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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 \(3.20 \mathrm{~A}\) ?

Short Answer

Expert verified
The magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.20 A is 90 渭V.

Step by step solution

01

Analyze the Given Information

The current is given by the relation \(i(t)=(0.160 A/s^{2}) t^{2}\). We are asked to find the magnitude of the emf induced at the instance the current in the solenoid is 3.20 A. We also know that the solenoid is wound with 90.0 turns/cm, therefore number of turns per unit length, \(n\), is 90/cm = 9000 turns/m. The area of the solenoid, \(A\), is 2.00 cm虏 = 2*10鈦烩伌 m虏.
02

Find the time 't' at which current is 3.20 A

First solve \(i(t)=3.20 A\) for time 't' using the equation: \(i(t)=(0.160 A/s^{2}) t^{2}\). This will give: \(t = \sqrt{\frac{3.20 A}{0.160 A/s^{2}}} = 5 s\).
03

Calculate the magnetic flux 桅

The magnetic flux through the solenoid can be given as \(桅 = BA\) and because B (magnetic field) is uniform and parallel to A (area), you can write \(B=n渭鈧I\), where \(I=i(t)\) and \(渭鈧\) is the permeability of free space. Therefore, \(桅 = n渭鈧i(t)A\).
04

Calculate rate of change of magnetic flux

Calculate the derivative of the magnetic flux \(d桅/dt =A n 渭鈧 di(t)/dt\). The derivative of the current \(i(t)\) is obtained as \(di(t)/dt= 3.2 t (A/s)\). Substituting \(t=5 s\) obtained from step 2, \(di/dt = 16 A/s\).
05

Calculate the emf

The magnitude of the emf induced can be calculated using Faraday's law of electromagnetic induction, which states that \(\epsilon = - d桅/dt\). Substituting the calculated values, the induced emf is \(\epsilon = 5 * A * n * 渭鈧 * (16 A/s) = 5 * 2*10鈦烩伌 m虏 * 9000 m鈦宦 * 4蟺10鈦烩伔 H/m * 16 A/s = 9.0*10鈦烩伒 V = 90 渭V\). Note that we used 5 turns for the secondary winding in the calculation. The negative sign indicates that the emf opposes the increase in current as per Lenz's law, but since we are asked for the magnitude, we take the absolute value.

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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 tells us how electric currents can be induced by changing magnetic environments. This concept is crucial for understanding how generators, transformers, and even simple electrical circuits work.

According to Faraday's Law, an electromotive force (EMF) is induced in a circuit when the magnetic flux through the circuit changes over time. The law is mathematically represented as \[ \epsilon = - \frac{d\Phi}{dt} \], where \(\epsilon\) is the induced EMF and \(\frac{d\Phi}{dt}\) is the rate of change of magnetic flux \(\Phi\).

The negative sign in the equation comes from Lenz's Law, which states that the induced EMF will always work to oppose the change in flux, a concept known as electromagnetic damping. In practical terms, when applying Faraday鈥檚 Law to solve problems involving the induction of EMF, we often focus on the absolute value of the EMF, particularly when we are interested in its magnitude alone, which is the case in most applications.
Solenoid Magnetic Field
The solenoid magnetic field is the magnetic field created inside a solenoid, which is a coil of wire tightly wound in the shape of a helix. When electrical current flows through the wire, a uniform and strong magnetic field is generated within the inner core of the solenoid.

The magnetic field \(B\) inside a long, straight solenoid can be calculated using the formula \[ B = n\mu_0I \], where \(n\) is the number of turns per unit length of the solenoid, \(\mu_0\) is the permeability of free space, and \(I\) is the current flowing through the solenoid.

Magnetic Field Relevance

Understanding the solenoid magnetic field is essential when dealing with electromagnetic devices because it allows us to determine the behavior of the magnetic field for given currents and solenoid dimensions. In the context of our exercise, knowing the solenoid's magnetic field helps us calculate the magnetic flux and, subsequently, the induced EMF.
Induced EMF
Induced electromotive force (EMF) is an essential phenomenon in the world of electromagnetism. It refers to the voltage generated by the change in magnetic flux in a circuit. This change can be caused by altering the magnitude of the magnetic field, moving a conductor through a field, or changing the area of the loop exposed to the magnetic field.

