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| ### Statement |
| ### Statement |
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| $11.3.11.$ [Insert the problem statement] |
| $11.3.11$ |
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| Find the inductance per unit length of a two-wire line. The line consists of |
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| two parallel straight wires of radius r, the distance between the centerlines |
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| of which are h ≫ r. Through the wires flow equal in modulus, but oppositely |
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| directed currents. There is no magnetic field inside the wires. |
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| ### Solution |
| ### Solution |
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| Magnetic field between the two wires |
| Magnetic field between the two wires |
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| Each straight wire produces an azimuthal magnetic field on the outside. | | Each straight wire produces an azimuthal magnetic field on the outside. |
| For a wire carrying current I, the field at a distance$ \rho $from its center is: | | For a wire carrying current I, the field at a distance$ \rho $from its center is: |
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| $B = \frac{\mu_0 I}{2\pi\rho} \quad (\rho > r)$ | | $B = \frac{\mu_0 I}{2\pi\rho} \quad (\rho > r)$ |
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| Between the two wires (separation h), the fields add constructively because the opposite currents generate fields in the same direction in that region. If we place wire $1 at x = 0$ and wire 2 at$ x = h$, the total field at a point $x \in (r, h-r) $is: | | Between the two wires (separation h), the fields add constructively because the opposite currents generate fields in the same direction in that region. If we place wire $1 at x = 0$ and wire 2 at$ x = h$, the total field at a point $x \in (r, h-r) $is: |
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| $B(x) = \frac{\mu_0 I}{2\pi x} + \frac{\mu_0 I}{2\pi (h-x)}$ | | $B(x) = \frac{\mu_0 I}{2\pi x} + \frac{\mu_0 I}{2\pi (h-x)}$ |
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| Magnetic flux per unit length | | Magnetic flux per unit length |
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| The flux through the rectangle of width $h-2r$ and unit height is: | | The flux through the rectangle of width $h-2r$ and unit height is: |
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| $\Phi' = \int_{r}^{h-r} B(x)\,dx | | $\Phi' = \int_{r}^{h-r} B(x)\,dx |
| = \frac{\mu_0 I}{2\pi} \int_{r}^{h-r} \left( \frac{1}{x} + \frac{1}{h-x} \right) dx$ | | = \frac{\mu_0 I}{2\pi} \int_{r}^{h-r} \left( \frac{1}{x} + \frac{1}{h-x} \right) dx$ |
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| Each integral gives $\ln\frac{h-r}{r}$ Adding them: | | Each integral gives $\ln\frac{h-r}{r}$ Adding them: |
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| $\Phi' = \frac{\mu_0 I}{2\pi} \cdot 2\ln\frac{h-r}{r} | | $\Phi' = \frac{\mu_0 I}{2\pi} \cdot 2\ln\frac{h-r}{r} |
| = \frac{\mu_0 I}{\pi} \ln\frac{h-r}{r}$ | | = \frac{\mu_0 I}{\pi} \ln\frac{h-r}{r}$ |
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| Since $h \gg r fwe can approximate $h-r \approx h$ | | Since $h \gg r fwe can approximate $h-r \approx h$ |
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| $\Phi' \approx \frac{\mu_0 I}{\pi} \ln\frac{h}{r}$ | | $\Phi' \approx \frac{\mu_0 I}{\pi} \ln\frac{h}{r}$ |
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| Inductance per unit length | | Inductance per unit length |
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| By definition$ L' = \Phi'/I$ | | By definition$ L' = \Phi'/I$ |
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| $\boxed{L' = \frac{\mu_0}{\pi} \ln\frac{h}{r}}$ | | $\boxed{L' = \frac{\mu_0}{\pi} \ln\frac{h}{r}}$ |
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