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Chapter 30: Problem 43

One solenoid is centered inside another. The outer one has a length of 50.0 \(\mathrm{cm}\) and contains 6750 coils, while the coaxial inner solenoid is 3.0 \(\mathrm{cm}\) long and 0.120 \(\mathrm{cm}\) in diameter and contains 15 coils. The current in the outer solenoid is changing at 49.2 A/s. (a) What is the mutual inductance of these solenoids? (b) Find the emf induced in the innner solenoid.

Short Answer

Expert verified
(a) 0.00369 H, (b) -0.1815 V

Step by step solution

01

Understand the Concept of Mutual Inductance

Mutual inductance ( M ) between two coils relates the emf induced in one coil due to the change of current in the other. It depends on the number of turns, the geometry, and magnetic permeability.
02

Calculate the Magnetic Field of Outer Solenoid

The magnetic field inside a solenoid is given by the formula: \( B = \mu_0 \cdot n \cdot I \), where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\) is the permeability of free space, \( n \) is the number of turns per unit length and \( I \) is the current. For the outer solenoid:\[ n = \frac{6750}{0.5} = 13500 \, \text{turns/m} \]Hence, \[ B = \mu_0 \cdot 13500 \cdot I \].
03

Calculate Flux Through Inner Solenoid

The magnetic flux \( \Phi \) through the inner solenoid due to the outer solenoid's field is: \[ \Phi = B \cdot A \cdot N_{\text{inner}} \], where\( A \) is the cross-sectional area of the inner solenoid and \( N_{\text{inner}} = 15 \) is its number of turns. With radius \( r = 0.0012 \text{ m} \), we have: \[ A = \pi r^2 \approx 4.52 \times 10^{-6} \text{ m}^2 \], so \[ \Phi = (\mu_0 \cdot 13500 \cdot I) \cdot 4.52 \times 10^{-6} \cdot 15 \].
04

Derive the Formula for Mutual Inductance

The mutual inductance \( M \) is given by: \[ M = \frac{\Phi}{I} \]. Using the flux expression from Step 3: \[ M = \mu_0 \cdot 13500 \cdot 4.52 \times 10^{-6} \cdot 15 \approx 0.00369 \, \text{H} \].
05

Find Emf Induced in Inner Solenoid

The induced emf \( \varepsilon \) due to changing current is given by:\( \varepsilon = -M \frac{dI}{dt} \). Using the mutual inductance from Step 4 and \( \frac{dI}{dt} = 49.2 \, \text{A/s} \), we get:\( \varepsilon = -0.00369 \times 49.2 \approx -0.1815 \, \text{V} \).

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Key Concepts

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

Solenoids
A solenoid is essentially a coil of wire that is tightly wound, usually in the shape of a cylinder. When an electric current passes through it, a magnetic field is created. Solenoids are important in the study of electromagnetism because they are foundational components in many electromagnetic devices. The strength of the magnetic field produced inside a solenoid depends on several factors:
  • The number of coils or turns of the wire.
  • The current flowing through the wire.
  • The presence of a material core that can increase the magnetic field strength.
In the context of the exercise given, two solenoids are placed within one another. This arrangement allows for the potential study of mutual inductance, where changes in current in one solenoid can induce electromagnetic effects in the other.
Electromagnetic Induction
Electromagnetic induction is a principle where a change in magnetic field within a closed loop of wire induces a voltage, or electromotive force (emf), in the wire. This is a foundational concept in electromagnetism, vital for understanding how many electrical devices work.
  • Faraday's Law of Induction states that the induced emf in any closed circuit is equal to the time rate of change of the magnetic flux through the circuit.
  • This means if the magnetic field changes due to a changing current, the change will induce an emf in nearby wires or coils, like our solenoids.
In our exercise, the changing current in the outer solenoid changes the magnetic field. This change induces an emf in the inner solenoid according to the laws of electromagnetic induction.
EMF Calculation
The electromotive force (emf) is calculated using the principle of mutual inductance, whereby we determine how the change in current of one solenoid affects another. The formula we used to calculate the mutual inductance M is:\[ M = \frac{\Phi}{I} \]where \( \Phi \) is the magnetic flux due to the outer solenoid's field, and \( I \) is the current. In practice:
  • The magnetic field created by the outer solenoid is dependent on its turns per unit length and current.
  • The total magnetic flux \( \Phi \) is then dependent on this field, the area of the inner solenoid, and its number of coils.
The emf in the inner solenoid is then given by:\[ \varepsilon = -M \frac{dI}{dt} \]where \( \frac{dI}{dt} \) denotes the rate of change of current. This helps us determine the voltage induced in the solenoid due to changes in magnetic flux.

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Most popular questions from this chapter

An \(L-C\) circuit consists of a \(60.0-\mathrm{mH}\) inductor and a \(250-\mu \mathrm{F}\) capacitor. The initial charge on the capacitor is 6.00\(\mu \mathrm{C}\) , and the initial current in the inductor is zero. (a) What is the maximum voltage across the capacitor? (b) What is the maximum current in the inductor? (c) What is the maximum energy stored in the inductor? (d) When the current in the inductor has half its maximum value, what is the charge on the capacitor and what is the energy stored in the inductor?

A \(7.00-\mu \mathrm{F}\) capacitor is initially charged to a potential of 16.0 \(\mathrm{V}\) . It is then connected in series with a 3.75 -mH inductor. (a) What is the total energy stored in this circuit? (b) What is the maximum current in the inductor? What is the charge on the capacitor plates at the instant the current in the inductor is maximal?

A coil has 400 turns and self-inductance 4.80 \(\mathrm{mH}\) The current in the coil varies with time according to \(i=(680 \mathrm{mA}) \cos (\pi t / 0.0250 \mathrm{s})\) . (a) What is the maximum emf induced in the coil? (b) What is the maximum average flux through each turn of the coil? (c) At \(t=0.0180\) s, what is the magnitude of the induced emf?

In an \(L-C\) circuit, \(L=85.0 \mathrm{mH}\) and \(C=3.20 \mu \mathrm{F} .\) During the oscillations the maximum current in the inductor is 0.850 \(\mathrm{mA}\) . (a) What is the maximum charge on the capacitor? (b) What is the magnitude of the charge on the capacitor at an instant when the current in the inductor has magnitude 0.500 \(\mathrm{mA}\) ?

An \(L \cdot R-C\) series circuit has \(L=0.450 \mathrm{H}, C=2.50 \times \)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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