/*! 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 31 A long, thin solenoid has 400 tu... [FREE SOLUTION] | 91Ó°ÊÓ

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A long, thin solenoid has 400 turns per meter and radius 1.10 \(\mathrm{cm}\) . The current in the solenoid is increasing at a uniform rate dildt. The induced electric field at a point near the center of the solenoid and 3.50 \(\mathrm{cm}\) from its axis is \(8.00 \times 10^{-6} \mathrm{V} / \mathrm{m}\) . Calculate dildt.

Short Answer

Expert verified
\(\frac{di}{dt} = 5.71 \, \text{A/s}\)

Step by step solution

01

Understand the Problem Context

The problem involves a solenoid where the current is increasing at a steady rate. As a result, an induced electric field is produced at a given distance from the solenoid's axis. We need to find the rate of change of current (\( \frac{di}{dt} \)).
02

Identify the Relevant Formula

The induced electric field \( E \) in a solenoid can be linked to the rate of change of current through Faraday's Law of Induction. The formula for the induced electric field in a point outside the solenoid is:\[ E = - \frac{d \Phi_B}{dt} \]The magnetic flux \( \Phi_B \) inside a solenoid is \( \Phi_B = \mu_0 n A i \), where \( \mu_0 \) is the permeability of free space, \( n \) is the number of turns per unit length, \( A \) is the cross-sectional area of the solenoid, and \( i \) is the current.
03

Use Cross-sectional Area

Since the solution requires the rate of change of the magnetic flux, we'll first need the expression for the flux linked with the solenoid considering a circular path around the solenoid. The area for the circular path is \( A = \pi r^2 \) with \( r = 0.011 \, \text{m} \).
04

Solve for Rate of Change of Flux

Relate the rate of change of magnetic flux to the rate of change of current.The magnetic flux \(\Phi_B\) is:\[ \Phi_B = \mu_0 n \pi (0.011^2) i \]Using Faraday's Law and differentiating:\[ E(2 \pi (0.035)) = \mu_0 n \pi (0.011^2) \frac{di}{dt} \]
05

Plug in Known Values

Plug in the known values:- Induced electric field, \( E = 8.00 \times 10^{-6} \, \text{V/m} \).- Number of turns per unit length, \( n = 400 \, \text{turns/m} \).- Permeability of free space, \( \mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A} \).- \( r = 0.035 \, \text{m} \) for the loop around the solenoid.Substitute and simplify:\[ 8.00 \times 10^{-6} (2 \pi (0.035)) = 4\pi \times 10^{-7} (400) \pi (0.011^2) \frac{di}{dt} \]
06

Solve for \( \frac{di}{dt} \)

Rearrange the equation to solve for \( \frac{di}{dt} \):\[ \frac{di}{dt} = \frac{8.00 \times 10^{-6} \times 2 \pi \times 0.035}{4\pi \times 10^{-7} \times 400 \pi \times 0.011^2} \]Calculate to find \( \frac{di}{dt} \): \[ \frac{di}{dt} \approx 5.71 \, \text{A/s} \]
07

Finalize the Answer

The required rate of change of current for the given solenoid configuration is approximately \( 5.71 \, \text{A/s} \).

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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 explains how a change in magnetic field or flux can induce an electromotive force (emf). This principle is fundamental in understanding many electromagnetic phenomena. According to Faraday's Law, the induced emf in a closed loop equals the negative rate of change of magnetic flux through the loop.
In simpler terms, if you have a loop of wire in a changing magnetic field, a current will be induced in that loop. The formula representing this is:
  • \[ ext{emf} = - \frac{d \Phi_B}{dt} \]
Here, \( \Phi_B \) is the magnetic flux, and \( \frac{d \Phi_B}{dt} \) is its rate of change over time. The negative sign indicates Lenz's Law, which states that the induced current's direction will oppose the change in magnetic flux.
This law is crucial in electrical engineering and physics, as it underpins the operation of transformers, electric generators, and inductive sensors. Without Faraday's Law, our modern electrical systems wouldn't function as they do today.
Solenoid
A solenoid is a long coil of wire designed to create a uniform magnetic field when an electric current flows through it. Solenoids are often used in experiments and applications that require precise magnetic fields.
The magnetic field inside an ideal solenoid is given by the formula:
  • \[ B = \mu_0 n i \]
Where \( B \) is the magnetic field, \( \mu_0 \) is the permeability of free space, \( n \) is the number of turns per unit length, and \( i \) is the current flowing through the solenoid.
A key characteristic of a solenoid is that the field inside is strong and uniform, while the field outside is much weaker. This makes solenoids useful in devices such as electromagnets, MRI machines, and even household applications like doorbells and starters in vehicles.
Understanding how solenoids work helps in harnessing magnetic fields for practical applications, showcasing yet again the wonders of electromagnetic principles in everyday life.
Induced Electric Field
An induced electric field occurs when a changing magnetic field generates a force that moves charges in a conductor. It can arise outside of a solenoid when the current within the solenoid changes, as demonstrated by Faraday's Law of Induction.
The relationship between the induced electric field \( E \) and the rate of change of the magnetic flux \( \Phi_B \) is given by:
  • \[ E = - \frac{d \Phi_B}{dt} \]
Unlike electrostatic fields, induced electric fields exist even in the absence of conductors; they are generated due to changes in the magnetic field.
This concept is particularly important in the operation of electric generators, where a rotating magnetic field induces an electric field and thereby generates electric current.
Developing a clear understanding of induced electric fields allows one to appreciate the intricate balance and interaction between electric and magnetic fields—a cornerstone of electromagnetism.
Magnetic Flux
Magnetic flux, denoted by \( \Phi_B \), measures the total magnetic field that passes through a particular area. It is an essential concept in electromagnetism, especially when dealing with Faraday's Law of Induction.
The mathematical expression for magnetic flux is:
  • \[ \Phi_B = B \cdot A \cdot \cos\theta \]
Here, \( B \) represents the magnetic field, \( A \) is the area the field passes through, and \( \theta \) is the angle between the magnetic field lines and the perpendicular to the surface.
Magnetic flux helps quantify the magnetic field's effect over a surface, which is especially useful when analyzing systems with many coils or turns, like solenoids.
It's also fundamental in understanding how transformers and inductors work—components that are vital in converting and regulating voltages in our electric grids.
Grasping the concept of magnetic flux aids in the comprehension of many real-world electromagnetic systems and innovations.

