A long coaxial cable carries current I (the current flows down the surface of the inner cylinder, radius a, and back along the outer cylinder, radius b) as shown in Fig. 7.40. Find the magnetic energy stored in a section of length l.

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Accepted Answer

According to Ampère’s law, the field between the cylinders is

[latex]\pmb{B}=\frac{\mu _{0}I}{2\pi s}\hat{\phi }. [/latex]

Elsewhere, the field is zero. Thus, the energy per unit volume is

[latex]\frac{1}{2\mu _{0}}\left(\frac{\mu _{0}I}{2\pi s} \right)^2=\frac{\mu _{0}I^{2}}{8\pi^{2}s^{2}}. [/latex]

The energy in a cylindrical shell of length l, radius s, and thickness ds, then, is

[latex]\left(\frac{\mu _{0}I^{2}}{8\pi^{2}s^{2}}\right)=2\pi lsds=\frac{\mu _{0}I^{2}l}{4\pi}\left(\frac{ds}{s} \right) . [/latex]

Integrating from a to b, we have:

[latex]W=\frac{\mu _{0}I^{2}l}{4\pi}\ln\left(\frac{b}{a} \right) . [/latex]

By the way, this suggests a very simple way to calculate the self-inductance of the cable. According to Eq. 7.30, the energy can also be written as [latex]\frac{1}{2} LI^{2}[/latex]. Comparing the two expressions,[latex]^{19}[/latex]

[latex]W=\frac{1}{2} LI^{2}.[/latex] (7.30)

[latex]L=\frac{\mu _{0}l}{2\pi}\ln\left(\frac{b}{a} \right) . [/latex]

This method of calculating self-inductance is especially useful when the current is not confined to a single path, but spreads over some surface or volume, so that different parts of the current enclose different amounts of flux. In such cases, it can be very tricky to get the inductance directly from Eq. 7.26, and it is best to let Eq. 7.30 define L.

[latex]\phi = L I. [/latex] (7.26)

[latex]^{19}[/latex] Notice the similarity to Eq. 7.28—in a sense, the rectangular toroid is a short coaxial cable, turned on its side

[latex]L=\frac{\mu _0N^2h}{2\pi}ln\left(\frac{b}{a} \right) [/latex] (7.28)

Question 7.13: A long coaxial cable carries current I (the current flows do...

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