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Chapter 25: Problem 35

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 \(d i / d t\). 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 \(d i / d t\)

Short Answer

Expert verified
The rate of change of current (\(di/dt\)) is approximately \(0.144 A/s\).

Step by step solution

01

Identify Applicable Formulas

The formula to calculate the magnetic field inside a solenoid is \(B=\mu n i\), where \(\mu\) is the magnetic permeability, \(n\) is the number of turns per unit length, and \(i\) is the current. For a solenoid, Faraday's law is given by \(\varepsilon=-\frac{d\Phi_B}{dt}\), where \(\varepsilon\) is the EMF (related to the electric field), and \(\Phi_B=AB\) is the magnetic flux, dependent on the area \(A\) and the magnetic field \(B\).
02

Apply Faraday's Law

We can express Faraday's law in terms of E-field for a circular path of radius \(r\) in a solenoid: \(E=\frac{-1}{2\pi r}\frac{d\Phi_B}{dt}=-n\frac{di}{dt}\frac{o \mu r^2}{2}\). Given that \(E=8.00 \times 10^{-6} \mathrm{~V} / \mathrm{m}\), and \(r=3.50 \mathrm{~cm}\), we can solve for \(\frac{di}{dt}\) to find the rate that the current must be changing at.
03

Calculate \(\frac{di}{dt}\)

Rearrange the formula from Step 2 to solve for \(\frac{di}{dt}\): \(\frac{di}{dt}=-\frac{2E}{n\mu r^2}\). Substitute the known values: \(E = 8.00 \times 10^{-6} \mathrm{V/m}\), \(n = 400 \mathrm{turns/m}\), \(\mu = 4\pi \times 10^{-7}\), and \(r = 0.035 \mathrm{m}\). This will give the desired rate of change of current.

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Key Concepts

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

Faraday's Law
Faraday's Law is a fundamental principle that governs how electric currents and magnetic fields interact. Formulated by Michael Faraday, it posits that a change in magnetic environment of a coil of wire will induce an electromotive force (EMF), or voltage, in the coil. The induced EMF is directly proportional to the rate at which the magnetic flux through the coil is changing with time.

The formula that encapsulates this law is \[ \varepsilon = -\frac{d\Phi_B}{dt} \] where \( \varepsilon \) is the induced EMF and \( \Phi_B \) is the magnetic flux. Notice the negative sign, which is a reflection of Lenz's Law, indicating that the induced EMF will always work to oppose the change in flux that produced it. For students trying to solve problems using Faraday's Law, it's essential to understand not just the magnitudes but also the direction of the induced currents and their effects.
Solenoid Magnetic Field
A solenoid is essentially a coil of wire designed to create a uniform magnetic field in its interior. When an electric current passes through it, it behaves like a bar magnet, with distinct north and south poles. This magnetic field in a solenoid is given by \[ B = \mu n i \] where \( B \) is the magnetic field strength, \( \mu \) is the magnetic permeability of the medium, \( n \) is the number of turns of wire per unit length of the solenoid, and \( i \) is the current.

Practical Understanding in Solenoids

Understanding the creation of a magnetic field in a solenoid is essential for applications in electromagnets, transformers, and inductors. Students should pay attention to the role of each component in the formula; for instance, more turns or higher current will result in a stronger magnetic field.
Magnetic Flux
Magnetic flux, denoted as \( \Phi_B \), is a measure of the total magnetic field that passes through a given area. In a more technical sense, it's the product of the magnetic field and the area of the surface it penetrates, together with the angle between the magnetic field lines and the perpendicular to the surface. It is expressed as \[ \Phi_B = AB\cos(\theta) \] where \( A \) represents the area through which the field is passing and \( \theta \) is the angle between the field and the normal to the surface.

Understanding Magnetic Flux

For a solenoid, the magnetic field is uniform and the angle is 0 degrees, meaning cos(0) = 1. This simplifies our flux calculation to \( \Phi_B = AB \). Comprehending the concept of magnetic flux is vital for grasping Faraday's Law, as it is the change in this flux that results in electromagnetic induction.
Magnetic Permeability
Magnetic permeability, symbolized as \( \mu \), is a property that characterizes the ability of a material to support the formation of a magnetic field within itself. It is an indication of the ease with which magnetic field lines can pass through a medium. The value thus affects the strength of the magnetic field in materials like the core of a solenoid.