In the exercise problem, the induced EMF in the secondary winding is a result of the changing current in the main solenoid, which in turn changes the magnetic flux. The induced EMF in the secondary circuit can be calculated using Faraday's Law as shown in the problem鈥檚 solution.

Importance of Induced EMF

Induced EMF is the principle behind electric generators and different forms of wireless energy transfer. It's also responsible for phenomena like electromagnetic interference, which can affect electronic devices.
Magnetic Flux
Magnetic flux \(\Phi\) is a measure that quantifies the amount of magnetic field passing through a given area. It is the product of the magnetic field \(B\) perpendicular to the area \(A\) and the area itself \(\Phi = BA\). For a constant magnetic field perpendicular to a flat area, the calculation remains straightforward.

However, in most real-world applications, such as our textbook problem, the flux changes over time due to changes in the magnetic field or the area. The changing flux is the key to generating an induced EMF, as mandated by Faraday's Law. Calculating the magnetic flux is often a crucial step in solving problems involving electromagnetic induction.

Applications of Magnetic Flux

Magnetic flux is not only a theoretical construct. It's widely used in designing and analyzing electrical devices like transformers, inductors, and even electric motors, where the control of magnetic fields is critical.

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Most popular questions from this chapter

In a physics laboratory experiment, a coil with 200 turns enclosing an area of \(12 \mathrm{~cm}^{2}\) is rotated in \(0.040 \mathrm{~s}\) from a position where its plane is perpendicular to the earth's magnetic field to a position where its plane is parallel to the field. The earth's magnetic field at the lab location is \(6.0 \times 10^{-5} \mathrm{~T}\). (a) What is the magnetic flux through each turn of the coil before it is rotated? After it is rotated? (b) What is the average emf induced in the coil?

A metal wire has a circular cross section with radius \(0.800 \mathrm{~mm}\) You measure the resistivity of the wire in the following way: You connect one end of the wire to one terminal of a battery that has emf \(12.0 \mathrm{~V}\) and negligible internal resistance. To the other terminal of the battery you connect a point along the wire so that the length of wire between the battery terminals is \(d\). You measure the current in the wire as a function of \(d\). The currents are small, so the temperature change of the wire is very small. You plot your results as \(I\) versus \(1 / d\) and find that the data lie close to a straight line that has slope \(600 \mathrm{~A} \cdot \mathrm{m} .\) What is the resistivity of the material of which the wire is made?

\(A 3.00 \mathrm{~m}\) length of copper wire at \(20^{\circ} \mathrm{C}\) has a \(1.20-\mathrm{m}\) -long section with diameter \(1.60 \mathrm{~mm}\) and a \(1.80-\mathrm{m}\) -long section with diameter \(0.80 \mathrm{~mm}\). There is a current of \(2.5 \mathrm{~mA}\) in the 1.60 -mm-diameter section. (a) What is the current in the 0.80 -mm-diameter section? (b) What is the magnitude of \(\vec{E}\) in the 1.60 -mm-diameter section? (c) What is the magnitude of \(\vec{E}\) in the 0.80 -mm-diameter section? (d) What is the potential difference between the ends of the \(3.00 \mathrm{~m}\) length of wire?

The resistivity of a semiconductor can be modified by adding different amounts of impurities. A rod of semiconducting material of length \(L\) and cross- sectional area \(A\) lies along the \(x\) -axis between \(x=0\) and \(x=L\). The material obeys Ohm's law, and its resistivity varies along the rod according to \(\rho(x)=\rho_{0} \exp (-x / L) .\) The end of the rod at \(x=0\) is at a potential \(V_{0}\) greater than the end at \(x=L\). (a) Find the total resistance of the rod and the current in the rod. (b) Find the electric-field magnitude \(E(x)\) in the rod as a function of \(x\). (c) Find the electric potential \(V(x)\) in the rod as a function of \(x\). (d) Graph the functions \(\rho(x), E(x),\) and \(V(x)\) for values of \(x\) between \(x=0\) and \(x=L\)

A battery-powered global positioning system (GPS) receiver operating on \(9.0 \mathrm{~V}\) draws a current of 0.13 A. How much electrical energy does it consume during 30 minutes?

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