/*! 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 35 A flat circular coil carrying a ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 27: Problem 35

A flat circular coil carrying a current of 8.80 A has a magnetic dipole moment of \(0.194 \mathrm{~A} \cdot \mathrm{m}^{2}\) to the left. Its area vector \(A\) is \(4.0 \mathrm{~cm}^{2}\) to the left. (a) How many turns does the coil have? (b) An observer is on the coil's axis to the left of the coil and is looking toward the coil. Does the observer see a clockwise or counterclockwise current? (c) If a huge \(45.0 \mathrm{~T}\) external magnetic field directed out of the paper is applied to the coil, what torque (magnitude and direction) results?

Short Answer

Expert verified
a) The coil has 550 turns. b) The observer looking toward the coil would see a counterclockwise current. c) A torque of 8.73 N.m directed into the paper would result when a 45.0 T external magnetic field is applied to the coil.

Step by step solution

01

Calculate the Number of Turns in the Coil

The formula that relates the magnetic dipole moment (\(m\)) to the current (\(I\)), area (\(A\)), and number of turns (\(n\)) in the coil is given by \(m = nAI\). The number of turns can therefore be calculated by dividing the magnetic dipole moment by the product of the current and the area. Given \(m = 0.194 A.m2, I = 8.8 A\), and \(A = 4.0 cm2 = 4.0 \times 10^{-4} m2 \), we can find \(n = \frac{m}{AI}\) = 550 turns.
02

Determine the Direction of the Current

To determine the direction of the current, think about the area vector and the current direction. The right-hand rule for magnetic fields states that if you curve your fingers in the direction of the current, your thumb points in the direction of the magnetic field. Since the observer sees the area vector to the left and is looking toward the coil, the fingers would have to curl counterclockwise (when looking at the coil) for the thumb to point to the left. So, the observer would see a counterclockwise current.
03

Calculate Torque

The torque \(\tau\) on a coil in a magnetic field is given by the formula \(\tau = mB\sin \theta\). Given that \(m = 0.194 A.m^2\), \(B = 45.0 T\), and \(\theta = 90^\circ\) (since the external magnetic field is out of the page, the angle between the magnetic dipole moment and the magnetic field is 90 degrees), the torque can be calculated as \(\tau = (0.194 A.m^2)(45.0 T)\sin 90 = 8.73 N.m\). The direction of the torque is perpendicular to the plane formed by the magnetic field and the magnetic moment (since the torque is maximum when these directions are perpendicular). Since the torque tends to align the magnetic moment with the magnetic field, the torque's direction is into the paper.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Magnetic Field
A magnetic field is a fundamental concept in physics, indicating the presence of a magnetic influence that can attract or repel magnetic materials such as iron, and affect moving charges. This invisible field is represented by field lines emanating from north poles and merging at south poles. Visualizing it practically, a magnetic field can be detected using a compass; its needle aligns with local magnetic field lines.

The behavior of a magnetic field is characterized by its strength, measured in Teslas (T), and its direction, which is the way it would influence the north pole of a magnet. For a current-carrying coil, like in the exercise, the magnetic field created runs perpendicular to the flow of current and can be deduced by the right-hand rule.

Magnetic fields are not only fundamental in academic studies but also have practical applications in various technologies including MRI machines, maglev trains, and electronic devices.
Torque on a Coil
When a current-carrying coil is placed in a magnetic field, it experiences a force on each segment of the loop. This force causes a torque which can make the coil rotate until it aligns with the magnetic field. The torque on the coil, represented by the Greek letter \(\tau\), depends on the magnetic dipole moment, the strength of the magnetic field, and the angle between them.

The magnitude of the torque is given by the equation \(\tau = mB\sin(\theta)\), where \(m\) is the magnetic dipole moment, \(B\) is the magnetic field strength, and \(\theta\) is the angle between the coil's magnetic moment and the magnetic field. The direction of the torque can be found using the right-hand rule. If this torque is not zero鈥攚hich occurs when the coil is not aligned with the magnetic field鈥攊t will cause the coil to rotate and can be harnessed in devices like electric motors and galvanometers.
Right-Hand Rule
The right-hand rule is a handy mnemonic device used in physics to determine the direction of parameters such as magnetic fields, forces, and torque. For current-carrying conductors, if you grip the conductor with your right hand with your thumb pointing in the direction of the current, your fingers will curl in the direction of the induced magnetic field.

In the context of a magnetic coil, place your right thumb in the direction of the current's conventional flow (positive to negative), and your curled fingers will indicate the direction of the magnetic field inside the coil. Similarly, when determining torque, align your fingers along the line connecting the magnetic field \(B\) and the magnetic moment \(m\), and your thumb will then point towards the torque's direction. This method ensures a consistent, visual way to solve related problems, especially for those who are visual learners.
Number of Turns in a Coil
The number of turns in a coil, often represented by the variable \(n\), is crucial in determining the strength of the magnetic field generated by a current flowing through the coil. The relationship between the magnetic dipole moment \(m\), current \(I\), area \(A\), and number of turns \(n\) is given by the formula \(m = nAI\).

