Question

Inductance \(\quad\) The inductance \(L\) (in microhenrys) of a straight nonmagnetic wire in free space is \(L=0.00021\left(\ln \frac{2 h}{r}-0.75\right)\) where \(h\) is the length of the wire in millimeters and \(r\) is the radius of a circular cross section. Approximate \(L\) when \(r=2 \pm \frac{1}{16}\) millimeters and \(h=100 \pm \frac{1}{100}\) millimeters.

Short answer

The approximate change in inductance \(\Delta L\) when \(r=2 \pm \frac{1}{16}\) millimeters and \(h=100 \pm \frac{1}{100}\) millimeters is \(-0.02817\) microhenrys.

Step-by-step solution

Identify the Variables

First, identify the given variables. Here the length of the wire \(h = 100 \pm 0.01\) mm and the radius \(r = 2 \pm \frac{1}{16}\) mm.

Calculate the Differentials

Next, let's calculate the differentials \(dh\) and \(dr\). For \(h\), \(dh = 0.01\) mm and for \(r\), \(dr = \frac{1}{16}\) mm.

Apply the Differential

Now, differentiate the formula for \(L\) with respect to both \(h\) and \(r\) and evaluate them. This can be done using the Chain rule of differentiation. Here is how it will be for \( \frac{dL}{dr} \): \( \frac{dL}{dr} = 0 - \frac{0.00021}{2r} \). Plugging in \( r = 2 \), we will get \( \frac{dL}{dr} = -0.0000525 \) microhenrys per millimeter. Similarly for \( \frac{dL}{dh} \): \( \frac{dL}{dh} = \frac{0.00021}{h} \). Plugging in \( h = 100 \), we will get \( \frac{dL}{dh} = 0.0000021 \) microhenrys per millimeter.

Linear Approximation Expression

Express the linear approximation expression as \( \Delta L \approx \frac{dL}{dh} \cdot \Delta h + \frac{dL}{dr} \cdot \Delta r \). Substituting the differentials and deltas calculated before, we will get \( \Delta L \approx \frac{0.0000021}{0.01} - \frac{0.0000525}{\frac{1}{16}} \approx -0.02817 \) microhenrys.