Update Content - 2024-12-17

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2024-12-17 16:38:55 +01:00
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@@ -182,7 +182,7 @@ And we see that \\(H\\) has units of amperes per meter.
The magnetic field of a solenoid is shown in [Figure 7](#figure--fig:morrison16-solenoid).
The field intensity inside the solenoid is nearly constant, while outside its intensity falls of rapidly.
Using Ampere's law <eq:ampere_law>:
Using Ampere's law \eqref{eq:ampere\_law:}
\begin{equation}
\oint H dl \approx n I l
@@ -244,12 +244,12 @@ V = n^2 A k \mu\_0 \frac{dI}{dt} = L \frac{dI}{dt}
where \\(k\\) relates to the geometry of the coil.
Equation <eq:inductance_coil> states that if \\(V\\) is one volt, then for an inductance of one henry, the current will rise at the rate of one ampere per second.
Equation \eqref{eq:inductance\_coil} states that if \\(V\\) is one volt, then for an inductance of one henry, the current will rise at the rate of one ampere per second.
### The energy stored in an inductance {#the-energy-stored-in-an-inductance}
One way to calculate the work stored in a magnetic field is to use Eq. <eq:inductance_coil>.
One way to calculate the work stored in a magnetic field is to use Eq. \eqref{eq:inductance\_coil}.
The voltage \\(V\\) applied to a coil results in a linearly increasing current.
At any time \\(t\\), the power \\(P\\) supplied is equal to \\(VI\\).
Power is the rate of change of energy or \\(P = d\bm{E}/dt\\) where \\(\bm{E}\\) is the stored energy in the inductance.