Why is the b field inside a thick toroid largest along inner edge

[Maxwellkid]

Why is the b field inside a thick toroid largest along inner edge and smallest along outer edge?

 

[Bob S]

The B field inside an air-filled toroid is given by

B = u₀NI/L

where NI = amp turns, and L is circumference of field line, which is shorter around the inner edge.

[Edit] Look at it this way. The turns on the inner side of the toroid are closer together than the turns on the outer side. So the number of turns per meter is larger on the inner side.

 

[Maxwellkid]

The B field inside an air-filled toroid is given by

B = u₀NI/L

where NI = amp turns, and L is circumference of field line, which is shorter around the inner edge.

If u investigate insde the toroid, don't the outer edge cancel out the inner edge so that the b field is equal both along inner and outer edge?

 

[Bob S]

If u investigate insde the toroid, don't the outer edge cancel out the inner edge so that the b field is equal both along inner and outer edge?

You are correct that the average B field is the same as the central field in the toriod. The lower B field at the outer edge averages out with the higher B field at the inner edge.

 

[Maxwellkid]

You are correct that the average B field is the same as the central field in the toriod. The lower B field at the outer edge averages out with the higher B field at the inner edge.

then, why does it state in my book that the b field along inner edge of the toroid is greater than the b field along outer edge?

 

[gabbagabbahey]

If u investigate insde the toroid, don't the outer edge cancel out the inner edge so that the b field is equal both along inner and outer edge?

B(s)=\frac{\mu_0 N I}{2\pi s} represents the strength of the field due to the entire toroid, at a radial distance where s from the axis of the toroid, inside the toroid.

It is true that the field outside the toroid is zero, and hence, outside the toroid, the contributions to the field due to the current elements on the inner edge will cancel the contributions to the field due to the current elements on the outer edge. But, B(s)=\frac{\mu_0 N I}{2\pi s} does not tell you the values of thes contributions, it only tlls you the total field at each point inside the toroid, not the field due to just the current elements at that point.

 

[Bob S]

Hi maxwell Kid

From Bob S
The B field inside an air-filled toroid is given by

B = u0NI/L

where NI = amp turns, and L is circumference of field line, which is shorter around the inner edge.

[Edit] Look at it this way. The turns on the inner side of the toroid are closer together than the turns on the outer side. So the number of turns per meter is larger on the inner side.

then, why does it state in my book that the b field along inner edge of the toroid is greater than the b field along outer edge?

This is a good question. Already I stated in an earlier post that the azimuthal B field in the toroid is given by

B_θ = u₀NI/(2 pi r)

where r is the radial distance from the major axis of the toroid. So B_θ has a radial dependence, specifically 1/r. Because curl B = 0, this may mean that some other component may not be zero. However, from symmetry, nothing has an azimuthal (θ) dependence, including B_r. So we have to conclude that B has only an r dependence. If we examine Curl B in cylindrical coordinates we would see that this is indeed true.

In fact, if you look at the magnetic field around an infinitely long straight wire, B also has only a 1/r dependence.

Bob S

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