/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 In this problem, you derive the ... [FREE SOLUTION] | 91影视

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Chapter 20: Problem 49

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.

Short Answer

Expert verified
Question: Calculate the self-inductance of a long solenoid with n turns per unit length, a length l, and a radius r. Answer: The self-inductance (L) of a long solenoid can be calculated using the formula $$L= \mu_0 n^2 \ell \pi r^2$$, where 碌_0 is the permeability of space, n is the number of turns per unit length, l is the length of the solenoid, and r is the radius 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 (渭_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 (桅) 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 (桅_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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Most popular questions from this chapter

The armature of an ac generator is a rectangular coil \(2.0 \mathrm{cm}\) by \(6.0 \mathrm{cm}\) with 80 turns. It is immersed in a uniform magnetic field of magnitude 0.45 T. If the amplitude of the emf in the coil is $17.0 \mathrm{V},$ at what angular speed is the armature rotating?
In Section \(20.9,\) in order to find the energy stored in an inductor, we assumed that the current was increased from zero at a constant rate. In this problem, you will prove that the energy stored in an inductor is \(U_{\mathrm{L}}=\frac{1}{2} L I^{2}-\) that is, it only depends on the current \(I\) and not on the previous time dependence of the current. (a) If the current in the inductor increases from \(i\) to \(i+\Delta i\) in a very short time \(\Delta t,\) show that the energy added to the inductor is $$\Delta U=L i \Delta i$$ [Hint: Start with \(\Delta U=P \Delta t .]\) (b) Show that, on a graph of \(L i\) versus \(i,\) for any small current interval \(\Delta i,\) the energy added to the inductor can be interpreted as the area under the graph for that interval.(c) Now show that the energy stored in the inductor when a current \(I\) flows is \(U=\frac{1}{2} L I^{2}\)
A step-down transformer has 4000 turns on the primary and 200 turns on the secondary. If the primary voltage amplitude is \(2.2 \mathrm{kV},\) what is the secondary voltage amplitude?
A transformer with 1800 turns on the primary and 300 turns on the secondary is used in an electric slot car racing set to reduce the input voltage amplitude of \(170 \mathrm{V}\) from the wall output. The current in the secondary coil is of amplitude \(3.2 \mathrm{A}\). What is the voltage amplitude across the secondary coil and the current amplitude in the primary coil?
A de motor has coils with a resistance of \(16 \Omega\) and is connected to an emf of \(120.0 \mathrm{V}\). When the motor operates at full speed, the back emf is \(72 \mathrm{V}\). (a) What is the current in the motor when it first starts up? (b) What is the current when the motor is at full speed? (c) If the current is \(4.0 \mathrm{A}\) with the motor operating at less than full speed, what is the back emf at that time?
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