/*! 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 61 A closely wound, circular coil w... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A closely wound, circular coil with radius 2.40 \(\mathrm{cm}\) has 800 turns. What must the current in the coil be if the magnetic field at the center of the coil is 0.0580 \(\mathrm{T}\) ?

Short Answer

Expert verified
The current must be approximately 2.77 A.

Step by step solution

01

Understanding the Given Values

The problem provides specific values: the radius of the coil, \( r = 2.40 \, \text{cm} = 0.0240 \, \text{m} \), the number of turns in the coil, \( N = 800 \), and the magnetic field at the center, \( B = 0.0580 \, \text{T} \). We need to find the current, \( I \), that creates this magnetic field.
02

Applying the Magnetic Field Formula for a Circular Coil

The magnetic field at the center of a closely wound, circular coil is given by the formula \( B = \frac{\mu_0 N I}{2R} \), where \( B \) is the magnetic field, \( \mu_0 = 4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A} \) is the permeability of free space, \( N \) is the number of turns, \( I \) is the current, and \( R \) is the radius of the coil in meters.
03

Rearranging the Formula to Find the Current

To find the current \( I \), rearrange the formula to solve for \( I \): \( I = \frac{2BR}{\mu_0 N} \). By plugging in the values given in the problem, \( I = \frac{2 \times 0.0580 \, \text{T} \times 0.0240 \, \text{m}}{4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A} \times 800} \).
04

Calculating the Current

Substitute the values into the rearranged formula: \( I = \frac{2 \times 0.0580 \times 0.0240}{4\pi \times 10^{-7} \times 800} \). Calculate the numerator: \( 2 \times 0.0580 \times 0.0240 = 0.002784 \). Calculate the denominator: \( 4\pi \times 10^{-7} \times 800 = 1.00530965 \times 10^{-3} \). Divide to find \( I = \frac{0.002784}{1.00530965 \times 10^{-3}} \).
05

Finding the Value of the Current

Now, simplify the division to find that \( I = 2.77 \). Therefore, the current required in the coil to produce a magnetic field of 0.0580 T at its center is approximately 2.77 A.

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 91Ó°ÊÓ!

Key Concepts

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

Circular Coil
A circular coil is essentially a wire that is wound in a circular shape, often multiple times, forming loops or turns. These loops enhance the magnetic field generated due to the current flowing through the wire.

In the context of electromagnetism, the more turns (or loops) a coil has, the stronger the magnetic field it can produce for the same amount of current. For a coil with a given radius, the setup allows for an easy, yet impactful demonstration of how electricity can be used to induce magnetism. In our example, a closely wound coil with a radius of 2.40 cm with 800 turns is used to achieve a particular magnetic field.

- **Radius**: The distance from the center of the coil to any point on its circular edge, crucial for magnetic calculations. - **Number of Turns (N)**: Indicates how many times the wire loops around in the coil, affecting the magnetic field strength.

When considering coils in electromagnetic applications, the concept of turns and the specific shape (circular here) plays a crucial role in determining the outcome of induced magnetic fields.
Magnetic Field Formula
The magnetic field within a circular coil is central to electromagnetic studies. This field is primarily determined by the number of turns, the current passing through the coil, and its radius. The magnetic field formula is given by:

\[B = \frac{\mu_0 N I}{2R}\]

Where:
  • \( B \): Magnetic field at the center of the coil, measured in Tesla ( \( \mathrm{T} \) )
  • \( \mu_0 \) : Permeability of free space, a constant approximately equal to \( 4\pi \times 10^{-7} \, \mathrm{T} \cdot \mathrm{m/A} \)
  • \( N \): Number of turns
  • \( I \): Current through the coil in Amperes ( \( \, \mathrm{A} \) )
  • \( R \): Radius of the coil in meters
Understanding this formula is crucial. It demonstrates that the magnetic field is directly proportional to the number of turns and the current, and inversely proportional to the coil's radius. This means, with a fixed radius, increasing the number of turns or current will enhance the magnetic field strength.
Current Calculation
To find the unknown current needed to generate a specific magnetic field at the center of the circular coil, we need to rearrange the magnetic field formula. Let's isolate the current (\( I \)) on one side of the equation:

\[I = \frac{2BR}{\mu_0 N}\]

- Insert our given values: \( B = 0.0580 \) T, \( R = 0.0240 \) meters, \( N = 800 \).- Permeability \( \mu_0 \) is a universal constant: \( 4\pi \times 10^{-7} \mathrm{T} \cdot \mathrm{m/A} \).

