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BenRW
BenRW
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Current, I, is given by the formula:

$$ {I=\frac{V}{R}} $$

So, for a current of 1000 A, you need:

$$ {1000 A = \frac{volts}{resistance}} $$

Resistance is proportional to the length of a wire, and inversely proportional to the cross-section of a wire. If you have more turns, you increase the resistance, which reduces the current at any given voltage. As you're interested in making an electromagnet, it's worth saying that this reduces the current by exactly the same ratio that increasing the number of turns strengthens the magnetic field.

So, for your example, you either need a thicker wire, or you need more volts. Copper has a resistivity of $1.68\cdot10^{-8}\Omega/m$, so a wire with a cross section of $1mm^2$ and a length of $1m$ would have a resistance of $0.01678 \Omega$, and a 1kA current flowing through that 1m wire would require a voltage of

$$ V = 1000A \cdot 0.01678 \Omega = 16.78 volts $$

However, it would get very hot. Power is $P=VI$, here $16.78V \cdot 1000A = 16.78 kilowatts$. Normally this sort of power demand is engineered away by using low current and high voltage, but you don't want that, so instead you will need to lower the voltage by making the conductor thicker. Make it twice the cross-section to halve the resistance, and halve your voltage, and halve your power use. A solid copper bar with a $1cm^2$ cross section, being 100 times the cross section of your wire, would reduce the power use to 167.8 W, which is far less likely to melt when you switch it on.

And remember, with these numbers, this is what it looks like for each $1m$ long section of the coil. A 15m long, 1m radius cylinder with 2 turns/meter would be close to 190 such sections, needing 190 times the voltage and dissipating 190 times the power. For the $1cm^2$ copper bar example, that's ~31.9 volts and 31,882 watts for 1kA. (Which wouldn't be anything like enough for 1-2 T field, but that scales up in a nice clean way so I won't go into it).

Current, I, is given by the formula:

$$ {I=\frac{V}{R}} $$

So, for a current of 1000 A, you need:

$$ {1000 A = \frac{volts}{resistance}} $$

Resistance is proportional to the length of a wire, and inversely proportional to the cross-section of a wire. If you have more turns, you increase the resistance, which reduces the current at any given voltage. As you're interested in making an electromagnet, it's worth saying that this reduces the current by exactly the same ratio that increasing the number of turns strengthens the magnetic field.

So, for your example, you either need a thicker wire, or you need more volts. Copper has a resistivity of $1.68\cdot10^{-8}\Omega/m$, so a wire with a cross section of $1mm^2$ and a length of $1m$ would have a resistance of $0.01678 \Omega$, and a 1kA current flowing through that 1m wire would require a voltage of

$$ V = 1000A \cdot 0.01678 \Omega = 16.78 volts $$

However, it would get very hot. Power is $P=VI$, here $16.78V \cdot 1000A = 16.78 kilowatts$. Normally this sort of power demand is engineered away by using low current and high voltage, but you don't want that, so instead you will need to lower the voltage by making the conductor thicker. Make it twice the cross-section to halve the resistance, and halve your voltage, and halve your power use. A solid copper bar with a $1cm^2$ cross section, being 100 times the cross section of your wire, would reduce the power use to 167.8 W, which is far less likely to melt when you switch it on.

Current, I, is given by the formula:

$$ {I=\frac{V}{R}} $$

So, for a current of 1000 A, you need:

$$ {1000 A = \frac{volts}{resistance}} $$

Resistance is proportional to the length of a wire, and inversely proportional to the cross-section of a wire. If you have more turns, you increase the resistance, which reduces the current at any given voltage. As you're interested in making an electromagnet, it's worth saying that this reduces the current by exactly the same ratio that increasing the number of turns strengthens the magnetic field.

So, for your example, you either need a thicker wire, or you need more volts. Copper has a resistivity of $1.68\cdot10^{-8}\Omega/m$, so a wire with a cross section of $1mm^2$ and a length of $1m$ would have a resistance of $0.01678 \Omega$, and a 1kA current flowing through that 1m wire would require a voltage of

$$ V = 1000A \cdot 0.01678 \Omega = 16.78 volts $$

However, it would get very hot. Power is $P=VI$, here $16.78V \cdot 1000A = 16.78 kilowatts$. Normally this sort of power demand is engineered away by using low current and high voltage, but you don't want that, so instead you will need to lower the voltage by making the conductor thicker. Make it twice the cross-section to halve the resistance, and halve your voltage, and halve your power use. A solid copper bar with a $1cm^2$ cross section, being 100 times the cross section of your wire, would reduce the power use to 167.8 W, which is far less likely to melt when you switch it on.

And remember, with these numbers, this is what it looks like for each $1m$ long section of the coil. A 15m long, 1m radius cylinder with 2 turns/meter would be close to 190 such sections, needing 190 times the voltage and dissipating 190 times the power. For the $1cm^2$ copper bar example, that's ~31.9 volts and 31,882 watts for 1kA. (Which wouldn't be anything like enough for 1-2 T field, but that scales up in a nice clean way so I won't go into it).

Source Link
BenRW
BenRW
  • 890
  • 4
  • 19

Current, I, is given by the formula:

$$ {I=\frac{V}{R}} $$

So, for a current of 1000 A, you need:

$$ {1000 A = \frac{volts}{resistance}} $$

Resistance is proportional to the length of a wire, and inversely proportional to the cross-section of a wire. If you have more turns, you increase the resistance, which reduces the current at any given voltage. As you're interested in making an electromagnet, it's worth saying that this reduces the current by exactly the same ratio that increasing the number of turns strengthens the magnetic field.

So, for your example, you either need a thicker wire, or you need more volts. Copper has a resistivity of $1.68\cdot10^{-8}\Omega/m$, so a wire with a cross section of $1mm^2$ and a length of $1m$ would have a resistance of $0.01678 \Omega$, and a 1kA current flowing through that 1m wire would require a voltage of

$$ V = 1000A \cdot 0.01678 \Omega = 16.78 volts $$

However, it would get very hot. Power is $P=VI$, here $16.78V \cdot 1000A = 16.78 kilowatts$. Normally this sort of power demand is engineered away by using low current and high voltage, but you don't want that, so instead you will need to lower the voltage by making the conductor thicker. Make it twice the cross-section to halve the resistance, and halve your voltage, and halve your power use. A solid copper bar with a $1cm^2$ cross section, being 100 times the cross section of your wire, would reduce the power use to 167.8 W, which is far less likely to melt when you switch it on.