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The magnetic flux density around a long straight conductor is given, B = $\dfrac{\mu_0\cdot I}{2\pi R}$

I have integrated this expression (along a conductor of length l) in order to calculate the flux contained within distance R, $\phi = \dfrac{\mu_0 \cdot I \cdot l}{2\pi }( ln(R)- ln(0))$

However $ln(0)$ is undefined and therefore this expression cannot be evaluated. This comes from the flux density being infinite when $R = 0$. Have I made a mistake here?

Additionally, is it possible to use this formula to calculate the total flux (i.e. integrating from $R = 0$ to $R = infinity$ ). This formula would imply that the total flux around the conductor is infinite.

Is there an alternative way to calculate the total flux around a long straight conductor? Perhaps something along the lines of $\phi = I \cdot L$ (where I is the current in the conductor and L its inductance)

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  • $\begingroup$ I've added the homework-and-exercises tag. In the future, please use that tag on this type of question. $\endgroup$ – Ben Crowell Oct 2 '18 at 15:27
  • $\begingroup$ Why are you integrating with respect to $R$? $\endgroup$ – Aaron Stevens Oct 2 '18 at 15:27
  • $\begingroup$ I'm integrating with respect to R as I'm trying to calculate the flux by summing the flux density along distance R $\endgroup$ – Chris Brown Oct 2 '18 at 15:33
  • $\begingroup$ I'm trying to create a model for something I have built in the lab so I didn't use the homework-and-exercises tag @BenCrowell $\endgroup$ – Chris Brown Oct 2 '18 at 15:34
  • $\begingroup$ What you have forgotten is that inside the wire the magnetic field drops linearly to zero at the centre. $\endgroup$ – Farcher Oct 2 '18 at 16:06
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I assume that you want to calculate flux $\Phi=\int BdS$ across the rectangle with dimensions $R$ and constant $l$ which is outside of the conductor. Then $\Phi=\int_{R_1}^{R_2}\frac{\mu_0 I}{2\pi R}ldR$.

If this is what you've done then your result is correct.

I don't think there is any problem with infinity. If $R$ grows, the $\Phi$ should grow too, although very slowly (which can be seen by looking at the logarithm at your result.

P.S. If it's that $1/R$ in $B$ bothering you, try looking at this: $\sum\limits_{i=1}^\infty \frac{1}{n} = \infty$ Sometimes even sum of infinite series with decreasing numbers can be infinite.

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  • $\begingroup$ Thanks for the clarification but I'm still unsure how you would calculate the total flux around a conductor (this is a useful number as I can then use it to work out the proportion of flux which is coupled to another conductor) $\endgroup$ – Chris Brown Oct 3 '18 at 7:42
  • $\begingroup$ I probably don't understand what you mean by 'total flux around a conductor'. What is the area you want to use for integration? $\endgroup$ – Andrej Oct 3 '18 at 11:04
  • $\begingroup$ For conductor of length $l$, I want to know the total flux around it. Therefore the area is $l *2*\pi*R$ (surface of a cylinder) and this should be integrated for $R=0$ to $R=infinity$ $\endgroup$ – Chris Brown Oct 3 '18 at 13:05
  • $\begingroup$ Well then the total flux is zero. $\vec B$ will look like in this picture google.com/…: and the cylinder you are talking about is going to be parallel to $\vec B$ at every point of its surface. Normal vector of the cylinder is going to be perpendicular to $B$ everywhere and therefore $\Phi=\int \vec{B} d \vec{S} = 0$ $\endgroup$ – Andrej Oct 3 '18 at 13:34

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