/*! 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 12 A long solenoid, of radius \(a\)... [FREE SOLUTION] | 91Ó°ÊÓ

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A long solenoid, of radius \(a\), is driven by an alternating current, so that the field inside is sinusoidal: \(\mathbf{B}(t)=B_{0} \cos (\omega t) \hat{z}\). A circular loop of wire, of radius \(a / 2\) and resistance \(R\), is placed inside the solenoid, and coaxial with it. Find the current induced in the loop, as a function of time.

Short Answer

Expert verified
The induced current in the loop is \( I(t) = \frac{B_0 \pi a^2 \omega}{4R} \sin(\omega t) \).

Step by step solution

01

Understand Faraday's Law of Induction

Faraday's Law of Induction states that the emf (electromotive force) induced in a circuit is equal to the negative rate of change of magnetic flux through the circuit. Mathematically, this is expressed as \( \text{emf} = - \frac{d\Phi_B}{dt} \), where \( \Phi_B \) is the magnetic flux.
02

Calculate Magnetic Flux Through the Loop

The magnetic flux through the loop is given by \( \Phi_B = \int \mathbf{B} \cdot d\mathbf{A} \). Since the field is uniform within the solenoid and the loop is circular of radius \( \frac{a}{2} \), \( \Phi_B = B(t) \cdot \pi \left( \frac{a}{2} \right)^2 \). Substitute \( B(t) = B_0 \cos(\omega t) \) to find: \[ \Phi_B(t) = B_0 \pi \left( \frac{a}{2} \right)^2 \cos(\omega t). \]
03

Calculate the Induced Emf Using Faraday's Law

Differentiate the expression for magnetic flux with respect to time to find the induced emf: \[ \text{emf} = -\frac{d}{dt} \left( B_0 \pi \left( \frac{a}{2} \right)^2 \cos(\omega t) \right). \] Use the chain rule: \[ \text{emf} = B_0 \pi \left( \frac{a}{2} \right)^2 \omega \sin(\omega t). \]
04

Calculate the Induced Current

Ohm's Law relates current to emf and resistance: \( I(t) = \frac{\text{emf}}{R} \). Substitute the expression for emf to get: \[ I(t) = \frac{B_0 \pi \left( \frac{a}{2} \right)^2 \omega \sin(\omega t)}{R} = \frac{B_0 \pi a^2 \omega}{4R} \sin(\omega t). \]

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Key Concepts

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

Magnetic Flux
Magnetic flux is a fundamental concept related to Faraday's Law of Induction and is denoted by the symbol \( \Phi_B \). It represents the amount of magnetic field passing through a given area. In practical terms, it helps us understand how magnetic fields interact with physical objects, such as coils and loops. For a uniform magnetic field, magnetic flux through a surface can be calculated with the formula: \( \Phi_B = \int \mathbf{B} \cdot d\mathbf{A} \), where:
  • \( \mathbf{B} \) is the magnetic field vector.
  • \( d\mathbf{A} \) is a differential area vector perpendicular to the surface.
In our problem, the magnetic field is sinusoidal inside a solenoid, described by \( \mathbf{B}(t) = B_0 \cos(\omega t) \hat{z} \). With a circular loop inside the solenoid, the area \( \frac{a^2 \pi}{4} \) is constant, allowing us to express the magnetic flux as \( \Phi_B(t) = B_0 \pi \left( \frac{a}{2} \right)^2 \cos(\omega t) \). This time-varying flux is key for calculating the electromotive force (emf).
Electromotive Force (emf)
Electromotive force, commonly referred to as emf, is the energy provided per charge by a source of electrical energy. According to Faraday's Law of Induction, the emf induced in a circuit is equal to the negative rate of change of magnetic flux through the circuit. This can be mathematically expressed as \( \text{emf} = - \frac{d\Phi_B}{dt} \). To determine the emf in our exercise:
  • We differentiate the expression for magnetic flux \( \Phi_B = B_0 \pi \left( \frac{a}{2} \right)^2 \cos(\omega t) \) with respect to time \( t \).
  • Using the chain rule, the derivative of \( \cos(\omega t) \) with respect to time is \( -\omega \sin(\omega t) \).
  • This gives us: \( \text{emf} = B_0 \pi \left( \frac{a}{2} \right)^2 \omega \sin(\omega t) \).
Through this process, we convert a time-varying magnetic flux into an induced emf, which can drive current through a loop or circuit.
Ohm's Law
Ohm’s Law is a cornerstone of basic electronics and physics, establishing the relation between voltage (\text{emf), current \((I)\), and resistance \((R)\). It is represented by the equation: \( I = \frac{\text{emf}}{R} \). In simple terms, it tells us that the current flowing through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance of the conductor.
  • In the context of our exercise, we've calculated the induced emf due to the changing magnetic flux.
  • By applying Ohm’s Law, we determine the induced current in the loop: \( I(t) = \frac{B_0 \pi \left( \frac{a}{2} \right)^2 \omega \sin(\omega t)}{R} \).
  • Simplifying further, \( I(t) = \frac{B_0 \pi a^2 \omega}{4R} \sin(\omega t) \), shows how resistance, the geometry of the loop, and the magnetic field dynamics influence the current.
This highlights Ohm's Law's utility in connecting the concepts of electromotive force and magnetic flux to tangible electrical characteristics like current and resistance.

