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Chapter 29: Problem 70

Falling Square Loop. A vertically oricnted, square loop of copper wire falls from a region where the field \(\overrightarrow{\boldsymbol{B}}\) is horizontal. uniform. where the field is zero. The loop is released from rest and initially is entirely within the magnetic-field region. Let the side length of the loop be \(s\) and let the diameter of the wire be \(d\) . The resistivity of copper is \(\rho_{R}\) and the density of copper is \(\rho_{m}\) . If the loop reaches its terminal speed while its upper segment is still in the magnetic- field region, find an expression for the terminal speed.

Short Answer

Expert verified
The terminal velocity \( v_t \) is given by: \[ v_t = \frac{\rho_m \cdot g \cdot d \cdot \pi (d/2)^2}{B^2 \cdot \rho_R} \].

Step by step solution

01

Understanding the Problem

We are dealing with a square loop of copper wire falling through a magnetic field, inducing a current due to changing magnetic flux. Resistive forces will counteract gravitational forces until a terminal velocity is reached.
02

Magnetic Flux Calculation

The magnetic flux through the loop is given by \( \Phi = B \cdot s^2 \), where \( B \) is the magnetic field strength and \( s \) is the side length of the square loop. As the loop falls, the change in flux will induce an electromotive force (emf) according to Faraday's law of induction.
03

Electromotive Force and Induced Current

Faraday's law states that the emf \( \epsilon \) is equal to the change in magnetic flux over time: \( \epsilon = -\frac{d\Phi}{dt} \). For a moving loop, \( \epsilon = B \cdot s \cdot v_t \), where \( v_t \) is the terminal velocity. The induced current \( I \) can be calculated using Ohm's law: \( I = \frac{\epsilon}{R} \).
04

Resistive Force on the Loop

The current flowing in the magnetic field experiences a force given by \( F = B \cdot I \cdot s \). This force opposes the motion of the loop and balances the gravitational force at terminal velocity.
05

Expression for Resistance

The resistance of the loop, \( R \), can be determined using the resistivity of copper: \( R = \frac{\rho_R \cdot 4s}{\pi (d/2)^2} \). This accounts for the entire loop being a circle made of copper wire.
06

Balancing Forces at Terminal Velocity

At terminal velocity, the gravitational force \( mg \), where \( m = \rho_m \cdot s^2 \cdot d \), is balanced by the magnetic force \( F = B \cdot I \cdot s \). Setting \( mg = B \cdot I \cdot s \) gives us \( \rho_m \cdot s^2 \cdot d \cdot g = B \cdot \left( \frac{B \cdot s \cdot v_t}{R} \right) \cdot s \).
07

Solve for Terminal Velocity

Substitute the expression for resistance and solve for \( v_t \):\[ v_t = \frac{\rho_m \cdot g \cdot d \cdot \pi (d/2)^2}{B^2 \cdot \rho_R} \]. This is the terminal velocity of the falling loop.

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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 is a fundamental principle in electromagnetism. It describes how a changing magnetic field can induce an electromotive force (emf) in a conductor. This law is the cornerstone of electromagnetic induction and has practical applications in many electrical devices.

The law states that the induced emf in a closed loop is equal to the negative rate of change of magnetic flux through the loop. Mathematically, it can be expressed as:
  • \(\epsilon = -\frac{d\Phi}{dt}\)
Where:
  • \(\epsilon\) is the induced emf,
  • \(\Phi\) is the magnetic flux,
  • \(t\) is time.
In the context of the falling square loop in a magnetic field, as the loop moves, the flux through it changes. This changing flux is what causes an emf to be generated according to Faraday's law. This induced emf is crucial because it leads to the generation of an induced current, which interacts with the magnetic field to counteract the loop's motion at terminal velocity.
Ohm's Law
Ohm's Law is a foundational concept in electronics and physics, establishing the relationship between voltage, current, and resistance in an electrical circuit. It is represented by the formula:
  • \(V = I \cdot R\)
Where:
  • \(V\) is voltage,
  • \(I\) is current,
  • \(R\) is resistance.
In our square loop of copper wire, Ohm's Law helps us to determine the current flowing through the loop as a result of the induced emf from Faraday's law. The formula to find the current is:
  • \(I = \frac{\epsilon}{R}\)
This tells us that the current \(I\) is equal to the induced emf \(\epsilon\) divided by the resistance \(R\) of the loop. The resistance can be calculated from the resistivity of the copper wire, its length, and cross-sectional area. Ohm's Law is used to find how much current flows due to the induced emf, affecting the dynamics as the loop reaches terminal velocity.
Magnetic Flux
Magnetic Flux is a measure of the quantity of magnetism, representing the total number of magnetic field lines passing through a given area. It is a critical concept for understanding electromagnetic induction.

