Electromagnetism;Ampere's Law Application for finding magnetic field strength(B) inside a current carrying solenoid

Question is that why we multiply the current in one loop to the number of turns(enclosed in amperian rectangular loop) ALTHOUGH the current flowing(charges flowing per unit time) is SAME through all the loops??? There is a SINGLE complete circuit If there would be more than one circuits comprising each loop, then i think we should add all currents in individual loops BUT in this situation there is a single circuit... I hope my question is clear


Answer by @Frecher is correct! I will try to give physical essence to your question.

Each loop of current generates its own magnetic field regardless of whether they are part of the same circuit or not. Think of two loops which are in the same circuit and have the same current flowing through them. Now separate the two loops very very far away from each other (you have very long wire!). Now they don't interact with each other but they still generate the magnetic field at their own places. This means each loop must be generating magnetic field even though they are part of the same circuit and the same current flows through them!

All this is actually incorporated automatically in Ampere's law.

  • $\begingroup$ Thanks for the help... Now I understand why the current enclosed is the sum of the individual currents in loops... So actually the net magnetic field of the solenoid is the sum total of all the individual magnetic fields... $\endgroup$ Aug 20 '18 at 17:17

An Amperian loop $abcd$ is shown in the diagram below.

enter image description here

So you perform the integration $\displaystyle \int_{\rm abcda} \vec B \cdot d \vec s$ around the closed loop $abcda.

That closed Amperian loop defines a surface (shown in pink) and you need to sum the total current passing through that surface $\vec I_{\rm enclosed}$ which you will see is the same magnitude current passing repeatedly through the Amperian surface.

That is what Ampere's law demands that you do.

You could have the Amperian loop just encircle one wire and that would give the same result for the magnetic field $B$ because the path for the line integral would be correspondingly shorter.

Source of unmodified diagram.

  • $\begingroup$ Sir can you explain what is wrong in taking the amperial loop as a circle of a radius $r$, where $r$ is the distance from the axis of solenoid? B should be same due to circular symmetry, but since there would be no current enclosed in that circle (because current flows through the wires on the surface), so that should give $0$ magnetic field. $\endgroup$ Mar 31 at 10:11
  • $\begingroup$ No. I enclosed equals zero but so does $ \int_{\rm loop} \vec B \cdot d \vec s$ as the magnetic field is at right angles to the loop. $\endgroup$
    – Farcher
    Mar 31 at 11:25

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