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Most popular questions from this chapter

As a lecture demonstration a short cylindrical bar magnet is dropped down a vertical aluminum pipe of slightly larger diameter, about 2 meters long. It takes several seconds to emerge at the bottom, whereas an otherwise identical piece of unmagnetized iron makes the trip in a fraction of a second. Explain why the magnet falls more slowly. \({ }^{12}\)

The magnetic field of an infinite straight wire carrying a steady current \(I\) can be obtained from the displacement current term in the Ampère/Maxwell law, as follows: Picture the current as consisting of a uniform line charge \(\lambda\) mov. ing along the \(z\) axis at speed \(v\) (so that \(I=\lambda v\) ), with a tiny gap of length \(\epsilon\), which reaches the origin at time \(t=0\). In the next instant (up to \(t=\epsilon / v\) ) there is no real current passing through a circular Amperian loop in the \(x y\) plane, but there is a displacement current, due to the "missing" charge in the gap. (a) Use Coulomb's law to calculate the z component of the clectric field, for points in the \(x y\) plane a distance \(s\) from the origin, due to a segment of wire with uniform density \(-\lambda\) extending from \(z_{1}=v t-\epsilon\) to \(z_{2}=v t\). (b) Determine the flux of this electric field through a circle of radius \(a\) in the \(x y\) plane. (c) Find the displacement current through this circle. Show that \(I_{d}\) is equal to \(I\), in the limit as the gap width \((\epsilon)\) goes to zero. 35

Assuming that "Coulomb"s law" for magnetic charges \(\left(q_{m}\right)\) reads $$ \mathbf{F}=\frac{\mu_{0}}{4 \pi} \frac{q_{m_{1}} q_{m_{2}}}{{2}^{2}} \varepsilon_{1} $$ work out the force law for a monopole \(q_{n}\) moving with velocity \(v\) through electric and magnetic fields \(\mathbf{E}\) and \(\mathbf{B}\). \({ }^{26}\)

Electrons undergoing cyclotron motion can be sped up by increasing the magnetic ficld; the accompanying electric field will impart tangential acceleration. This is the principle of the betatron. One would like to keep the radius of the orbit constant during the process. Show that this can be achieved by designing a magnet such that the average field over the area of the orbit is twice the field at the circumference (Fig. 7.53). Assume the electrons start from rest in zero field, and that the apparatus is symmetric about the center of the orbit. (Assume also that the electron velocity remains well below the speed of light, so that nonrelativistic mechanics applies.)

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