# Energy vs. Time in a Magnetic Field due to a Current Carrying Wire

Gold Member
I want to know the total energy contained in a magnetic field due to a long wire (just consider a 1m segment) as a 1amp current is turned on starting at time zero. I'm assuming zero turn-on time for convenience. At t=0 the cylindrical field is formed and I wish to know the total energy as a function of time, (or equivalent distance ct) where t is in nanoseconds. Is it just the volume integral of the field squared times mu? Thanks.

Is it just the volume integral of the field squared times mu?
Yes.

Of course, it won't turn on in zero time, that what inductance is all about. With your assumption of zero turn on time, there will just be two values, zero before it is turned on, and the final value after it is turned on.

• bob012345
Mentor
Also, if the wire is infinitely long then the energy will be infinite.

• bob012345
Gold Member
Thanks for the answers! I'm figuring a turn on time on the order of 100ps and I ignored it for convenience in the OP. I found the self inductance of a straight wire to be about 50nH independent of gauge. Interestingly, one has to add resistance to the circuit for faster switching as the natural resistance per meter would make the L/R time constant too big.

Basically, in cylindrical coordinates, the total field energy grows as the natural log of the distance r away from the wire per unit length of the wire. If the wire were turned on then off say after 1ns, we would see a pulse propagating radially, in cylindrical coordinates, for a long wire, at the speed of light with the pulse shape following the ever diminishing field strength going as 1/r.

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Also, if the wire is infinitely long then the energy will be infinite.
True, but I don't have that kind of money. :)

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Ok, since the field expands at the speed of light, and the source needs to be on else the field goes away, do we say the field propagates or is "static"? In other words, is energy continuously being invested to maintain the field or not? Or is the energy necessary to maintain the source an independent consideration?

Thanks.

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Gold Member
The stored magnetic energy for a steady current should be straightforward to integrate. A time-dependent case is another matter altogether. You have pointed out that the current front propagates down the wire. Realize also that part of the energy radiates away. You'll need a treatment with full Maxwell equations, including retarded potentials. You should use a realistic rise time (e.g., include both inductance and resistance per unit length). Note that the problem simplifies if your wire is part of a parallel wire transmission line, since the behavior of transmission lines is well understood.

• bob012345
Gold Member
The stored magnetic energy for a steady current should be straightforward to integrate. A time-dependent case is another matter altogether. You have pointed out that the current front propagates down the wire. Realize also that part of the energy radiates away. You'll need a treatment with full Maxwell equations, including retarded potentials. You should use a realistic rise time (e.g., include both inductance and resistance per unit length). Note that the problem simplifies if your wire is part of a parallel wire transmission line, since the behavior of transmission lines is well understood.

Thanks. I'll leave the time dependent situation for a separate question. I'm still seeking clarification of my last post regarding the physical understanding of the magnetic field around the wire in steady state with regards to the energy. I understand the calculation regarding how much energy is in the field as function of volume in space but I don't understand the physics such as does it take a steady infusion of energy to maintain the field or not? If so, what is the power? Is there a minimum power required to maintain the field around a 1 amp current. Questions like that.

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The power supplied is just that required to replace ohmic losses. A steady DC current flowing in a superconducting wire would dissipate no power.

• bob012345 and Dale
Gold Member
The power supplied is just that required to replace ohmic losses. A steady DC current flowing in a superconducting wire would dissipate no power.

Thanks. I would have thought that since signals can't go faster than light it takes a long time to fill all space so energy has to come from somewhere to do that even if a very small amount.

Mentor
Sure, but that isn’t “maintaining” a field, it is expanding it. It takes no energy to maintain the field.

• bob012345
Gold Member
Sure, but that isn’t “maintaining” a field, it is expanding it. It takes no energy to maintain the field.
Thanks. I see what you're saying. It's easier to visualize that inside a solid held together by electric forces than visualizing a field not needing "replenishing" around a wire (for me) but I get it. And if another wire is around, Lorentz forces happen yet no energy is used unless those forces are allowed to do work. Then the currents must supply that energy. Thanks again.

• Dale