# Fundamental Physics/Electromagnetic heat

## Electromagnetic heat

It is found that electromagnetic conductor generates electromagnetic heat when connect with electric current

## Conductor and electromagnetic heat energy

### With a straight wire conductor

Electromagnetic heat is in the form of Electromagnetic heat energy, ${\displaystyle E_{R}}$

${\displaystyle B_{R}=Li={\frac {\mu }{2\pi r}}i}$
${\displaystyle R(T)=R_{o}+nT}$
${\displaystyle R(T)=R_{o}e^{nT}}$
${\displaystyle E_{R}=i^{2}R(T)=mC\Delta T=pv}$
${\displaystyle m={\frac {E_{R}}{C\Delta T}}}$
${\displaystyle v={\frac {mC\Delta T}{p}}}$

In general,

${\displaystyle pv=mC\Delta T}$

### With a magnetic coil conductor

Electromagnetic heat is in the form of Radiant photon heat energy, ${\displaystyle E_{o}}$

${\displaystyle B_{o}=Li=\mu _{o}{\frac {Ni}{l}}}$
${\displaystyle E_{o}=hf_{o}}$
${\displaystyle f_{o}={\frac {\omega _{o}}{\lambda _{o}}}={\frac {C}{\lambda _{o}}}}$
${\displaystyle h=p\lambda _{o}={\frac {pC}{f_{o}}}}$

In general,

${\displaystyle pC=hf_{o}}$

### Electric photon heat energy

Electromagnetic heat is in the form of Electric photon heat energy , ${\displaystyle E}$

${\displaystyle B=Li=\mu {\frac {Ni}{l}}}$
${\displaystyle E=hf}$
${\displaystyle f={\frac {\omega }{\lambda }}={\frac {C}{\lambda }}}$
${\displaystyle h=p\lambda ={\frac {pC}{f}}}$

In general,

${\displaystyle pC=hf}$

With

${\displaystyle f>f_{o}}$
${\displaystyle f_{o}={\frac {C}{\lambda _{o}}}={\frac {300\times 10^{6}}{400-700nm}}}$