1
$\begingroup$

enter image description here

They used the ampere law to calculate magnetic field by a toroid ( assuming perfectly circular coils provinding symmetry and neglecting effects due to helical nature) whis is $μ_0ni$(where $n$ is no of loops per unit length and $i$ is the current flowing in toroid) inside the toroid and elsewhere 0. i can easily verify the law for a straight long wire but not able to do it for a ring. enter image description here

$\endgroup$
  • 1
    $\begingroup$ Think about what symmetry is present in your torroid field that isn't true here $\endgroup$ – Aaron Stevens Aug 20 at 13:55
  • $\begingroup$ Toroid radius should be very greater than radius if circular ring $\endgroup$ – yuvraj singh Aug 20 at 13:57
  • $\begingroup$ Does the Amperian loop you have drawn match the usual Amperial loop used to determine the field for the torroid? $\endgroup$ – Aaron Stevens Aug 20 at 13:59
  • $\begingroup$ my doubt is only that can we apply ampere law for a current carrying ring and loop as shown above . Whats written under the image explain only that why i ended up with this doubt. $\endgroup$ – aryan bansal Aug 20 at 14:11
1
$\begingroup$

The current carrying loop that you have shown is different from the loop in the book as in current flows along the loop in your picture whereas it flows around the loop (like a curved helix) in the second picture. Ampere's law is valid for both the cases.

$\endgroup$
  • $\begingroup$ can u prove it for first picture(my reputation is under 15 right now). $\endgroup$ – aryan bansal Aug 20 at 14:32
  • $\begingroup$ Check this $\endgroup$ – Akshat Sharma Aug 20 at 14:54
  • $\begingroup$ i m saying to prove that the law is valid for ring by biot stewart law $\endgroup$ – aryan bansal Aug 20 at 14:58
1
$\begingroup$

Ampere's Law is always valid (ignoring displacement current for now). However, you can only use it to determine the field of a configuration when certain symmetries are present. This is why you only see it used on

  • Cylinders (Like straight wires)
  • Planes (Current sheets)
  • Straight solenoids
  • Torroidal solenoids

The reason we can only use Amepere's law to determine the field in these specific systems is because we can make symmetry arguments that allows us to move the field term $B$ out of the line integral.

The problem with your drawing of an Amperian loop that is concentric with a section of the current loop is that the field is not behaving in a nice way around your Amperian loop. It changes magnitude as well as direction relative to the loop. Compare this to the torroidal solenoid where we can argue that the field has a constant magnitude and always points in the same direction of our Amperian loop as we move around the loop. In your wire case, there would be no way to take the $B$ term out of your line integral then. Ampere's law cannot be used to determine the field along your Amerpian loop here.

$\endgroup$
  • $\begingroup$ if you take a finite length straight wire (magnetic field strength is not equal to μ0i/2πr) the loop integration would not come out be μ0i. Which means there should some condition where law is valid. therefore i was asking is it valid for current carrying ring. $\endgroup$ – aryan bansal Aug 20 at 14:42
  • $\begingroup$ For Ampere's law to work you must also assume that $div J=0$, which is not the case for a finite piece of wire. $\endgroup$ – hyportnex Aug 20 at 14:58
  • $\begingroup$ @aryanbansal I am not sure if I understand what you are talking about here. My answer shows that you cannot use Ampere's law to determine the field in this scenario. It does not mean it is not valid in this scenario, you just can't use it in this way. $\endgroup$ – Aaron Stevens Aug 20 at 15:05
  • $\begingroup$ Hyportnex could u please elaborate $\endgroup$ – aryan bansal Aug 21 at 11:46
  • $\begingroup$ @aryanbansal Ampere's law in differential form is $$\nabla\times\mathbf B=\mu_0\mathbf J$$ Therefore $$\nabla\cdot\mathbf J=\nabla\cdot(\nabla\times\mathbf B)=0$$ Once you add in displacement current then you can fix Ampere's law. $\endgroup$ – Aaron Stevens Aug 21 at 12:03

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.