Magnetic flux through a surface is given by:
  • \(\Phi = B \cdot A \cdot \cos \theta\)
Where:
  • \(\Phi\) is the magnetic flux,
  • \(B\) is the magnetic field strength,
  • \(A\) is the area the field lines pass through,
  • \(\theta\) is the angle between the magnetic field and the normal to the surface.
In the falling loop scenario, the loop initially experiences a uniform magnetic field. As it moves, the area through which magnetic field lines pass鈥攅ffectively making up the magnetic flux鈥攃hanges. This change results in an induced emf as described by Faraday's law. By understanding magnetic flux, the nuances of how the loop interacts with the magnetic field as it falls can be better comprehended.

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Most popular questions from this chapter

An airplane propeller of total length \(L\) rotates around its center with angular spced \(\omega\) in a magnctic ficld that is perpcndicular to the plane of rotation. Modeling the propeller as a thin, uniform bar, find the potential difference between (a) the center and either end of the propeller and (b) the two ends. (c) If the field is the earth's field of 0.50 \(\mathrm{G}\) and the propeller turns at 220 \(\mathrm{rpm}\) and is 2.0 \(\mathrm{m}\) long, what is the potential difference between the middle and either end? It this large enough to be concemed about?

A rod of pure silicon (resistivity \(\rho=2300 \Omega \cdot \mathrm{m} )\) is carry-ing a current. The electric field varies sinusoidally with time according to \(E=E_{0} \sin \omega t,\) where \(E_{0}=0.450 \mathrm{V} / \mathrm{m}, \omega=2 \pi f,\) and the frequency \(f=120 \mathrm{Hz}\) (a) Find the magnitude of the maximum conduction current density in the wire. (b) Assuming \(\epsilon=\epsilon_{0}\) , find the maximum displacement current density in the wire, and compare with the result of part (a). (c) At what frequency \(f\) would the maximum conduction and displacement densitics become equal if \(\epsilon=\epsilon_{0}\) (which is not actually the case)? (d) At the frequency determined in part (c), what is the relative phase of the conduction and displacement currents?

Displacement Current in a Wire. A long, straight, copper wire with a circular cross-scctional area of 2.1 \(\mathrm{mm}^{2}\) carries a current of 16 \(\mathrm{A}\) . The resistivity of the material is \(20 \times 10^{-8} \Omega \cdot \mathrm{m}\) . (a) What is the uniform electric field in the material? (b) If the cur- rent is changing at the rate of 4000 \(\mathrm{A} / \mathrm{s}\) , at what rate is the electric field in the material changing? (c) What is the displacement current density in the material in part (b)? (Hint: Since \(K\) for copper is very close to \(1,\) use \(\epsilon=\epsilon_{0} . )\) (d) If the current is changing as in part (b), what is the magnitude of the magnetic field 6.0 \(\mathrm{cm}\) from the center of the wire? Note that both the conduction current and the displacement current should be included in the calculation of \(B\) . Is the contribution from the displacement current significant?

Make a Generator? You are shipwrecked on a deserted tropical island. You have some electrical devices that you could uperate using a generalor but you have nu maguels. The eardis magnetic field at your location is horizontal and has magnitude 8.0 \(\times 10^{-5} \mathrm{T}\) , and you decide to try to use this field for a generator by rotating a large circular coil of wire at a high rate. You need to produce a peak emf of 9.0 \(\mathrm{V}\) and estimate that you can rotate the coil at 30 \(\mathrm{rpm}\) by turning a crank handle. You also decide that to have an acceptable coil resistance, the maximum mumber of turns the coil can have is 2000 . (a) What area must the coil have? (b) If the coil is circular, what is the maximum translational speed of a point on the coil as it rotates? Do you this device is feasible? Explain.

A circular wire loop of radius \(a\) and resistance \(R\) initially has a magnetic flux through it due to an external magnetic field. The extermal field then decreases to zero. A current is induced in the loop while the external field is changing; however, this current does not stop at the instant that the external field stops changing. The reason is that the current itself generates a magnetic field, which gives rise to a flux through the loop. If the current changes, the flux through the loop changes as well, and an induced emf appears in the loop to oppose the change. (a) The magnetic field at the center of the loop of radius a produced by a current i in the loop is given by \(B=\mu_{0} i / 2 a\) . If we use the crode approximation that the field has this same value at all points within the loop, what is the flux of this field through the loop? (b) By using Faraday's law, Eq. \((29.3),\) and the relationship \(\mathcal{E}=i R,\) show that after the external field has stopped changing, the current in the loop obeys the differential equation $$ \frac{d i}{d t}=-\left(\frac{2 R}{\pi \mu_{0} a}\right) i $$ (c) If the current has the value \(i_{0}\) at \(t=0\) , the instant that the external field stops changing. snive the equation in part \((b)\) to find \(i\) as a function of time for \(t>0 .\) (Hint: In Section 26.4 we encountered a similar differential equation, Eq. \((26.15),\) for the quantity \(q .\) This equation for \(i\) may be solved in the same way. (d) If the loop has radius \(a=50 \mathrm{cm}\) and resistance \(R=0.10 \Omega,\) how long after the external field stops changing will the current be equal to 0.010 \(\mathrm{o}\) (that is, \(\frac{1}{100}\) of its initial value)? (e) In solving the examples in this chapter, we ignored the effects described in this problem. Explain why this is a good approximation.
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