For free space, magnetic permeability is a constant, termed the permeability of free space, and is denoted as \( \mu_0 \), with a value of approximately \( 4\pi \times 10^{-7} \)henry per meter (H/m).

Applications in Induction and Solenoids

For students and engineers, understanding magnetic permeability is critical when designing magnetic circuits, inductors, or transformers since it dictates how a given material will respond to a magnetic field and thereby influences the induction processes.

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Most popular questions from this chapter

The current in a wire varies with time according to the relationship \(I=55 \mathrm{~A}-\left(0.65 \mathrm{~A} / \mathrm{s}^{2}\right) t^{2} .\) (a) How many coulombs of charge pass a cross section of the wire in the time interval between \(t=0\) and \(t=8.0 \mathrm{~s} ?(\mathrm{~b}) \mathrm{What}\) constant current would transport the same charge in the same time interval?

BIO A person with body resistance between his hands of \(10 \mathrm{k} \Omega\) accidentally grasps the terminals of a \(14 \mathrm{kV}\) power supply. (a) If the internal resistance of the power supply is \(2000 \Omega,\) what is the current through the person's body? (b) What is the power dissipated in his body? (c) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be \(1.00 \mathrm{~mA}\) or less?

The voltage drop \(V_{a b}\) across each of resistors \(A\) and \(B\) was measured as a function of the current \(I\) in the resistor. The results are shown in the table: $$ \begin{array}{l|llll} \text { Resistor } A & & & & \\ I(\mathrm{~A}) & 0.50 & 1.00 & 2.00 & 4.00 \\ V_{a b}(\mathrm{~V}) & 2.55 & 3.11 & 3.77 & 4.58 \\ & & & & \\ \begin{array}{l} \text { Resistor } B \\ I(\mathrm{~A}) \end{array} & 0.50 & 1.00 & 2.00 & 4.00 \\ V_{a b}(\mathrm{~V}) & 1.94 & 3.88 & 7.76 & 15.52 \end{array} $$ (a) For each resistor, graph \(V_{a b}\) as a function of \(I\) and graph the resistance \(R=V_{a b} / I\) as a function of \(I\). (b) Does resistor \(A\) obey Ohm's law? Explain. (c) Does resistor \(B\) obey Ohm's law? Explain. (d) What is the power dissipated in \(A\) if it is connected to a \(4.00 \mathrm{~V}\) battery that has negligible internal resistance? (e) What is the power dissipated in \(B\) if it is connected to the battery?

\(\mathrm{A}\) cell phone or computer battery has three ratings marked on it: a charge capacity listed in mAh (milliamp-hours), an energy capacity in Wh (watt-hours), and a potential rating in volts. (a) What are these three values for your cell phone? (b) Convert the charge capacity \(Q\) into coulombs. (c) Convert the energy capacity \(U\) into joules. (d) Multiply the charge rating \(Q\) by the potential rating \(V,\) and verify that this is equivalent to the energy capacity \(U\). (e) If the charge \(Q\) were stored on a parallel-plate capacitor with air as the dielectric, at the potential \(V,\) what would be the corresponding capacitance? (f) If the energy in the battery were used to heat \(1 \mathrm{~L}\) of water, estimate the corresponding change in the water temperature? (The heat capacity of water is \(4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} .)\)

25.80 If the conductivity of the thread results from the aqueous coating only, how does the cross-sectional area \(A\) of the coating compare when the thread is \(13 \mathrm{~mm}\) long versus the starting length of \(5 \mathrm{~mm} ?\) Assume that the resistivity of the coating remains constant and the coat- (b) \(\frac{1}{4} A_{5 \mathrm{~mm}}\) ing is uniform along the thread. \(A_{13 \mathrm{~mm}}\) is about (a) \(\frac{1}{10} A_{5 \mathrm{~mm}}\) (c) \(\frac{2}{5} A_{5 \mathrm{~mm}} ;\) (d) the same as \(A_{5 \mathrm{~mm}}\).
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