Reduced to base SI units, one henry is the equivalent of one kilogram meter squared per second squared per ampere squared (kg m2 s-2 A-2).

This does not look like the standard F = ma formula for a force but there is a mass term. What does the mass term represent? I am trying to understand how the units gel together.

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    \$\begingroup\$ Does it help to think of it as \$\frac{J}{A^2}\$? \$\endgroup\$ – Hearth Dec 10 '18 at 22:42
  • \$\begingroup\$ you mean joule over ampere squared? \$\endgroup\$ – Sedumjoy Dec 10 '18 at 22:54
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    \$\begingroup\$ You could say that \$1\:J =1\: V \:A \:s\$, but it's also \$1\: kg\: m^2 \:s^{-2}\$. Just use the appropriate units for the task at hand. \$\endgroup\$ – Chu Dec 11 '18 at 2:18
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    \$\begingroup\$ Magnetic charge (the Weber) is a measure of angular momentum (Joule-s) per Coulomb; and magnetic charge per amp is the Henry. Magnetic moment and angular momentum are related. Revolving charged particles have both mass and charge and magnetic moment and angular momentum both increase with the rate. (The ratio of these is the gyromagnetic ratio. Note there is also a difference in the angular momentum of electron spin and the angular momentum of the electron in an atomic orbit, so to speak.) \$\endgroup\$ – jonk Dec 11 '18 at 2:33
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    \$\begingroup\$ Inductance is volts x time divided by current. Do you know the SI unit for volts? \$\endgroup\$ – Andy aka Dec 11 '18 at 10:31

Let \$P\$ be power in watts, \$I\$ be current in amps, \$W\$ be work in Joules,

\$A\$ Acceleration in meters per \$\text{second}^2\$ \$D\$ distance in meters, \$M\$ Mass in kg.

\$T\$ Time in seconds, \$F\$ Force in newtons and \$V\$ voltage in volts.

We know \$ P = V \cdot I\$ so \$V = \dfrac{P}{I}\$.

Basic physics should tell you Power is Work divided by time \$P = \dfrac{W}{T}\$.

Work is Force times distance \$W = F \cdot D\$

Force is mass times Acceleration \$F = M \cdot A\$.

Putting all this together we see.

\$ V = \dfrac{P}{I} = \dfrac{W}{I \cdot T} = \dfrac{F \cdot D}{I \cdot T} = \dfrac{M \cdot A \cdot D}{I \cdot T} = \dfrac{M \cdot D \cdot D}{I \cdot T \cdot T^2} = \dfrac{M \cdot D^2}{I \cdot T^3}\$

Using standard SI units the volt is therefore \$\dfrac{\mathrm{kg} \cdot \mathrm{m}^2}{\mathrm{A} \cdot \mathrm{s}^3}\$

Now we know \$ V = L \cdot \dfrac{\text{d}I}{\text{d}t}\$ Now dimensionally \$ L = \dfrac{\text{volts}}{\text{amps}} \cdot \text{time}\$

Using standard SI units the henry is therefore \$\dfrac{\mathrm{kg} \cdot \mathrm{m}^2}{\mathrm{A} \cdot \mathrm{s}^3} \cdot \dfrac{\text{s}}{\text{A}} = \dfrac{\mathrm{kg} \cdot \mathrm{m}^2}{\mathrm{A}^2 \cdot \mathrm{s}^2}\$


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