Peak Magnetic Field Calculation

1. From Circuit Parameters to Peak Current

The first step is to solve the differential equation we derived previously to find an expression for the current, $I(t)$.

The General Solution

The equation $L\ddot{I}+R\dot{I}+\frac{1}{C}I=0$ is a standard second-order linear homogeneous ODE. As you know from your mathematics courses, we assume a solution of the form $I(t)=e^{\lambda t}$, which leads to the characteristic equation:
$L{\lambda }^2+R\lambda +\frac{1}{C}=0$
The roots are given by the quadratic formula:
$\lambda =-\frac{R}{2L}\pm \sqrt{{\left(\frac{R}{2L}\right)}^2-\frac{1}{LC}}$
For pulsed magnets, the resistance $R$ is kept as low as possible to maximize the current. This ensures the circuit is in the underdamped regime, where $(R/2L{)}^2<1/LC$. This means the term inside the square root is negative, and the solution will be oscillatory.

Let's define the damping factor $\alpha =\frac{R}{2L}$ and the natural resonant frequency ${\omega }_0=\frac{1}{\sqrt{LC}}$. The roots become $\lambda =-\alpha \pm i{\omega }_d$, where ${\omega }_d=\sqrt{{\omega }_0^2-{\alpha }^2}$ is the damped angular frequency.

The general solution for the current is then:
$I(t)=e^{-\alpha t}(A\cos ({\omega }_{d}t)+B\sin ({\omega }_{d}t))$

To find the constants $A$ and $B$, we apply the initial conditions at the moment the switch is closed ($t=0$):

1. The initial current is zero: $I(0)=0$.
2. The capacitor, charged to $V_0$, is initially connected across the inductor and resistor. Since $I(0)=0$, there is no voltage drop across the resistor, so the full capacitor voltage appears across the inductor: $V_L(0)=L\frac{dI}{dt}{|}_{t=0}=V_0$. Thus, $\frac{dI}{dt}{|}_{t=0}=\frac{V_0}{L}$.

Applying these conditions (a standard exercise you can verify) yields the specific solution:
$I(t)=\frac{V_0}{{\omega }_{d}L}{e}^{-\alpha t}\sin ({\omega }_{d}t)$
This equation describes the decaying sinusoidal current pulse that generates the magnetic field.

The Low-Loss Approximation for Peak Current

To find the peak current, $I_{peak}$, we would need to find the time $t_{peak}$ where $\frac{dI}{dt}=0$ and substitute it back into $I(t)$. This is algebraically complex. However, we can make a very effective approximation. In a well-designed pulsed magnet, the resistance $R$ is very small compared to the other circuit impedances ($R\ll \sqrt{L/C}$). This is the low-loss (or high-Q) limit.

In this limit:

• The damping $\alpha =R/2L$ is small.
• The damped frequency ${\omega }_d$ is very close to the natural frequency ${\omega }_0$.
• The exponential decay term $e^{-\alpha t}$ changes very little during the first quarter cycle as the current rises to its peak. The peak occurs at approximately $t_{peak}\approx \frac{\pi }{2{\omega }_0}$, and at this time $e^{-\alpha t_{peak}}\approx 1$.

With these approximations, the current equation simplifies to that of a pure LC circuit:
$I(t)\approx \frac{V_0}{{\omega }_{0}L}\sin ({\omega }_{0}t)$
The peak current is simply the amplitude of this sine wave:
$I_{peak}\approx \frac{V_0}{{\omega }_{0}L}=\frac{V_0}{L/\sqrt{LC}}=V_0\sqrt{\frac{C}{L}}$

This is a powerful result. We can make it even more intuitive by expressing it in terms of the initial stored energy in the capacitor, $E=\frac{1}{2}C{V}_0^2$.
From this, $V_0=\sqrt{\frac{2E}{C}}$. Substituting into our expression for $I_{peak}$:
$I_{peak}\approx \sqrt{\frac{2E}{C}}\sqrt{\frac{C}{L}}=\sqrt{\frac{2E}{L}}$

So, we arrive at the central relationship for estimating the peak current:
$I_{peak}\approx \sqrt{\frac{2E}{L}}$

This formula embodies a key principle of energy conservation: in an ideal, lossless circuit, all the electrical energy $E$ initially stored in the capacitor is converted into magnetic energy $E_L=\frac{1}{2}L{I}_{peak}^2$ at the moment of peak current. Equating these gives our result directly.

2. A Practical Calculation Method

While deriving the peak current from the differential equation is instructive, in a laboratory setting, one often works with directly measurable quantities. A fantastic resource for this practical approach is a paper from MIT's National Magnet Laboratory on pulsed magnetic metal forming. It outlines a simple method based on the pulse's half-period, $T$, which can be read from an oscilloscope.

The method described in the paper can be summarized as follows:

1. Determine the total charge stored: $Q=C{V}_0$.
2. Measure the pulse half-period, $T$. The current rises from zero to its peak in the first half of this half-period, i.e., the rise time is $t_{rise}=T/2$.
3. Calculate the average current during the rise: $I_{avg}=Q/t_{rise}=2Q/T$.
4. For a sinusoidal pulse, the peak value is $\pi /2$ times the average value over the first half-cycle. Thus, $I_{peak}=\frac{\pi }{2}{I}_{avg}=\frac{\pi Q}{T}$.