The more turns a coil has, the stronger the magnetic field it produces because each loop of wire cumulatively adds to the total magnetic field. In the provided exercise, by dividing the magnetic dipole moment by the product of current and the coil area, we calculated that the coil has 550 turns. Practical applications are evident in the design of electromagnets, transformers, and inductors where the number of turns is adjusted to achieve desired electrical properties and performance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You are using a type of mass spectrometer to measure charge-to-mass ratios of atomic ions. In the device, atoms are ionized with a beam of electrons to produce positive ions, which are then accelerated through a potential difference \(V\). (The final speed of the ions is great enough that you can ignore their initial speed.) The ions then enter a region in which a uniform magnetic field \(\vec{B}\) is perpendicular to the velocity of the ions and has magnitude \(B=0.250 \mathrm{~T}\). In this \(\overrightarrow{\boldsymbol{B}}\) region, the ions move in a semicircular path of radius \(R .\) You measure \(R\) as a function of the accelerating voltage \(V\) for one particular atomic ion: $$ \begin{array}{l|lllll} \boldsymbol{V}(\mathbf{k} \mathbf{V}) & 10.0 & 12.0 & 14.0 & 16.0 & 18.0 \\ \hline \boldsymbol{R}(\mathrm{cm}) & 19.9 & 21.8 & 23.6 & 25.2 & 26.8 \end{array} $$ (a) How can you plot the data points so that they will fall close to a straight line? Explain. (b) Construct the graph described in part (a). Use the slope of the best-fit straight line to calculate the charge-to-mass ratio \((q / m)\) for the ion. \((\mathrm{c})\) For \(V=20.0 \mathrm{kV},\) what is the speed of the ions as they enter the \(\vec{B}\) region? (d) If ions that have \(R=21.2 \mathrm{~cm}\) for \(V=12.0 \mathrm{kV}\) are singly ionized, what is \(R\) when \(V=12.0 \mathrm{kV}\) for ions that are doubly ionized?

A dc motor with its rotor and field coils connected in series has an internal resistance of \(3.2 \Omega .\) When the motor is running at full load on a \(120 \mathrm{~V}\) line, the emf in the rotor is \(105 \mathrm{~V}\). (a) What is the current drawn by the motor from the line? (b) What is the power delivered to the motor? (c) What is the mechanical power developed by the motor?

A conducting bar with mass \(m\) and length \(L\) slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current \(I\) in the rails and bar, and a constant, uniform, vertical magnetic field \(\vec{B}\) fills the region between the rails (Fig. \(\mathbf{P 2 7 . 5 9}\) ). (a) Find the magnitude and direction of the net force on the conducting bar. Ignore friction, air resistance, and electrical resistance. (b) If the bar has mass \(m,\) find the distance \(d\) that the bar must move along the rails from rest to attain speed \(v\). (c) It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth \((11.2 \mathrm{~km} / \mathrm{s}) .\) Let \(B=0.80 \mathrm{~T}, I=2.0 \times 10^{3} \mathrm{~A}, m=25 \mathrm{~kg}\) and \(L=50 \mathrm{~cm} .\) For simplicity assume the net force on the object is equal to the magnetic force, as in parts (a) and (b), even though gravity plays an important role in an actual launch in space.

An electromagnet produces a magnetic field of \(0.550 \mathrm{~T}\) in a cylindrical region of radius \(2.50 \mathrm{~cm}\) between its poles. A straight wire carrying a current of 10.8 A passes through the center of this region and is perpendicular to both the axis of the cylindrical region and the magnetic field. What magnitude of force does this field exert on the wire?

One method for determining the amount of corn in early Native American diets is the stable isotope ratio analysis (SIRA) technique. As corn photosynthesizes, it concentrates the isotope carbon-13, whereas most other plants concentrate carbon-12. Overreliance on corn consumption can then be correlated with certain diseases, because corn lacks the essential amino acid lysine. Archaeologists use a mass spectrometer to separate the \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\) isotopes in samples of human remains. Suppose you use a velocity selector to obtain singly ionized (missing one electron) atoms of speed \(8.50 \mathrm{~km} / \mathrm{s}\), and you want to bend them within a uniform magnetic field in a semicircle of diameter \(25.0 \mathrm{~cm}\) for the \({ }^{12} \mathrm{C}\). The measured masses of these isotopes are \(1.99 \times 10^{-26} \mathrm{~kg}\left({ }^{12} \mathrm{C}\right)\) and \(2.16 \times 10^{-26} \mathrm{~kg}\left({ }^{13} \mathrm{C}\right) .\) (a) What strength of magnetic field is required? (b) What is the diameter of the \({ }^{13} \mathrm{C}\) semicircle? (c) What is the separation of the \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\) ions at the detector at the end of the semicircle? Is this distance large enough to be easily observed?
See all solutions

Recommended explanations on Physics Textbooks

Medical Physics

Read Explanation

Fluids

Read Explanation

Turning Points in Physics

Read Explanation

Solid State Physics

Read Explanation

Fields in Physics

Read Explanation

Work Energy and Power

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.