0
$\begingroup$

We have an AB conductor of length $2L$ and an infinite one coming out of the screen as shown in the picture (a is the distance between the 2nd conductor and the middle of the finite conductor)

enter image description here

Horizontal conductor carries a current $I$ and the other one a current $I'$. I found dF to be : $$dF=\frac{μ_0\cdot I\cdot I'}{2π}\cdot \frac{x}{x^2+a^2}dx$$ My notes say that I can find torque using this : $$dT=dF\times2x$$ and integrating from $0$ to $L$.That gives this result :$$T=\frac{μ_0\cdot I\cdot I'}{π}\cdot \frac{L}{L^2+a^2}a[\frac{L}{a}-arctan(\frac{L}{a})]$$

I tried a different way myself. I integrated to find the force for half the conductor and considered that it's exerted at the middle point between 0 and L(this might be wrong). Then to find torque: $T=F\cdot 2\frac{L}{2}=FL$ since the forces are couples I multiplied by 2. However, this gave me something different (integrating F instead of T) $$T=\frac{μ_0\cdot I\cdot I'}{4π}\cdot ln(L^2+a^2)\cdot L$$ What led to a wrong result?

$\endgroup$
  • $\begingroup$ You do understand what you did with the first calculation, right? You calculated all of the dF forces which are vectors directed either out of or into the plane of your paper. Those forces result in a torque about the "a" axis that will try to turn the bottom wire so that it is parallel to the top wire. What you're doing in the second calculation, though, doesn't seem to make much sense. It's like you're just taking any force and length values that you can find without thinking about the physics of what's going on and then just multiplying them together. $\endgroup$ – user93237 Jun 5 '17 at 18:31
  • $\begingroup$ I do think about the physics , it's just that there are some things I'm not sure about. I try to imagine that there is one force at $x=-L/2$ and another at $x=L/2$. So integrating from 0 to L , all these little dFs will give me a net force F which would be exerted on the centre of mass of the right half of the conductor , that is $L/2$. The same applies for the other side , so we have a force couple each acting at $-L/2$ and $L/2$ respectively. For the torque , I use the formula for couples $T=F2d $ where d is the distance from the centre of mass of the whole conductor. $\endgroup$ – John Katsantas Jun 5 '17 at 18:37
  • 1
    $\begingroup$ OK, now I understand what you were trying to do with your second calculation. The problem is that one can't use a "force times average length" or "center of mass type reasoning" to calculate the torque in this way. The center of mass is a very useful concept, but in practice the problem with it is that using it seems so simple and natural that people often automatically assume that "center of mass type reasoning" can be carried over to other types of calculations but it can't. $\endgroup$ – user93237 Jun 5 '17 at 18:47
  • $\begingroup$ That's why I was confused. I used it in a similar problem and the result was just fine. I hope someone gives a good explanation as to why it doesn't work here. $\endgroup$ – John Katsantas Jun 5 '17 at 18:51
  • $\begingroup$ It's not so much a question of "why doesn't it work here" as it is a question of "why does it work" with certain problems in mechanics involving linear and angular momentum and the balancing of objects due to gravity. You don't have to go through the derivations in detail, but it might be a worthwhile use of your time to briefly skim through the derivations to see why "center of mass" is a useful concept in those cases. $\endgroup$ – user93237 Jun 5 '17 at 19:23

Your Answer

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

Browse other questions tagged or ask your own question.