# Determining magnetic flux density of a closely wound toroidal coil

I am using David K. Chengs's Fundamentals of Engineering Electromagnetics book. In chapter 5/question 5.2, there is a given closely wound toroidal coil, and it asks what is the magnetic flux density at the inside.

This question is okay but at the end it says;

"It is apparent B=0 for r<(b-a) and r>(b+a) since the net total current enclosed by a contour constructed in these two regions is zero."

I understand, at r<(b-a) region, there is no current therefore B=0, but I did not understand the other part which is r>(b+a) how total net current is zero here?

Please see the picture of a question to understand better.  • "electromagnetic flux density"??? – Andy aka Dec 13 '18 at 10:10

For r<(b-a), there is no current therefore B=0 (as you already mentioned.)

For r>(b+a), you have to sum up the currents according to their direction. The number of currents/wires pointing into the plot plane is equal to the number of currents pointing out of the surface. The total current is therfore zero.

• Yes, I thought that, but then I realized that current pointing into the surface and pointing out of the surface is one same current. That why it was pointless to me adding those two currents. Anyway, thank you. – Erdem Uysal Dec 13 '18 at 8:17

r>(b-a) is the volume outside of the toroid. (the entire universe outside the donut)

r<(b+a) is the volume inside the toroid. (donut hole if you will)

In an ideal toroid, where the turns are perfectly circular and balanced, magnetic field at the center and outside the toroid is zero. Inside the toroid there is no current flowing. Outside it, if you draw an Amperian Loop Ampere's law you will see that current at the outer and inner circumference of the toroid are in opposite directions and will cancel each other out. Since the net current is zero, the magnetic field outside of the toroid is also zero.

In practice, the coil is helical and a small magnetic field will exist outside the toroid.