# Inductance not working in simple circuit

1. Jul 24, 2012

### JeffMandell

I want to create a simple circuit consisting of a power source and an inductor.

I have a simple circuit consisting of a battery and a wire wound into a solenoid, then separated by a thin layer of air is another solenoid that is connected to a multimeter. Both solenoids have a similar diameter to a quarter and are about 40 turns. The battery is 9 Volts. The first solenoid gets hot when the circuit is closed so I know that the battery works.

My problem is that I am not seeing a reading on the multimeter.

My concern is that the duracell 9 Volt battery is a form of direct current and I think that I need to be using alternating current. If this is the case what forms of alternating current can be used for such an experiment. My second concern is that the secondary solenoid does not complete a circuit (both ends of the coil attach to a multimeter). Where should the ends of the solenoid go to complete the circuit.

Here's a cool link to show how awesome inductance can be:

Last edited by a moderator: Sep 25, 2014
2. Jul 24, 2012

### cepheid

Staff Emeritus
Yes, you need AC. It is a *changing* magnetic field that induces an electric field (and hence a voltage)

If there is a voltage across the secondary, the multimeter will measure it.

3. Jul 24, 2012

### 256bits

You noticed the difference between your two inductors in the video and the ones you have made? For one the daimeter and secondly presumably the number of turns.

Simply, for an air coil inductor, L = μ$_{o}$ N$^{2}$ A / l
where
μ$_{o}$ is the permeabitlity of free space ( 4∏ x 10exp(-7) ) H/m
N is the number of turns
A is the cross sectional area in square meters
l is the length of the air coil

Different formulas apply for differennt variations of how the coil is wound ( flat , short etc )

So, with your small size ( about a quarter ) , low turns (40 ), and short length , the inductance of your coil is quite small.

In addition, you also noticed in the video that when the individual moved the coils farther apart or at right angles to one another, the sound from the speaker became more faint. What is happening is that the mutual inductance between the two coils became less and and less wherby less lines of magnetism from the first coil found their way through the coil of the second. Two coils at right angles to one another theoretically have no coupling.

You can increase the inductance by winding the coil on a material of high permeability and the mutual inductance by winding both coils on the same. Iron is one such material.

The formula for the voltage across the ideal inductor is v(t) = L di/dt. Theoretically connecting an ideal voltage DC source to an ideal inductor would give a continious rate of change of current so that the inductance voltage would equal the supply voltage,

The catch is the word ideal. Voltage sources have internal resistance and, such as your 9 v battery, have a limiting amount of current they can produce. Real inductors have a resistance of the wire ( which applis more to your setup ).

In your case, as you connect the inductor to your 9 v battery the voltage drop is split between the inductance of the coil and its resistance as they are both in series, As time progresses and the di/dt ( rate of change of current ) across the inductor causes the current i to increase, the i r voltage drop across the resistance increases. Eventually the current becomes so large that the ir drop equals 9 v and the di/dt becomes zero.

For your setup this should have in milliseconds and your mutlimeter would record only a glitch of the needle movement or none at all. An oscilliscope setup to record the glitch for you to see the voltage or current in the second inductor might be a option.

Last edited by a moderator: Sep 25, 2014