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I need to produce about 200 Gauss magnetic field by power up a copper coil (as shown in picture) using 9V battery.

hen I connected to 9V battery, Voltage across the coils are very low (~ 0.7V) and unable to give sufficient current to produce the magnetic field I wanted.

When I connected to 9V battery, Voltage across the coils are very low (~ 0.7V) and unable to give sufficient current to produce the magnetic field I wanted.

The Coil:

Turn = 120

Wire = AWG #16 (1.2mm Diameter)

Diameter = 8cm

Thickness = 2-2.5cm

Question:

  1. I don't understand why the voltage become so low when I connected to the coil (Because of short circuit?)

  2. How to increase my current (to 4 - 6A) so that I could produce enough magnetic feild.**

Thanks!

** Sorry for the missing info.

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    \$\begingroup\$ "as shown in picture" - did you forget to upload your image? \$\endgroup\$ – Kev Jan 22 '14 at 8:55
  • \$\begingroup\$ @Kev I think new users with reputation 1 are not allowed to attach images \$\endgroup\$ – alexan_e Jan 22 '14 at 9:24
  • \$\begingroup\$ New users can link to images, and someone will usually edit it inline. \$\endgroup\$ – Phil Frost Jan 22 '14 at 13:35
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Generating 200 Gauss using a 9 Volt battery is a pretty tall order.

The formula for magnetic flux density for a long solenoid coil is:

B = n x I / (2.02 x L),  where 
    B = flux density in Gauss
    L = coil length in inches
    I = current in Amperes
    n = number of turns in the coil

*Applicable only if length of coil >> diameter, by a factor of 5 or more

Since the coil, battery and current specifications are not provided, taking some assumptions here:

The Lithium battery can supply a maximum rated current of 1 Ampere (see linked datasheet). Other 9 Volt batteries will deliver less than this.

Number of turns needed per inch:

n = (B x 2.02 x L) / I
  = 404

Thus, your coil would need to have at least 404 turns per inch to approximately generate 200 Gauss at the center of the coil, and the coil itself must be less than 0.2 inches in diameter. That is evidently impractical for a self-made coil.

So let us approach the problem from the opposite direction: How much current is needed for a coil of 100 turns per inch to achieve 200 Gauss?

Flipping the above formula around, we get a required current of 4.0424 Amperes per inch of coil length. If the coil is 2 inches long, that figure doubles.

Again, it is impractical to draw anywhere close to 4.0425 Amperes from a 9 Volt battery.

In addition, what wire gauge would you be using, to allow 4 Amperes of current to flow through without heating up the coil or melting the insulation / enamel?


The reason the battery voltage is dropping is the internal resistance of the battery:

It simply is not designed to support the kind of current you are trying to draw from it, and this is causing the voltage drop due to current across that internal resistance, dropped voltage V = I x R(internal).


Solution:

Revisit your requirements, including how much current the wire can sustain, wind a fresh coil of suitable turn density, and then use a high current power supply, not a 9 Volt battery.

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  • \$\begingroup\$ Thank you very much. This really help! If I use the high current power supply such as LIPO Battery, would there be "voltage drop problem" just like using the 9V battery? Anyway to counter it? adding a small resistance? \$\endgroup\$ – Yvonne Jan 23 '14 at 7:17
  • \$\begingroup\$ @user35922 If you reduce current, the Gauss (B) value in the formula above also decreases - regardless of whether current is reduced through a resistor or through the coil's resistance. While a suitable LiPo or LiFePo4 battery would be able to sustain several amperes of continuous current, you need to revisit your coil itself first: How many turns per inch do you have, and what is the resistance of the coil? I seriously doubt that someone can hand-craft a copper coil at 400+ turns per inch, or even 100 turns per inch, that would not overheat due to coil resistance. \$\endgroup\$ – Anindo Ghosh Jan 23 '14 at 7:27
  • \$\begingroup\$ Another thing, after seeing the images you have added of the coil, that coil will not work: Length of coil needs to be at least 5 times the diameter, yours is way way too short and wide: Try 100 turns per inch and 5 inches, on a 1 inch diameter spool. \$\endgroup\$ – Anindo Ghosh Jan 23 '14 at 7:30
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    \$\begingroup\$ DO NOT USE A LIPO BATTERY FOR THESE EXPERIMENTS - it's not safe to use easily damaged high capacity batteries for experiments with low impedance, poorly characterized loads. Do your experiments with a bench supply having adjustable current limit and overload protection. \$\endgroup\$ – Chris Stratton Jan 23 '14 at 16:17
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The coil of wire you have made (picture not actually showing) is likely to have a resistance that is too low to sustain 9V across it from a small 9V battery. Try using a power supply or winding more turns. Have you calculated what current needs to flow to get the required magnetic flux? Have you looked at the data sheet for the battery to see how much current it can supply whilst still maintaining a decent terminal voltage?

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