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)

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.