# Heating ring in Solenoid

Discussion in 'Physics' started by janalexand, Jan 29, 2015.

1. ### janalexand Thread Starter New Member

Jan 29, 2015
2
0
Hello,

I try to model the following problem: I have a coil with in the centre of that coil a metal ring. The metal ring is aligned with the axis of the coil. I try to induce a current in the ring by powering the coil.

I tested this, it works. I cannot find the same results in test as in simulation.
The coil is primer transformer side, the ring is secondary side.
I know the Resistance and the inductance of the ring. There is 10W dissipated in the ring.
N1 = 3
l1 = 5cm
N2 = 1
R2 = 0,4 ohm
H2 = 64 . 10^-9 H

the result of my simulation is about 400% off. So I presume I made a mistake. Could someone have a look at it?

Thank you guys,

Jan

2. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
You havn't provided full details but why have you divided the field (B) by what I take it is the length of the generating coil (not the xsection area) and does this coil really only have three turns?

3. ### janalexand Thread Starter New Member

Jan 29, 2015
2
0
That is the standard formula to calculate the field in an ideal solenoid. It is derived directly from the Ampère law.

the coil only has 3 windings. Indeed.

4. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
Your formula is only approximate and gets more accurate for long solenoids (those where the length (l) is much greater than the radius (R)

For your solenoid of only 3 turns the correct formula is

$B = \frac{{\mu NI}}{{\sqrt {4{R^2} + {l^2}} }}$