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A toroidal solenoid with mean radius \(r\) and cross-sectional area \(A\) is wound uniformly with \(N_1\) turns. A second toroidal solenoid with \(N_2\) turns is wound uniformly on top of the first, so that the two solenoids have the same cross-sectional area and mean radius. (a) What is the mutual inductance of the two solenoids? Assume that the magnetic field of the first solenoid is uniform across the cross section of the two solenoids. (b) If \(N_1 = 500\) turns, \(N_2 = 300\) turns, \(r = 10.0\) cm, and \(A = 0.800 \, \mathrm{cm}^2\), what is the value of the mutual inductance?

Short Answer

Expert verified
The mutual inductance is \(4.8 \times 10^{-4}\) H.

Step by step solution

01

Understanding Mutual Inductance

Mutual inductance, denoted as \( M \), is the measure of how much voltage is induced in one coil due to the change in current in another coil. For two solenoids, it is given by \( M = \frac{\mu_0 \cdot N_1 \cdot N_2 \cdot A}{l} \), where \( \mu_0 \) is the permeability of free space and \( l \) is the length of the solenoid.
02

Expressing Length in Terms of Geometry

For a toroidal solenoid, the length \( l \) is the circumference of the solenoid, which can be calculated using \( l = 2\pi r \) where \( r \) is the mean radius of the toroid.
03

Calculating the Mutual Inductance Formula

Substitute the expression for \( l = 2\pi r \) into the mutual inductance formula: \[ M = \frac{\mu_0 \cdot N_1 \cdot N_2 \cdot A}{2\pi r} \] This formula is used to calculate the mutual inductance based on the given parameters.
04

Substitute Given Values

Given that \( N_1 = 500 \), \( N_2 = 300 \), \( r = 10.0 \) cm = 0.1 m, and \( A = 0.800 \) cm\(^2\) = 0.00008 m\(^2\), and knowing that \( \mu_0 = 4\pi \times 10^{-7} \) H/m, substitute these into the formula:\[ M = \frac{4\pi \times 10^{-7} \cdot 500 \cdot 300 \cdot 0.00008}{2\pi \times 0.1} \]
05

Simplifying the Expression

Simplify the calculation:\[ M = \frac{4 \times 10^{-7} \cdot 500 \cdot 300 \cdot 0.00008}{0.1} \]Continue simplifying:\[ M = \frac{4 \times 10^{-7} \times 12000}{0.1} \] \[ M = 4.8 \times 10^{-4} \] The mutual inductance is \( 4.8 \times 10^{-4} \) H.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Toroidal Solenoid
A toroidal solenoid is a coil of wire shaped like a doughnut. Unlike a straight solenoid, which is cylindrical, the toroidal solenoid's core is circular. This distinctive shape is highly effective in confining the magnetic field within its core. This confinement is a result of its circular geometry, which ensures that the field lines are closed loops and do not extend outside the toroid.

When we have a toroidal solenoid, its length isn't measured in the way we would a straight solenoid. Instead, it's the circumference of the circular path, calculated using the formula for circumference, giving us: \[ l = 2\pi r \] where \( r \) is the 'mean radius' of the toroid. Understanding this property helps us when calculating the mutual inductance, as it prevents errors arising from incorrect length assumptions.
Magnetic Field
The magnetic field inside a toroidal solenoid is unique due to its geometry. This field is concentrated within the doughnut shape, offering a uniform and controlled area of magnetism. The value of the magnetic field inside a solenoid can be expressed by:\[ B = \frac{\mu_0 \cdot N \cdot I}{2\pi r} \]Here, \( B \) is the magnetic field, \( \mu_0 \) is the permeability of free space, \( N \) is the number of turns, and \( I \) is the current flowing through the coil. A critical feature of a toroidal solenoid is that the magnetic field strength depends directly on the number of wire turns and the current. The more coils it has, the stronger the magnetic field becomes, maintaining consistency across the solenoid's cross-section.
Permeability of Free Space
The permeability of free space, represented as \( \mu_0 \), is a fundamental physical constant vital for understanding magnetic phenomena. It represents how well a magnetic field can permeate a vacuum and is crucial for calculating inductance and fields in solenoids.

\( \mu_0 \) has a specific value of:\[ \mu_0 = 4\pi \times 10^{-7} \, \text{H/m} \] This constant appears in equations governing the behavior of inductors like solenoids, dictating the relationship between electric current and magnetic fields. In the context of toroidal solenoids, it helps us calculate how the change in current through one solenoid affects another, thereby determining mutual inductance.
Cross-Sectional Area
In the context of toroidal solenoids, the cross-sectional area \( A \) is the amount of space enclosed by the rings of wire in the toroid. This measurement plays a crucial role in determining the strength of the magnetic field within the solenoid and, thus, its inductance.

The formula for mutual inductance includes this area as:\[ M = \frac{\mu_0 \cdot N_1 \cdot N_2 \cdot A}{2\pi r} \] where \( M \) is the mutual inductance. Larger cross-sectional areas allow more field lines to pass through, enhancing the magnetic effect. In practical terms, doubling the cross-sectional area would double the amount of magnetic field produced, provided all other variables remain constant.

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Most popular questions from this chapter

A 10.0-cm-long solenoid of diameter 0.400 cm is wound uniformly with 800 turns. A second coil with 50 turns is wound around the solenoid at its center. What is the mutual inductance of the combination of the two coils?

An inductor is connected to the terminals of a battery that has an emf of 16.0 V and negligible internal resistance. The current is 4.86 mA at 0.940 ms after the connection is completed. After a long time, the current is 6.45 mA. What are (a) the resistance \(R\) of the inductor and (b) the inductance \(L\) of the inductor?

When the current in a toroidal solenoid is changing at a rate of 0.0260 A/s, the magnitude of the induced emf is 12.6 mV. When the current equals 1.40 A, the average flux through each turn of the solenoid is 0.00285 Wb. How many turns does the solenoid have?

It is proposed to store \(1.00 \, \mathrm{kW} \cdot \mathrm{h} = 3.60 \times 10^6\) J of electrical energy in a uniform magnetic field with magnitude 0.600 T. (a) What volume (in vacuum) must the magnetic field occupy to store this amount of energy? (b) If instead this amount of energy is to be stored in a volume (in vacuum) equivalent to a cube 40.0 cm on a side, what magnetic field is required?

(a) A long, straight solenoid has \(N\) turns, uniform cross sectional area \(A\), and length \(l\). Show that the inductance of this solenoid is given by the equation \(L = \mu_0 AN^2/l\). Assume that the magnetic field is uniform inside the solenoid and zero outside. (Your answer is approximate because \(B\) is actually smaller at the ends than at the center. For this reason, your answer is actually an upper limit on the inductance.) (b) A metallic laboratory spring is typically 5.00 cm long and 0.150 cm in diameter and has 50 coils. If you connect such a spring in an electric circuit, how much self-inductance must you include for it if you model it as an ideal solenoid?

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