r)^2\phi$ where $\phi$ is the magnetic flux through differential area $dA_x$.

Why does the flux linkage depend on the cross sectional area of the conductor, while the magnetic flux is not penetrating the front of the conductor?

Online, people have written that the flux links to the current. I also do not understand what they mean by this. If anyone could help shed some light on this problem your help would be greatly appreciated.

Here is the work leading up to d$\lambda$:

Assume we want to find the flux through differential surface $dA_x$ as shown in the figure. $dA_x$ has a width of $dx$ and length of 1m so $dA_x = dx$ such that the differential area vector is normal to the surface of the slice.

By Ampere's law, $\int{\vec{B}{\cdot}d\vec{l}} = \mu I_{en}$. Letting the dotted line in the figure represent the Amperian loop, and with $\vec{B}$ being parallel to $\vec{dl}$ in all locations the equation becomes \begin{align} B\int{dl} =& \mu I_{en} \\ B(2\pi x) =& \mu I_{en} \\ B =& \frac{\mu I_{en}}{2\pi x} \, . \end{align} Since we assume uniformly distributed current throughout the conductor we know thatJ_x$=$J_{tot}$. Therefore, $$\frac{I_x}{\pi x^2} = \frac{I}{\pi r^2} \, .$$ Solving for$I_x = (x/r)^2 I$, and substituting for$I_{en}$we see that$B_x = \mu x I/2 \pi r^2$. We need solve for the flux through$dA_x$: $$d\phi_x = \vec{B_x} \cdot \vec{dA_x} = \frac{\mu x I}{2 \pi r^2} dx \, .$$ Here is where I am confused though: There is no flux through the front of the conductor, so we cannot really perceive the front of the conductor to be a "turn" of wire since no flux will be linked to it if no magnetic field penetrates the front portion of the wire. Somehow I am supposed to get $$d\lambda = \frac{\pi x^2}{\pi r^2} d\phi = \frac{x^2}{r^2} d\phi \, .$$ I guess I am basically having difficulty understanding what the "turns" are in this problem. Perhaps my understanding of flux linkage is a bit off too. Does anyone know some good resources on this subject? I can't seem to find much of use on the internet • What is$d\lambda$? – Tea is life Apr 18 '15 at 21:40 •$d\lambda$is the flux linkage. – Kyle Tierney Apr 18 '15 at 22:24 ## 4 Answers http://nptel.ac.in/courses/Webcourse-contents/IIT-KANPUR/power-system/chapter_1/1_4.html "Note that the entire conductor cross section does not enclose the above flux. The ratio of the cross sectional area inside the circle of radius x to the total cross section of the conductor can be thought about as fractional turn that links the flux dφ . Therefore the flux linkage is " "as fractional turn" "as fractional turn" but i can't say that i understand that very good I think of flux linkage as $$N \cdot \phi$$, where $$N$$ is the number of times magnetic field pierce through the Riemann surface formed by the loops, which is roughly a stack of membranes. $$N$$ is usually referred to the number of loops in the coil, which can be hard to think in this problem. Think of the conductor as a bunch of loops, and see which loops are pierced through by that particular magnetic field line. Note that if magnetic field line pierce through a loop from both directions, then the effects are canceled. In the picture below. A is pierced from only one direction. B is pierced from both direction. C is not touched. Only loops A contributes to the flux linkage, and the fraction of loops like A (which is loops encompassed by the field line), is the fraction desired. The magnetic field only pierced through loop A. It took a tour into loop B and came out. This is my way of thinking. see whether it helps! think of the straight conductor in question is now turned to form a coil with one turn. Now, when we think of the flux linkage for the whole coil(with$\mathbf1$turn), we would first find the total flux(by the conductor coil, in perpendicular to it) and then multiply it by$\mathbf1$(No of turns). Note that, here we considered the (total) flux due to the total current in the conductor. but in, $$dϕ_x=\vec{Bx}⋅\vec{dAx}=\frac{μxI}{2πr^2}dx.$$ we considered the flux due to only$\textit {$I_x$}$. hence, it make sense to multiply by the fractional turns to get the flux linkage. you may think the fractional turns to be fractional volumes, thus fractional area$(πx^2/πr^2)\$(as length is same in top and bottom).

Therefore, incremental flux linkage is given by $$dλ=\frac{πx^2}{πr^2}dϕ=\frac{x^2}{r^2}dϕ.$$

• Please don't give explicitly complete solutions to homework questions. We are homework-solving service, after-all. – user36790 Sep 25 '16 at 4:15

No. Of turns simply gives the total current enclosed by or linked to the flux lines. As in case of coil with n turns and carrying current I ,the total current enclosed is ni thus the flux linkage is n times the flux. Here only a fractional part of current is enclosed by the flux lines thus we have fractional turn.