Since $Q=C{V}_0$ and the half-period $T\approx \pi \sqrt{LC}$ (for an underdamped oscillator), this formula becomes $I_{peak}\approx \frac{\pi C{V}_0}{\pi \sqrt{LC}}=V_0\sqrt{C/L}$, which is identical to the result we derived from the differential equation. This confirms both approaches are consistent.

3. From Peak Current to Peak Field

Once we know the peak current, calculating the peak magnetic field is straightforward. For a given coil geometry, the magnetic field produced is directly proportional to the current flowing through it.
$B=kI$
The proportionality constant, $k$, is often called the coil constant or transfer function, with units of Tesla per Ampere (T/A). For a simple, long solenoid with $n$ turns per unit length, $k={\mu }_{0}n$. For the complex geometries of real high-field magnets, $k$ must be calculated numerically (a topic for a later module), but it remains a constant for a fixed design.

Combining our results, the peak field is:
$B_{peak}=k{I}_{peak}\approx k\sqrt{\frac{2E}{L}}$
This is the main result of our lesson. It shows that for a given coil design (which fixes $k$ and $L$), the achievable peak field is determined by the energy stored in the capacitor bank. This is why high-field facilities invest in enormous, room-sized capacitor banks capable of storing MegaJoules of energy.

Worked Example

Let's apply this using the "Bank #5" example from the MIT paper (LINK):

• Given: $E=9\text{ kJ}$, $V=20\text{ kV}$, Half-period $T=6\mu s$. The coil used has a constant of $k\approx 70\text{ kG / 100 kA}=7\times 10^{-5}\text{ T/A}$.

Method 1: Using the Energy Formula

1. First, we need the inductance $L$. We can find it from the half-period. We need Capacitance first: $C=2E/V^2=2(9000)/(20000{)}^2=2.25\times 10^{-5}\text{ F}=22.5\mu \text{F}$.
   (Note: The paper calculates Q=0.9C from V=20kV and E=9kJ, implying C = 0.9/20k = 45uF. Let's use the paper's values to be consistent with its results.)
   Using the paper's implied capacitance of $45\mu \text{F}$:
   $T=\pi \sqrt{LC}⟹L=(T/\pi {)}^2/C=(6\mu s/\pi {)}^2/45\mu F\approx 81\text{ nH}$.
2. Now calculate $I_{peak}$: $I_{peak}\approx \sqrt{2E/L}=\sqrt{2(9000)/81\times 10^{-9}}\approx 471\text{ kA}$.
3. Calculate $B_{peak}$: $B_{peak}=k{I}_{peak}=(7\times 10^{-5})\times (471\times 10^3)\approx 33\text{ T}=330\text{ kG}$.

This matches the paper's result perfectly.

4. An Important Constraint: The Action Integral

Our formula $B_{peak}\approx k\sqrt{2E/L}$ seems to imply we can achieve any field by simply increasing the energy $E$. However, reality is constrained by material limits. Pushing too much current through the coil will cause it to overheat and fail.

A more advanced concept, which you will encounter in magnet design, is the action integral. It provides a thermal limit on the current pulse.

The key idea is that the total heating is determined by $\int j^2(t)dt$. For a given pulse shape (e.g., sine wave) and duration $\tau$, this integral is proportional to $j_{peak}^2\tau$. Since the allowable temperature rise of the conductor material is fixed, this sets a hard limit on the peak current density $j_{peak}$ for a given pulse duration.

This means that our ability to increase the field by increasing the capacitor energy $E$ is ultimately limited by the thermal and mechanical properties of the coil's conductor. We will explore these critical trade-offs in much more detail later in the course.

Conclusion

In this lesson, we made the crucial leap from a theoretical circuit model to a practical design equation. By solving the RLC circuit equation and applying a low-loss approximation, we derived a powerful formula for estimating the performance of a pulsed magnet.

Key Takeaways:

• The current in a pulsed magnet circuit is a decaying sinusoid given by $I(t)=\frac{V_0}{{\omega }_{d}L}{e}^{-\alpha t}\sin ({\omega }_{d}t)$.
• In the practical low-loss limit, the peak current can be estimated as $I_{peak}\approx \sqrt{2E/L}$, reflecting the conservation of energy between the capacitor and inductor.
• The peak magnetic field is directly proportional to the peak current: $B_{peak}=k{I}_{peak}$.
• The primary design equation for estimating peak field is therefore $B_{peak}\approx k\sqrt{2E/L}$.
• This peak field is not unlimited; it is constrained by material properties, particularly the thermal limit described by the action integral.

Next Lesson Preview:

We've now seen how $E$ and $L$ determine the peak field. But what determines the duration of the pulse? The pulse shape and length are critical for experiments. In the next lesson, we will investigate how to relate the desired pulse duration and shape to the system's inductance, capacitance, and resistance, diving deeper into the roles of $\alpha$ and ${\omega }_d$.