0
$\begingroup$

when we keep two wires near each other then they will experience same force is it then correct to explain by newton's 3rd law?

    -
Improve this question
$\endgroup$

3 Answers 3

Reset to default
1
$\begingroup$

when we keep two wires near each other then they will experience same force is it then correct to explain by newton's 3rd law?

Yes but... In the special case you are examining it's ok. But generally speaking when e.m. fields are involved you can't trust third law. A well known example is the following (very likely you'll find it on many e.m. books).

You're given two positive charges moving of uniform motion. Charge A along $+x$-axis, charge B along $+y$-axis. Consider the instant when B is in the origin and suppose at that instant $x_{\rm A}<0$.

The force A applies on B is purely electrical since A's magnetic field vanishes on $x$-axis. That force is directed along $+x$. We don't need to compute it.

The force B applies to A has an electric part, directed along $-x$ (that's not trivial). But there is also a magnetic force. In A's position B's magnetic field is directed along $+z$ so Lorentz' force on A is along $-y$.

Conclusion: Newton's third law doesn't hold in that case. The explanation is that there are electric and magnetic fields, having a non-vanishing momentum. Moreover that momentum is not constant because charges are moving and fields are varying in time. Only total (mechanical + em) momentum is conserved.

Improve this answer
$\endgroup$
0
$\begingroup$

Yes of course .they both suffer repulsion if they carry current in opposite direction . And attraction if current flows in same direction . Remember this action reaction pairs in exist in the form of field forces .

Improve this answer
$\endgroup$
0
$\begingroup$

This is a good question as its answer will surprise. When the wires are parallel, momentum is conserved. When not then Newton's third law is not obeyed and momentum is not conserved by the magnetic force.

Consider two charges moving at right angles to each other. The magnetic forces they exert on each other are the perpendicular to each other and do not comply with the third law. The conserved quantity is $\sum_i (m \vec v_i - q_i \vec A) $ where the sum is over the charges.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ I think you'd better elaborate your answer. Your first statement is wrong. If you have whatever wires carrying constant currents there is no electric field, then no e.m. momentum and third law applies. As to second statement it's sometimes right but not always. Your formula for the conserved quantity is obscure - I'd better say wrong. What does the product $\vec v_i \vec A$ mean? What is $\vec A\,$? If $\vec A$ is vector potential, it's a function of space and time. A conserved quantity involving e.m. field should be an integral over space. $\endgroup$
    – Elio Fabri
    Jun 23, 2019 at 15:33
  • $\begingroup$ @ElioFabri It is well known that the magnetic force does not conserve momentum. I corrected the typo in the formula. $\endgroup$
    – my2cts
    Jun 23, 2019 at 20:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.