Increasing number of turns in a Solenoid?

In summary, the conversation discusses the relationship between the number of turns in a solenoid, the magnetic field strength, and the power input from a battery. It is determined that the strength of the magnetic field is proportional to the number of turns and the current through the solenoid. The resistance of the solenoid also increases as the number of turns increases, affecting the current. The power delivered to the circuit is determined by the voltage and current, and the battery's discharge curve is a factor in determining the amount of power that can be drawn from it. Further study is suggested, particularly in the concepts of Amps and Amp-Hours.
  • #1
Wiz700
96
1
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?
 
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  • #2
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
Wiz700 said:
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?
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.

Wiz700 said:
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?
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
milesyoung said:
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.

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.


milesyoung said:
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?

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
Wiz700 said:
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.
Those are good assumptions :smile:

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

Wiz700 said:
I'm not sure how much power is delivered.
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?

Wiz700 said:
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!
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
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
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
Wiz700 said:
I can assume that the Solenoid MAX current draw is 2 Amps @ 1.5V
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
Woah... I have a lot of things to study!
 
  • #10
Wiz700 said:
Woah... I have a lot of things to study!
The difference between Amps and Amp-Hours would be a helpful place to start.
 

1. How does increasing the number of turns in a solenoid affect its magnetic field?

As the number of turns in a solenoid increases, the strength of its magnetic field also increases. This is because each turn of wire adds to the overall magnetic field, resulting in a stronger and more concentrated field.

2. What is the relationship between the number of turns and the inductance of a solenoid?

The number of turns in a solenoid is directly proportional to its inductance. This means that as the number of turns increases, so does the inductance, resulting in a higher resistance to changes in the current flowing through the solenoid.

3. Does increasing the number of turns in a solenoid affect its self-inductance?

Yes, increasing the number of turns in a solenoid also increases its self-inductance. This is because self-inductance is directly proportional to the number of turns in a solenoid, as well as the square of its radius and the permeability of the material within the solenoid.

4. What is the effect of increasing the number of turns in a solenoid on its inductive reactance?

Increasing the number of turns in a solenoid also increases its inductive reactance. This is because inductive reactance is directly proportional to the inductance of a solenoid, and as discussed previously, the inductance increases with the number of turns.

5. Are there any downsides to increasing the number of turns in a solenoid?

While increasing the number of turns in a solenoid can result in a stronger magnetic field and higher inductance, it can also lead to increased resistance and decreased efficiency. This is because the longer length of wire in the solenoid can cause more energy loss due to heat and increased voltage drops.

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