In this problem, you derive the expression for the self inductance of a long solenoid [Eq. \((20-15 a)] .\) The solenoid has \(n\) turns per unit length, length \(\ell,\) and radius \(r\) Assume that the current flowing in the solenoid is \(I\) (a) Write an expression for the magnetic field inside the solenoid in terms of \(n, \ell, r, I,\) and universal constants. (b) Assume that all of the field lines cut through each turn of the solenoid. In other words, assume the field is uniform right out to the ends of the solenoid-a good approximation if the solenoid is tightly wound and sufficiently long. Write an expression for the magnetic flux through one turn. (c) What is the total flux linkage through all turns of the solenoid? (d) Use the definition of self-inductance [Eq. \((20-14)]\) to find the self inductance of the solenoid.

Step by step solution

01

Find the Magnetic Field Inside the Solenoid

Using Ampere's Law, we can find the magnetic field (B) inside the solenoid. Ampere's Law states that the integral of the magnetic field (B) around a closed loop is equal to the current enclosed in this loop multiplied by the permeability of space (μ_0): $$\oint B \cdot dl = \mu_0 I_{enc}$$ As the solenoid is tightly wound and has n turns per unit length, the magnetic field is approximately constant inside and the enclosed current is: $$I_{enc} = n \cdot I \cdot \ell$$ For a solenoid, the magnetic field (B) can be expressed as: $$B = \mu_0 n I$$

02

Magnetic Flux Through One Turn

Magnetic flux (Φ) is defined as the product of the magnetic field and the area it penetrates, or: $$\Phi = B \cdot A$$ Since we know the magnetic field (B) inside the solenoid and the area of a turn can be computed (assuming a circular cross-section) as: $$A = \pi r^2$$ We can write the magnetic flux through one turn as: $$\Phi = \mu_0 n I \pi r^2$$

03

Total Flux Linkage Through All Turns

Now we need to find the total flux linkage (Φ_total) through all the turns of the solenoid. By definition, this is the sum of the magnetic flux through each turn, which can also be expressed as the product of the magnetic flux through one turn and the number of turns: $$\Phi_{total} = n \ell \Phi = n \ell \mu_0 n I \pi r^2$$

04

Calculate the Self-Inductance of Solenoid

Finally, we use the definition of self inductance (L) of a coil, given by: $$L = \frac{\Phi_{total}}{I}$$ Plugging in the expression for the total magnetic flux, we get: $$L = \frac{n \ell \mu_0 n I \pi r^2}{I}$$ Canceling the current (I) from the expression, we are left with the expression for the self-inductance of a solenoid: $$L = \mu_0 n^2 \ell \pi r^2$$

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