Como pode ser visto no arquivo em anexo, estou planejando a criação de um novo tipo de bateria que não usa energia química. Em vez disso, ela usa as propriedades de um indutor em um circuito sem resistência, para armazenar energia magnética.
Preciso de ajuda para discutir sobre a aplicabilidade dessa bateria.
Dúvidas:
Por não possuir resistência, o circuito onde a bateria carrega não dispersa energia de nenhuma maneira. Isso está correto, não é? Estamos desprezando alguma coisa?
Digamos que eu queira tirar o indutor carregado do circuito, a energia dele dispersaria se eu fizesse isso, ou continuaria sendo potencial?
Gustavo,
The energy density achievable with magnetic field storage is an order of magnitude (or two) lower than current state of the art batteries. For superconductors, you are limited by the critical field that the material can sustain.
Certainly, rapid discharges are possible - and this is the core notion behind homopolar generators, used to develop very high currents in, say, fusion machines. Or, in more extreme cases, EMP 'bomb'.
Thanks for bringing up these questions.
However, my question wasn't answered: what happens if I take the charged inductor from the circuit? Does it discharge?
Gustavo,
You will find that you cannot take an inductor out of circuit and keep the inductor 'charged' thereafter - the act of removing the inductor, and breaking the circuit means that the current in it ceases to flow. And the magnetic field will collapse - it is present only because there is a current. No current, no field.
You will want to stand back if you try this with a substantial current and a large solenoid. That stored energy will want to go somewhere.
This is unlike a capacitor, which is maintained by charge (not current). That can be taken out of circuit and the charges can be retained.
However, if you had a superconducting *ring*, you could induce currents into it externally. And then you could carry that ring (in a suitable environment!) around - and deploy the current at your (very cautious) leisure.
My apologies for a too-brief and insufficient answer the first time.
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