Applying these values gives:

\[I = \frac{2 \times 0.0580 \times 0.0240}{4\pi \times 10^{-7} \times 800}\]

Conducting the calculations step by step, first compute the numerator and then the denominator.- The numerator perturbation: \( 2 \times 0.0580 \times 0.0240 = 0.002784 \).- The denominator is: \( 4\pi \times 10^{-7} \times 800 = 1.00530965 \times 10^{-3} \).
Dividing these gives the current \( I \approx 2.77 \) A. This calculation reveals how much current is necessary to achieve the desired magnetic field.
Permeability of Free Space
The permeability of free space, denoted as \( \mu_0 \), is a fundamental physical constant essential in electromagnetism and appears in numerous equations related to magnetic fields. It represents the ability of a vacuum (free space) to support a magnetic field.

The value of \( \mu_0 \) is approximately:
  • \( \mu_0 = 4\pi \times 10^{-7} \, \mathrm{T} \cdot \mathrm{m/A} \)
In practical terms, this constant is used to quantify the relation between current, magnetic fields, and is used in the magnetic field formula. It helps establish a baseline for how magnetic fields are affected by and in the presence of electrical currents in a vacuum. By utilizing \( \mu_0 \), we can predict various magnetic phenomena precisely.

This constant reminds us of the deep linkage between electricity and magnetism, foundational to many modern technologies and scientific advancements.

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

(a) What is the speed of a beam of electrons when the simultaneous influence of an electric field of \(1.56 \times 10^{4} \mathrm{V} / \mathrm{m}\) and a magnetic field of \(4.62 \times 10^{-3} \mathrm{T}\) , with both fields normal to the beam and to each other, produces no deflection of the electrons? (b) In a diagram, show the relative orientation of the vectors \(\vec{\boldsymbol{v}}\) , \(\vec{\boldsymbol{E}}\) and \(\vec{\boldsymbol{B}}\) . (c) When the electric field is removed, what is the radius of the electron orbit? What is the period of the orbit?

Household magnetic fields. Home circuit breakers typically have current capacities of around 10 A. How large a magnetic field would such a current produce 5.0 \(\mathrm{cm}\) from a long- wire's center? How does this field compare with the strength of the earth's magnetic field?

Magnetic sensitivity of electric fish. In a problem dealing with electric fish in Chapter 19, we saw that these fish navigate by responding to changes in the current in seawater. This current is due to a potential difference of around 3.0 \(\mathrm{V}\) generated by the fish and is about 12 \(\mathrm{mA}\) within a centimeter or so from the fish. Receptor cells in the fish are sensitive to the current. since the current is at some distance from the fish, the sensitivity of these cells suggests that they might be responding to the magnetic field created by the current. To get some estimate of how sensitive the cells are, we can model the current as that of a long, straight wire with the receptor cells 2.0 \(\mathrm{cm}\) away. What is the strength of the magnetic field at the receptor cells?

\(\cdot\) A solenoid contains 750 coils of very thin wire evenly wrapped over a length of 15.0 \(\mathrm{cm} .\) Each coil is 0.800 \(\mathrm{cm}\) in diameter. If this solenoid carries a current of \(7.00 \mathrm{A},\) what is the magnetic field at its center?

\(\bullet\) Three long, straight electrical cables, running north and south, are tightly enclosed in an insulating sheath. One of the cables carries a 23.0 A current southward; the other two carry currents of 17.5 \(\mathrm{A}\) and 11.3 A northward. Use Ampere's law to calculate the magnitude of the magnetic field at a distance of 10.0 \(\mathrm{m}\) from the cables.

See all solutions

Recommended explanations on Physics Textbooks

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.