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Most popular questions from this chapter

It is impossible to have a uniform electric field that abruptly drops to zero in a region of space in which the magnetic field is constant and in which there are no electric charges. To prove this statement, use the method of contradiction: Assume that such a case is possible and then show that your assumption contradicts a law of nature. (a) In the bottom half of a piece of paper, draw evenly spaced horizontal lines representing a uniform electric field to your right. Use dashed lines to draw a rectangle abcda with horizontal side ab in the electric-field region and horizontal side \(c d\) in the top half of your paper where \(E=0 .\) (b) Show that integration around your rectangle contradicts Faraday's law, Eq. \((29.21) .\)

Shrinking Loop. A circular loop of flexible iron wire has an initial circunference of \(165.0 \mathrm{cm},\) but its circunference is decreasing at a constant rate of 12.0 \(\mathrm{cm} / \mathrm{s}\) due to a tangential pull on the wire. The loop is in a constant, viniform magnetic field oriented perpendicular to the plane of the loop and with magnitude 0.500 \(\mathrm{T}\) . (a) Find the emf induced in the loop at the instant when 9.0 s have passed. (b) Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.

Antenna emf. A satellite, orbiting the earth at the equator at an altitude of 400 \(\mathrm{km}\) , has an antenna that can be modeled as a \(2.0-\mathrm{m}-\) long rod. The antenna is oriented perpendicular to the earth's surface. At the cquator, the earth's magnetic field is cssentially horizontal and has a value of \(8.0 \times 10^{-5} \mathrm{T}\) ; ignore any changes in \(B\) with altitude. Assuming the orbit is circular, determine the induced emf between the tips of the antenna.

The magnetic field within a long, straight solenoid with a eireular cross section and radius \(R\) is increasing at a rate of \(d B / d t .\) (a) What is the rate of change of flux through a circle with radius \(r_{1}\) inside the solenoid, normal to the axis of the solenoid, and with center on the solenoid axis? (b) Find the magnitude of the induced electric field inside the solenoid, at a distance \(r_{1}\) from its axis. Show the direction of this field in a diagram. (c) What is the magnitude of the induced electric field outside the solenoid, at a distance \(r_{2}\) from the axis? (d) Graph the magnitude of the induced electric field as a function of the distance \(r\) from the axis from \(r=0\) to \(r=2 R\) . (e) What is the magnitude of the induced emf in circular turn of radius \(R / 2\) that has its center on the solenoid axis? (f) What is the magnitude of the indnced emf if the radins in part (e) is \(R ?(g)\) What is the induced emf if the radius in part \((e)\) is 2\(R ?\)

A capacitor has two parallel plates with area \(A\) separated by a distance \(d\) . The space between plates is filled with a material having dielectric constant \(K\) . The material is not a perfect insulator but has resistivity \(\rho\) . The capacitor is initially charged with charge of magnitude \(Q_{0}\) on each plate that gradually discharges by conduction through the dielectric. (a) Calculate the conduction current density \(j_{\mathrm{C}}(t)\) in the dielectric. (b) Show that at any instant the dis-placement current density in the diclectric is equal in magnitude to the oonduotion current density but opposite in direction, so the total current density is zero at every instant.

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