Problem 11.3.5 Solution | Mutual inductance. Inductance of conductors. Transformers

Statement

11.3.5. $$a.$$ What is the inductance of a solenoid of radius $$r$$ and length $l \gg r$ ? The number of turns per unit length of the solenoid is $$n$$ .

$$b^*$$ . Derive a formula for the inductance of the solenoid without neglecting the effect of the electron mass $$m_e$$ on the inductance. The cross-sectional area of the solenoid wire is $$S$$ , and the number of conduction electrons per unit volume of the conductor is $$n_e$$ . Can this effect on the inductance of coils used in radio engineering be neglected?

Solution

$$a.$$ The magnetic field in the solenoid is

$B=\mu_0nI$

The magnetic flux through it is

$\Phi=\mu_0nIN\pi r^2$

Then the inductance is

$L=\frac{d\Phi}{dI}=\frac{d(\mu_0nIN\pi r^2)}{dI}=\mu_0\pi r^2n^2l$

$$b^*$$ .(Savchenko's solution)

The equation of motion of an electron in the solenoid (from Newton's 2nd law) is

$e\left(E-\frac{LdI}{ldt}\right)=m_ea, l = 2\pi rN$

$I=\frac{dq}{dt}=en_eSv$

So the resulting equation is

$El = V = \left(E+\frac{m_el}{e^2n_eS}\right)\frac{dI}{dt}$

Then the total inductance is

$L_1 = L + \frac {m_el}{e^2n_eS}$

The second term is very small, which is easy to check by substituting numerical values, so it can be neglected and one can take $$L_1=L$$

Answer

$a. L = \mu_0\pi r^2n^2l$

$b^∗. L_1 = L + \frac {m_el}{e^2n_eS}. It can$

Formulas in this solution the whole sheet

m\vec a = \sum \vec F

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I = \frac{dq}{dt}, \qquad q = \int I\,dt

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тороид$B = \mu_0 n I, \qquad B = \frac{\mu_0 N I}{2\pi r}\ \text{(тороид)}$

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соленоид$L = \frac{\Phi}{I}, \qquad L = \mu_0 n^2 S l\ \text{(соленоид)}$

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L = \mu\mu_0 n^2 S l = \frac{\mu\mu_0 N^2 S}{l}, \qquad \frac{L}{l} = \frac{\mu\mu_0}{2\pi}\ln\frac{r_2}{r_1}

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