/*! 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 43 One solenoid is centered inside ... [FREE SOLUTION] | 91Ó°ÊÓ

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One solenoid is centered inside another. The outer one has a length of 50.0 \(\mathrm{cm}\) and contains 6750 coils, while the coarial 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 37.5 \(\mathrm{A} / \mathrm{s}\) . (a) what is the mutual inductance of these solenoids? (b) Find the emf induced in the innner solenoid.

Short Answer

Expert verified
(a) Calculate mutual inductance using the given formula. (b) Use the mutual inductance to compute the induced emf.

Step by step solution

01

Calculate the magnetic field of the outer solenoid

The magnetic field inside a solenoid is given by the formula \( B = \mu_0 n I \), where \( \mu_0 \) is the permeability of free space \( (4\pi \times 10^{-7} \, \mathrm{T\cdot m/A}) \), \( n \) is the number of turns per unit length, and \( I \) is the current. For the outer solenoid, \( n = \frac{6750}{0.5} \).\[ B = \mu_0 \cdot \left(\frac{6750}{0.5}\right) \cdot I \]
02

Calculate mutual inductance

The mutual inductance \( M \) is given by \( M = \frac{\Phi}{I} \), where \( \Phi \) is the flux through the inner solenoid due to the outer solenoid's field. The flux is given by \( \Phi = B \cdot A \cdot N \) where \( A \) is the cross-sectional area of the inner solenoid and \( N \) is the number of coils in the inner solenoid.\[ A = \pi \left(\frac{0.12}{2\times 100}\right)^2 \text{ m}^2 \]Compute \( M \) using \( M = \mu_0 \cdot n \cdot A \cdot N \).
03

Compute the induced EMF

The electromotive force (emf) induced in the inner solenoid is given by \( \mathrm{emf} = M \cdot \frac{dI}{dt} \), where \( \frac{dI}{dt} \) is the rate of change of current in the outer solenoid. Substitute \( M \) from Step 2 and \( \frac{dI}{dt} = 37.5 \, \mathrm{A/s} \) to find the emf.

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Key Concepts

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

Magnetic Field in Solenoid
A solenoid is a coil of wire designed to create a magnetic field when carrying an electric current. The strength of the magnetic field inside a long solenoid is uniform and is determined by the number of loops of wire and the current flowing through them. The formula to calculate the magnetic field ( \( B \) ) inside a solenoid is given by:\[ B = \mu_0 \cdot n \cdot I \]where:
  • \( B \) is the magnetic field
  • \( \mu_0 \) is the permeability of free space \( (4\pi \times 10^{-7} \text{ T·m/A}) \)
  • \( n \) is the number of turns per unit length
  • \( I \) is the current
The concept of a magnetic field in a solenoid is crucial because it allows solenoids to function as electromagnets. The strength of the magnetic field depends on the current and the density of turns in the coil, making solenoids very useful in situations where controlled magnetic fields are essential.
In our example, with the outer solenoid having 6750 coils over 50 cm, the calculation yields the number of turns per unit length \( n \ = \frac{6750}{0.5} \), which significantly impacts the magnetic field produced.
Induced EMF
Induced electromotive force (EMF) is a central concept in electromagnetism, especially in solenoids. When the magnetic environment of a conductor changes, it induces an electromotive force. Faraday’s Law of Electromagnetic Induction provides the basis for understanding this phenomenon. It states that a change in magnetic flux through a loop induces an EMF proportional to the rate of change of the flux.In the case of mutually inductive solenoids, the changing current in one solenoid affects the magnetic field in another, thereby inducing EMF. The induced EMF in the inner solenoid is determined by the mutual inductance ( \( M \) ) and the rate of change of current ( \( \frac{dI}{dt} \) ) in the outer solenoid:\[\text{emf} = M \cdot \frac{dI}{dt}\]This means the faster the current changes, the greater the induced EMF. In our exercise, a change of 37.5 A/s in the outer solenoid leads to an induced EMF in the smaller inner solenoid by interacting with mutual inductance calculated in earlier steps.
Electromagnetic Induction
Electromagnetic induction involves generating an electric current or EMF through a magnetic field. This principle is the cornerstone of many electrical devices like transformers and generators.
Understanding electromagnetic induction involves two critical ideas:
  • Faraday’s Law: which states that a time-varying magnetic field induces an EMF.
  • Lenz’s Law: which states that the direction of the induced EMF opposes the change in flux causing it.
In this context, electromagnetic induction is used to describe the process by which a changing magnetic field in the outer solenoid induces a current in the inner solenoid. As the current in the outer solenoid changes, it causes a time-varying magnetic field inside both solenoids, allowing us to use these foundational laws to calculate the effects, such as induced EMF, discussed previously.
Solenoid Calculations
Solenoid calculations allow you to determine critical values, such as magnetic fields and mutual inductance in solenoids.Let's break it down:
  • Number of Turns per Unit Length \( n \): This value is calculated as the total number of turns divided by the solenoid length.
  • Mutual Inductance \( M \): This measures how effectively a change in current in one solenoid will induce an EMF in another. Mathematically, \( M \ = \mu_0 \cdot n \cdot A \cdot N \), where \( A \) is the cross-sectional area of the inner solenoid and \( N \) is the number of turns in the inner solenoid.
  • Induced EMF: Useful for calculating the potential difference induced in the second solenoid due to changes in the magnetic field from the first one.
In a practical scenario within solenoid systems, these calculations are essential for predicting the operation of devices like transformers, which rely on mutual inductance to function efficiently.

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Most popular questions from this chapter

A \(18.0-\mu F\) capacitor is placed across a \(22.5-\mathrm{V}\) battery for several seconds and is then connected across a 12.0 -mH inductor that has no appreciable resistance. (a) After the capacitor and inductor are connected together, find the maximum current in the circuit. When the current is a maximum, what is the charge on the capacitor? (b) How long after the capacitor and inductor are connected together does it take for the capacitor to be completely discharged for the first time? For the second time? (c) Sketch graphs of the charge on the capacitor plates and the current through the inductor as functions of time.

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

Two toroidal solenoids are wound around the same form so that the magnetic field of one passes through the turns of the other. Solenoid 1 has 700 turns, and solenoid 2 has 400 turns. When the current in solenoid 1 is 6.52 A, the average flux through each turn of solenoid 2 is 0.0320 Wb. (a) What is the mutual inductance of the pair of solenoids? (b) When the current in solenoid 2 is 2.54 \(\mathrm{A}\) , what is the average flux through each turn of solenoid 1\(?\)

An \(L C\) cirruit containing an \(80.0-\mathrm{mH}\) inductor and a 1.25-nF capacitor oscillates with a maximum current of 0.750 \(\mathrm{A}\) . Calculate: (a) the maximum charge on the capacitor and (b) the oscillation frequency of the circuit. (c) Assuming the capacitor had its maximum charge at time \(t=0\) , calculate the energy stored in the inductor after 2.50 \(\mathrm{ms}\) of oscillation.

An Electromagnetic CarAlarm. Your latest invention is a car alarm that produces sound at a particularly annoying frequency of 3500 \(\mathrm{Hz}\) . To do this, the car-alarm circuitry must produce an altermating electric current of the same frequency. That's why your design includes an inductor and a capacitor in series. The maximum voltage across the capacitor is to be 12.0 \(\mathrm{V}\) (the same voltage as the car battery). To produce a sufficiently loud sound, the capacitor must store 0.0160 \(\mathrm{J}\) of energy. What values of capacitance and inductance should you choose for your car-alarm circuit?

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