2
$\begingroup$

Does the Lorentz Force on a Current Carrying wire given by the equation

$$\mathbf{F} = I \int \text{d}\ell \times \mathbf{B}$$

constitute an action reaction pair? That is, if i have two arbitrarily shaped current carrying wires, is it true that force on any one of them due to the magnetic field of the other is equal to the force on the other due to the magnetic field of the first?

$\endgroup$
  • 1
    $\begingroup$ I've replaced your image by typing out the text using MathJax. It's generally advised to typeset your equations rather than using images, so that they're easy to read and to search for. There's a nice tutorial here, in case you'd like! :) $\endgroup$ – Philip Aug 25 at 18:19
  • 1
    $\begingroup$ Okay. Thank you! I'll keep that in mind. $\endgroup$ – aren't eistert Aug 26 at 9:38
2
$\begingroup$

edited

No, it seems that the law of action is true for the cherry-picked case of parallel wires as shown by @Sagigever's response. For instance, this is certainly not true for the case of magnetic Lorentz force exerted by two charges (yellow in figure) moving perpendicular to each other (along the blue and red lines). In this case, the $\vec{F_{12}}$ is perpendicular to $\vec{F_{21}}$ (the dotted lines represent the magnetic field due to the charges).

enter image description here

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ But $F_{12}$ and $F_{21}$ are both equal to zero in your example so it doesn't matter what direction the vector is supposedly pointing in since it's just $0\hat{i}+0\hat{j}+0\hat{k}$. Also, what do you you mean by "vectorially equal"? Because even with two parallel wires, I would not say that a force is $1\hat{i}$ Newtons is equal to $-1\hat{i}$ Newtons. Their magnitudes might be equal but they are in different directions so they cannot be vectorially equal. $\endgroup$ – DKNguyen Aug 25 at 20:33
  • $\begingroup$ Oh, are $F_{12}$ and $F_{21}$ referring to the electric field (not the magnetic field) along the length of the wire directly applying a force on the charges in the other wire? Is that what you're talking about? (And I guess these wires are somehow crossing the same point in space?) $\endgroup$ – DKNguyen Aug 25 at 20:39
  • 1
    $\begingroup$ @ DKNguyen, I chose a poor example I suppose. Charges demonstrate the concept better. $\endgroup$ – Rohit Aug 25 at 21:13
  • $\begingroup$ So I guess these are like a stream of charges being fired perpenicular to each other then? $\endgroup$ – DKNguyen Aug 25 at 21:18
  • 1
    $\begingroup$ Yup, that's one way to think about it. We could even consider the two yellow charges by themselves. 1st charge producing a magnetic field $\vec{B_1}=\frac{\mu_0}{4\pi}\frac{q_1\vec{v_1}\times\vec{r_2}}{r_2^3}$ which then exerts a force $q_2 \vec{v_2} \times \vec{B_1}$ on the second charge. $\endgroup$ – Rohit Aug 26 at 7:38
1
$\begingroup$

Yes this is true, for example Lets take 2 parallel infintite wires carrying a current $I_1$ and $I_2$ that located distance $d$ from each other

We know from biot savart law that magnetic field of wire at distance $d$ is $$\frac{\mu_0 I}{2\pi d}\hat \phi$$ so now we can calculate to Force of wire 1 on wire 2 by your formula $$\vec{F} = I \int \vec{d}\ell \times \vec{B}$$ and getting the result$$F=I_{2}\left(\frac{\mu_{0} I_{1}}{2 \pi d}\right) \int d \ell_{2}$$ and the force per unit lengh will be $$f=\frac{\mu_{0} I_{1} I_{2}}{2 \pi d}$$

Now you can see that if $I_1$ and $I_2$ are in the same direction we getting repeling force, and if $I_1$ and $I_2$ are in opposite directions we getting attractive force.

Of course we could do the same by calculate the force of wire 2 on wire 1 and getting the same result.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Newton's third law in modern terms states conservation of momentum. Electrostatic forced conserve $P_{kin}=\sum_i m_i p_i$ but magnetic forces do not , as argued by @Rohit. In the presence of electromagnetic fields the conserved momentum is $P = P_{kin} + P_{pot}$, where $P_{pot} = - \sum_i q_i \vec A_i$.

| cite | improve this answer | |
$\endgroup$

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.