/*! 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 36 A long, thin solenoid has 900 tu... [FREE SOLUTION] | 91Ó°ÊÓ

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A long, thin solenoid has 900 turns per meter and radius \(2.50 \mathrm{~cm} .\) The current in the solenoid is increasing at a uniform rate of \(36.0 \mathrm{~A} / \mathrm{s}\). What is the magnitude of the induced electric field at a point near the center of the solenoid and (a) \(0.500 \mathrm{~cm}\) from the axis of the solenoid; (b) \(1.00 \mathrm{~cm}\) from the axis of the solenoid?

Short Answer

Expert verified
The magnitude of the induced electric field for the first position, 0.5 cm from the axis of the solenoid, will be computed following the steps. Similarly, the magnitude of the induced electric field for the second location, 1.00 cm from the axis of the solenoid will also be calculated. It’s crucial to keep track of signs in this type of exercise.

Step by step solution

01

Identify Given Data

In the exercise, the number of turns of the solenoid per unit length, denoted by \(n = 900\) turns/m, the rate of increase of current i.e \( \frac{di}{dt} = 36 \, A/s\), and the radius of the solenoid \(r = 2.5 \, cm\). The two distances from the axis of the solenoid are defined as \(r_1 = 0.5 \, cm\) and \(r_2 = 1.00 \, cm\). It is important to convert each value into the SI system, making \(r = 0.025 \, m\), \(r1 = 0.005 \, m\), \(r2 = 0.01 \, m\).
02

Calculate the Change in Magnetic Field

Using the formula for the magnetic field inside a solenoid, \( B = μ_0 (n i)\), where \( μ_0 \) is the permeability of free space, which is \(4π × 10^{-7} T.m/A\), and \(i\) is the current in the solenoid. The change in magnetic field over time becomes, \( \frac{dB}{dt}= μ_0 n \frac{di}{dt}\). Substitute the known values into this equation to compute \( \frac{dB}{dt}\).
03

Calculate the Induced Electric Field for \(r_1\) and \(r_2\)

Use Faraday’s law, \( ε = - \frac{dϕ}{dt}\), where \(ε\) is the induced electric field and \(ϕ\) is the magnetic flux. This relation can be modified as, \(ε = - r \frac{dB}{dt}\). Now plug in the values of \(r_1\) and \(r_2\) into this relation respectively to obtain the induced electric field at each of these locations.

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Key Concepts

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

Faraday's Law of Induction
Faraday's law of induction is a fundamental principle that describes how a changing magnetic environment can induce an electric field. When a magnetic field through a given area changes over time, it generates an electromotive force (emf) in any nearby conductive path. This is mathematically expressed as \( \text{emf} = -\frac{d\text{Φ}_B}{dt} \), where \( \text{Φ}_B \) is the magnetic flux through the area of interest. This negative sign is a consequence of Lenz's Law, which states that the induced emf will produce a current that opposes the change in magnetic flux that produced it. In the context of the solenoid example, as the current through the solenoid increases, it causes a change in the magnetic flux, which in turn induces an emf along the path, and hence an electric field in the vicinity of the solenoid. This induced electric field is a direct consequence of Faraday's law at work.
Magnetic Field Inside a Solenoid
The magnetic field inside a solenoid is quite uniform and parallel to the axis of the solenoid, especially near the center. This magnetic field can be calculated using the formula \( B = \text{μ}_0 n i \), with \( \text{μ}_0 \) being the permeability of free space, \( n \) the number of turns per unit length of the solenoid, and \( i \) the current passing through it. This formula is valid for an ideal solenoid where the length is much greater than its diameter. If the current through the solenoid changes, the magnetic field will also change. In our example, as the current rises, the magnetic field inside the solenoid will increase, creating a dynamic environment where Faraday’s law applies to induce an electric field around the solenoid.
Rate of Change of Current
The rate of change of current, denoted as \( \frac{di}{dt} \), is a measure of how fast the current through a circuit or component like a solenoid is changing over time. It plays a crucial role in electromagnetic induction because a time-varying current creates a time-varying magnetic field, which is the prerequisite for the induction of an electric field according to Faraday's law. In numerical problems, this rate is often given, as in our solenoid example with a rate of \( 36.0 A/s \). This change in current, when combined with the number of turns and the permeability of free space within Faraday's law, allows us to calculate the induced electric field for different points relative to the solenoid axis.
Permeability of Free Space
Permeability of free space, symbolized by \( \text{μ}_0 \), is a constant representing the extent to which a magnetic field can penetrate space. It is one of the fundamental constants of physics, indicating the capability of the vacuum to support the formation of a magnetic field. The value of \( \text{μ}_0 \) is approximately \( 4\text{π} \times 10^{-7} T\text{m/A} \), which plays an integral part in the formulas for the magnetic field in various configurations. In our exercise, the permeability of free space allows us to relate the physical properties of the solenoid—its number of turns per meter and the changing current—to calculate the magnetic field and therefore determine the induced electric field at various points near the solenoid.

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Most popular questions from this chapter

A very long, straight solenoid with a cross-sectional area of \(2.00 \mathrm{~cm}^{2}\) is wound with 90.0 turns of wire per centimeter. Starting at \(t=0\) the current in the solenoid is increasing according to \(i(t)=\left(0.160 \mathrm{~A} / \mathrm{s}^{2}\right) t^{2}\). A secondary winding of 5 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid. What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is \(3.20 \mathrm{~A}\) ?

A metal ring \(4.50 \mathrm{~cm}\) in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of \(1.12 \mathrm{~T}\) between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at \(0.250 \mathrm{~T} / \mathrm{s}\). (a) What is the magnitude of the electric field induced in the ring? (b) In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?

A battery has emf \(30.0 \mathrm{~V}\) and internal resistance \(r .\) A \(9.00 \Omega\) resistor is connected to the terminals of the battery, and the voltage drop across the resistor is \(27.0 \mathrm{~V}\). What is the internal resistance of the battery?

When a resistor with resistance \(R\) is connected to a \(1.50 \mathrm{~V}\) flashlight battery, the resistor consumes \(0.0625 \mathrm{~W}\) of electrical power. (Throughout, assume that each battery has negligible internal resistance.) (a) What power does the resistor consume if it is connected to a \(12.6 \mathrm{~V}\) car battery? Assume that \(R\) remains constant when the power consumption changes. (b) The resistor is connected to a battery and consumes \(5.00 \mathrm{~W}\). What is the voltage of this battery?

The magnetic flux through a coil is given by \(\Phi_{B}=\alpha t-\beta t^{3}\) where \(\alpha\) and \(\beta\) are constants. (a) What are the units of \(\alpha\) and \(\beta ?\) (b) If the induced emf is zero at \(t=0.500 \mathrm{~s},\) how is \(\alpha\) related to \(\beta ?\) (c) If the emf at \(t=0\) is \(-1.60 \mathrm{~V},\) what is the \(\mathrm{emf}\) at \(t=0.250 \mathrm{~s} ?\)

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