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Chapter 25: Problem 57

A slender rod, \(0.240 \mathrm{~m}\) long, rotates with an angular speed of \(8.80 \mathrm{rad} / \mathrm{s}\) about an axis through one end and perpendicular to the rod. The plane of rotation of the rod is perpendicular to a uniform magnetic field with a magnitude of \(0.650 \mathrm{~T}\). (a) What is the induced emf in the rod? (b) What is the potential difference between its ends? (c) Suppose instead the rod rotates at \(8.80 \mathrm{rad} / \mathrm{s}\) about an axis through its center and perpendicular to the rod. In this case, what is the potential difference between the ends of the rod? Between the center of the rod and one end?

Short Answer

Expert verified
The induced emf in the rod is -, the potential difference between the ends of the rod is -, and the potential difference when rotating about its center is - between both ends and - between center and one end.

Step by step solution

01

Calculate the induced emf in the rod

First calculate the maximum magnetic flux through the rod when it's perpendicular to the magnetic field. We can start by using the formula for magnetic flux, which is given by \( \Phi = B \cdot A \cdot \cos(\theta) \), where \( B \) is the magnetic field strength, \( A \) is the area, and \( \theta \) is the angle between the magnetic field lines and the normal (perpendicular) to the surface. Here, the area \( A \) is the path area that the rod sweeps out, which is a semicircle with radius \( r = L \) where \( L \) is the length of the rod. Hence \( A = 0.5 \cdot \pi \cdot L^2 \), and the magnetic flux \( \Phi = B \cdot 0.5 \cdot \pi \cdot L^2 \) as at maximum flux \( \theta = 0 \). Then we can calculate the rate of change of magnetic flux with respect to time, which is \( \Phi / T \), using \( T = 2 \cdot \pi / \omega \) where \( \omega \) is the angular speed, to get the induced emf from Faraday's law, \( \epsilon = -d\Phi / dt\). Substituting in our values and calculating we find the induced emf.
02

Determine the potential difference between its ends

The induced emf is the potential difference between the ends of the rod, which is what we just calculated in the previous step.
03

Calculate the potential difference when rotating about the center of the rod

If the rod is rotating about its center, the area swept out at one end is half that when it rotates around the end. So, the potential difference (induced emf) is half of the value we calculated in Step 1. The potential difference between the center of the rod and one end is just half of this new value.

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Key Concepts

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

Induced EMF
Electromotive force (EMF) is kind of like the electrical energy that's driving through a circuit. When something's moving through a magnetic field, like our rod spinning about its end, it generates this electrical energy. This happens due to changes in magnetic flux, or how much of the magnetic field is hitting our hypothetical area.
In the context of this exercise, as the rod rotates, it cuts through magnetic field lines and alters the magnetic field within its path, even though it's simply stuck in rotation. Using the formula that couples Faraday鈥檚 Law of Induction, the EMF can be given by:
  • \(\epsilon = -\frac{d\Phi}{dt}\)
The negative sign indicates Lenz's Law, suggesting the induced emf seeks to oppose the change in flux causing it.
Let's connect back to the practical: As the rod rotates, the magnetic flux changes, and it induces EMF across the rod. Calculating it uses the rate of change of this magnetic flux, which exploits angular speed and the physical sweep area a rotating rod describes.
Magnetic Flux
Magnetic flux represents the amount of magnetic field passing through a specified area. Think of it as counting how many lines of magnetic field threads through an area.
Understanding it begins with the formula:
  • \( \Phi = B \cdot A \cdot \cos(\theta) \)
Where:
  • \( B \) is the magnetic field strength.
  • \( A \) is the area the field lines pass through.
  • \( \theta \) is the angle between the magnetic field and perpendicular to the plane. Often, \( \theta = 0\) aligns the field perfectly perpendicular, maximizing the flux.
In our exercise, the rod sweeping out a semicircle when rotating about one end, embodies area \( A = 0.5 \cdot \pi \cdot L^2 \). This concept ensures we calculate accurately the movement and transformation of magnetic fields across our inventive rod.
Faraday's Law
Faraday's Law is the cornerstone of electromagnetic induction. It explains how changing magnetic fields can induce electricity.
The formula we often utilize in practice reflects this:
  • \( \epsilon = -\frac{d\Phi}{dt}\)
Faraday's Law suggests that the likelihood of induced voltage relates directly to how quickly the magnetic environment changes. Faster changes mean stronger induced EMF.
For our example here, imagine the rotation speed defining how speedily the magnetic flux changes. Faraday鈥檚 insight was to quantify this, guiding us into understanding that the quicker the change, the larger the induced EMF in anything, like a rod, finding itself part of this transformation. This principle is eminently practical, forming the basis for many electrical devices we encounter today.

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Most popular questions from this chapter

A long, straight solenoid with a cross-sectional area of \(8.00 \mathrm{~cm}^{2}\) is wound with 90 turns of wire per centimeter, and the windings carry a current of 0.350 A. A second winding of 12 turns encircles the solenoid at its center. The current in the solenoid is turned off such that the magnetic field of the solenoid becomes zero in \(0.0400 \mathrm{~s}\). What is the average induced emf in the second winding?

The region between two concentric conducting spheres with radii \(a\) and \(b\) is filled with a conducting material with resistivity \(\rho\). (a) Show that the resistance between the spheres is given by $$ R=\frac{\rho}{4 \pi}\left(\frac{1}{a}-\frac{1}{b}\right) $$ (b) Derive an expression for the current density as a function of radius, in terms of the potential difference \(V_{a b}\) between the spheres. (c) Show that the result in part (a) reduces to Eq. (25.10) when the separation \(L=b-a\) between the spheres is small.

A coil \(4.00 \mathrm{~cm}\) in radius, containing 500 turns, is placed in a uniform magnetic field that varies with time according to \(B=(0.0120 \mathrm{~T} / \mathrm{s}) t+\left(3.00 \times 10^{-5} \mathrm{~T} / \mathrm{s}^{4}\right) t^{4} .\) The coil is connected to a \(600 \Omega\) resistor, and its plane is perpendicular to the magnetic field. You can ignore the resistance of the coil. (a) Find the magnitude of the induced emf in the coil as a function of time. (b) What is the current in the resistor at time \(t=5.00 \mathrm{~s} ?\)

BIO Transmission of Nerve Impulses. Nerve cells transmit electric signals through their long tubular axons. These signals propagate due to a sudden rush of \(\mathrm{Na}^{+}\) ions, each with charge \(+e,\) into the axon. Measurements have revealed that typically about \(5.6 \times 10^{11} \mathrm{Na}^{+}\) ions enter each meter of the axon during a time of \(10 \mathrm{~ms}\). What is the current during this inflow of charge in a meter of axon?

A \(0.650-\mathrm{m}\) -long metal bar is pulled to the right at a steady \(5.0 \mathrm{~m} / \mathrm{s}\) perpendicular to a uniform, \(0.750 \mathrm{~T}\) magnetic field. The bar rides on parallel metal rails connected through a \(25.0 \Omega\) resistor (Fig. \(\mathbf{E} 29.28),\) so the apparatus makes a complete circuit. Ignore the resistance of the bar and the rails. (a) Calculate the magnitude of the emf induced in the circuit. (b) Find the direction of the current induced in the circuit by using (i) the magnetic force on the charges in the moving bar; (ii) Faraday's law; (iii) Lenz's law. (c) Calculate the current through the resistor.
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