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Energy required to demagnetize Iron. |
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| Jan18-13, 06:56 AM | #1 |
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Energy required to demagnetize Iron.
How would i go about calculating the AC current and overall energy required to demagnetize an Iron plate that has been magnetized by a permanent magnet?
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| Jan20-13, 08:35 PM | #2 |
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Welcome to Physics Forums!
The amount of current probably depends on how much resistance your plate offers and how long you're willing to wait. Unless you're degaussing the plate, you'd need to heat it to the Curie temperature (but based on the heat capacity of steel, you can figure out how much energy it'd take to do so). http://en.wikipedia.org/wiki/Curie_temperature |
| Jan21-13, 07:31 AM | #3 |
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Heating to the curie point would be easier to calculate using E = mcT, but the (T) temperature required to demagnetize is too high, that's why i want to use the method of running an AC current and gradually reducing current to zero. Hammering is another method.
Perhaps if the energy required is the same regardless of what method is used, i could use E from the heat formula and assume E = ItV where I is the current, measured in amperes [A]. t is the time period, measured in seconds [s]. V is electric potential or voltage in volts and solve for I using a V and t of my choice? But none of this takes into account the magnetic field B of the iron plate before the de-magnification efforts begin. The magnetic field generated by the current has to be big enough to change the direction of the magnetic dipole. And the current has to be gradually reduced to zero over time which further complicates the calculation. If i measure the magnetic field B of the iron plate could i use it to find I from B = (uI)/(2*pi*r), since the current will drop to zero i will use I/2 in the overall energy formula mcT = E = (I/2)tV and solve for t using a V of my choice? |
| Jan21-13, 09:25 PM | #4 |
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Energy required to demagnetize Iron.
It depends fundamentally on the alloy and its history (cold work, temper...). Between steel magnetically hard and a very soft alloy, the energy varies by >1,000,000.
Then, this energy can be computed from B and H but it's a minimum value that won't reflect the true expense. The minimum duration of each half-pulse is determined only by eddy currents, hence by the detailed shape of the part and little by the material. If you use coils, current will dissipate power over this duration, much more than degaussing needs. If you want to save power, a better way is to use permanent magnets to degauss the part. Move them alternately over the part at increasing distance. |
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