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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?

Short Answer

Expert verified
The mutual inductance of the two coils is approximately 0.493 \(\mu H\).

Step by step solution

01

Understand the Concept

Mutual inductance describes how the magnetic field created by one coil induces an electromotive force (EMF) in a secondary coil. It's important to recognize the mutual inductance formula connects properties of both coils involved.
02

Collect Known Data

The length of the solenoid is 10.0 cm or 0.10 m. Its diameter is 0.400 cm, so the radius is 0.0020 m. The number of turns in the solenoid is 800, and the second coil has 50 turns.
03

Use Solenoid Inductance Formula

The magnetic field inside an ideal solenoid is given by: \( B = \mu_0 \frac{n}{L} I \), where \( n \) is the number of turns per meter. Since \( n = \frac{800}{0.10} \), substitute into the equation.
04

Calculate the Magnetic Flux (Φ)

The magnetic flux through the second coil is: \( \Phi = B \cdot A \cdot N \), where \( A = \pi r^2 \) is the cross-sectional area of the solenoid and \( N \) is the number of turns in coil 2.
05

Calculate Mutual Inductance (M)

From the flux, we use the mutual inductance relationship: \( M = \frac{N_2 \Phi}{I} \), Where \( N_2 \) is the number of turns in the second coil.Substitute the expressions for \( \Phi \) and solve for \( M \).

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Key Concepts

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

Solenoid
A solenoid is a long coil of wire, tightly wound in a helical shape. It is often used to create a uniform magnetic field in a volume of space. When an electric current passes through a solenoid, a magnetic field is generated along its length. This magnetic field is linear, much like a bar magnet, with distinct north and south poles.

Solenoids are often utilized in experiments and technologies requiring uniform magnetic fields. The strength of the magnetic field inside a solenoid depends on several factors:
  • The number of turns of wire in the solenoid.
  • The amount of electric current passing through the wire.
  • The physical dimensions (length, diameter) of the solenoid.
When you wrap a second coil around a solenoid, as in the exercise, it can interact with the magnetic field of the solenoid to exhibit mutual inductance.
Magnetic Flux
Magnetic flux, represented by the symbol \(\Phi\), refers to the total magnetic field passing through a given area. It is a measure of the quantity of magnetism, accounting for the strength and the extent of a magnetic field.

To calculate magnetic flux, you can use the formula:\[ \Phi = B \cdot A \cdot \cos(\theta) \] where:
  • \(B\) is the magnetic field strength
  • \(A\) is the area the field lines pass through
  • \(\theta\) is the angle between the magnetic field and the normal (perpendicular) to the surface
For a solenoid, the magnetic flux through a second coil is affected by both the area enclosed by the coil and the strength of the magnetic field within the solenoid.
Electromotive Force (EMF)
Electromotive Force (EMF), although misleadingly named, is not actually a force but rather a potential difference. It's the energy provided by a battery or magnetic field to move a charge through an electric circuit. EMF is what pushes electrical current around a circuit, similar to how pressure pushes water through a hose.

In the context of mutual inductance and our exercise, EMF can be induced in a coil when there is a change in the magnetic flux through the coil. This induced EMF is given by Faraday's law of induction:
  • The induced EMF is proportional to the rate of change of flux linkage.
  • The direction of the induced EMF opposes the change in flux (Lenz's Law).
This principle is crucial for the functioning of transformers and inductors in circuits.
Coil Turns
A coil turn refers to a 360-degree wrap of wire around a core or cylinder. The number of turns in a coil is an important factor in determining the inductance and the strength of the generated magnetic field.

The more turns or loops a coil has, the greater its ability to induce an EMF:
  • In solenoids, the number of turns affects the intensity of the magnetic field created inside.
  • In our exercise, the solenoid has 800 turns, which contributes to generating a strong magnetic field.
  • The secondary coil has 50 turns and, when exposed to this field, can have an EMF induced in it.
This interplay between coil turns and magnetic fields is foundational in the concept of mutual inductance.
Magnetic Field
A magnetic field represents the influence a magnetic object exerts around itself. It can affect other magnetic materials and generate forces on moving charges, creating electric currents.

In solenoids, the magnetic field is concentrated and directed along the axis of the coil. This field is crucial in numerous applications like electromagnets, inductors, and magnetic resonance imaging (MRI).

For calculating the mutual inductance in the original exercise, understanding how to compute the magnetic field inside a solenoid is key. The formula for the magnetic field inside an ideal solenoid is:\[ B = \mu_0 \frac{n}{L} I \]where:
  • \(\mu_0\) is the permeability of free space.
  • \(n\) is the number of turns per unit length.
  • \(I\) is the current flowing through the solenoid.
Comprehending this helps determine how the solenoid's field induces an EMF in an adjacent coil.

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Most popular questions from this chapter

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?

(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?

A small solid conductor with radius \(a\) is supported by insulating, nonmagnetic disks on the axis of a thin-walled tube with inner radius \(b\). The inner and outer conductors carry equal currents \(i\) in opposite directions. (a) Use Ampere's law to find the magnetic field at any point in the volume between the conductors. (b) Write the expression for the flux \(d\Phi_B\) through a narrow strip of length \(l\) parallel to the axis, of width \(dr\), at a distance \(r\) from the axis of the cable and lying in a plane containing the axis. (c) Integrate your expression from part (b) over the volume between the two conductors to find the total flux produced by a current \(i\) in the central conductor. (d) Show that the inductance of a length \(l\) of the cable is $$L = l \frac{\mu_0}{2\pi} \mathrm{ln}(\frac{b}{a})$$ (e) Use Eq. (30.9) to calculate the energy stored in the magnetic field for a length l of the cable.

An air-filled toroidal solenoid has 300 turns of wire, a mean radius of 12.0 cm, and a cross-sectional area of 4.00 cm\(^2\). If the current is 5.00 A, calculate: (a) the magnetic field in the solenoid; (b) the self inductance of the solenoid; (c) the energy stored in the magnetic field; (d) the energy density in the magnetic field. (e) Check your answer for part (d) by dividing your answer to part (c) by the volume of the solenoid.

An \(L\)-\(R\)-\(C\) series circuit has \(L = 0.450\) H, \(C = 2.50 \times 10^{-5} \, \mathrm{F}\), and resistance \(R\). (a) What is the angular frequency of the circuit when \(R = 0\)? (b) What value must \(R\) have to give a 5.0\(\%\) decrease in angular frequency compared to the value calculated in part (a)?

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