# Increasing number of turns in a Solenoid?

1. May 6, 2013

### Wiz700

Hello!

I understand that by increasing the number turns the magnetic field is strengthened. But would that "extra" turns demand more power input from the power source?

2. May 6, 2013

### Wiz700

Another thing!

I'm using a double AA battery rated at 2A/H, so will my Solenoid use the battery's 1.5V and draw 2 Amps?
Thus power input will be 3W? Or it depends on the Solenoid's design?

3. May 6, 2013

### milesyoung

The strength of the magnetic field is proportional to N*I, where N is the number of turns of the solenoid and I is the current through it.

What happens to the resistance of the solenoid as you increase the number of turns on it? How will it affect the current?

If V is the voltage of your battery and I is the current through it, what is the power delivered to your circuit?

I don't know how familiar you are with RL circuits but let's assume we're talking DC only and disregard the initial transient state of your circuit.

It seems a bit odd that your battery should determine both the voltage across your circuit and the current through it. Shouldn't the load have something to say about it?

4. May 6, 2013

### Wiz700

Well I'm not sure what happens to R as N increases... I assume R increases. So I only can assume that I decreases... Based on Ohm's law.
I'm not sure how much power is delivered. I'm not familiar with RL circuits, yes its DC only.
I just has basic knowledge of circuits.

You're right. I just assumed the maximum output that the battery will deliver. I don't know how much the load will draw, and I can't measure it right now. I can only guess!

5. May 6, 2013

### milesyoung

Those are good assumptions

A solenoid is, after all, just a piece of conducting wire wound into an odd shape. Everything else being equal, you'd expect a longer wire to have a larger resistance:
http://en.wikipedia.org/wiki/Resistivity

The power delivered to your circuit is V*I. If you continue your reasoning, what happens to the power as you increase the number of turns on the solenoid?

If you hook up your battery, with know voltage, V, to a resistive load, what determines how much current, I, it will draw? You mentioned Ohm's law?

6. May 7, 2013

### Wiz700

I got my meter so I'll make a Solenoid and measure the resistance.
Then use Ohm's law to figure out how much current that coil is going to draw!

Thanks.

7. May 7, 2013

### Wiz700

I can assume that the Solenoid MAX current draw is 2 Amps @ 1.5V
The power = 3W
However, more turns = higher R = less current drawn out.

8. May 7, 2013

### milesyoung

No you can't. The 2 amp-hour rating does not mean your battery somehow limits the current you can draw from it to 2 amps. Your battery has an internal resistance (not by design) that is, in series with the resistance of your connecting wires and that of the solenoid itself, what limits the current you can draw from it.

Further, your battery is rated at 2 amp-hours, but this does not mean you can pull 2 amps from it at 1.5 V for one hour. You might have a hope of pulling out 1.5*2 Wh worth of energy if you discharge it at very low power, i.e. very low current. For an alkaline cell at 2 amps, its terminal voltage will drop like a rock and you'll probably only be able to pull out a fraction of those 1.5*2 Wh.

If available, the discharge curves of your battery might give you a much better idea of what to expect.

9. May 8, 2013

### Wiz700

Woah... I have a lot of things to study!

10. May 10, 2013

### Jiggy-Ninja

The difference between Amps and Amp-Hours would be a helpful place to start.