/*! 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 33 The magnetic field around the he... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The magnetic field around the head has been measured to be approximately 3.0 \(\times\) 10\(^{-8}\) G. Although the currents that cause this field are quite complicated, we can get a rough estimate of their size by modeling them as a single circular current loop 16 cm (the width of a typical head) in diameter. What is the current needed to produce such a field at the center of the loop?

Short Answer

Expert verified
The current needed is approximately 3.82 \\times 10^{-7} A.

Step by step solution

01

Understand the Formula for Magnetic Field

The magnetic field \( B \) at the center of a circular loop of current \( I \) and radius \( R \) is given by the formula: \[ B = \frac{\mu_0 \cdot I}{2R} \] where \( \mu_0 \) is the permeability of free space, \( 4\pi \times 10^{-7} \; \text{T}\cdot\text{m/A} \). We need to find the current \( I \) using this equation.
02

Calculate the Radius of the Loop

The diameter of the loop is given as 16 cm. To find the radius \( R \), divide the diameter by 2: \[ R = \frac{16}{2} = 8 \text{ cm} = 0.08 \text{ m} \] Convert this to meters since the magnetic field equation requires the radius in meters.
03

Rearrange the Formula to Solve for Current

Rearrange the formula for magnetic field to solve for the current \( I \): \[ I = \frac{2B \cdot R}{\mu_0} \] Now, we will substitute the known values into this formula.
04

Substitute the Known Values

Substitute the values of \( B = 3.0 \times 10^{-8} \; \text{G} = 3.0 \times 10^{-12} \; \text{T} \) (converting Gauss to Tesla), \( R = 0.08 \; \text{m} \), and \( \mu_0 = 4\pi \times 10^{-7} \; \text{T}\cdot\text{m/A} \) into the rearranged formula: \[ I = \frac{2 \times 3.0 \times 10^{-12} \times 0.08}{4\pi \times 10^{-7}} \]
05

Calculate the Current

Perform the calculation: \[ I = \frac{2 \times 3.0 \times 10^{-12} \times 0.08}{4 \times 3.1416 \times 10^{-7}} \] Simplify the expression to find \( I \): \[ I = \frac{4.8 \times 10^{-13}}{1.2566 \times 10^{-6}} \approx 3.82 \times 10^{-7} \; \text{A} \] Thus, the current needed is approximately \( 3.82 \times 10^{-7} \; \text{A} \).

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 Current Loop
A circular current loop is a simple, yet fascinating, way to understand how electricity can create magnetism. This concept involves a loop of wire through which an electric current flows. As current moves through this loop, it generates a magnetic field perpendicular to the plane of the loop.
This makes circular loops particularly interesting because they demonstrate the fundamental relationship between electricity and magnetism.
  • The strength of the magnetic field produced by a current loop depends on the current flowing through the wire.
  • The magnetic field is strongest at the center of the loop and weakens as we move away from the center.
  • Typically, this magnetic field is uniform inside the loop.
Imagine the magnetic field lines circling through the loop and extending outward. They form an invisible shell of sorts, showcasing the magnetic effects caused by the flow of electricity.
Permeability of Free Space
Permeability of free space, denoted as \( \mu_0 \), is a crucial constant in the understanding of magnetic fields. It represents the magnetic properties of the vacuum of space and acts as a measure of how much magnetic field can pass through space without any obstruction.
This constant is fundamental in physics, especially when applying the magnetic field formula to calculate the effects generated by current loops or wires.
  • The value of \( \mu_0 \) is approximately \( 4\pi \times 10^{-7} \; \text{T}\cdot\text{m/A} \), which is derived from experimental observations.
  • This constant helps determine the strength of the magnetic field produced by a given current in a loop.
  • In simple terms, it aids our comprehension of how vacuums influence magnetic field lines.
Utilizing \( \mu_0 \) is essential to solving problems involving magnetic fields created by electric currents, making it a cornerstone of electromagnetic theory.
Magnetic Field Formula
The magnetic field formula is a fundamental equation that relates several important physical quantities. It is used to calculate the magnetic field at the center of a circular current loop. This formula is expressed as:\[ B = \frac{\mu_0 \cdot I}{2R} \]Here, \( B \) is the magnetic field strength, \( I \) is the current flowing through the loop, and \( R \) is the radius of the loop. The equation tells us how each of these elements interplay to produce a magnetic field.
  • The magnetic field \( B \) is directly proportional to the current \( I \), meaning a higher current results in a stronger magnetic field.
  • It is inversely proportional to the radius \( R \), suggesting that a smaller loop results in a stronger magnetic field.
  • The presence of \( \mu_0 \), the permeability of free space, links magnetic properties to the vacuum through which the field is generated.
Applying this formula allows physicists and engineers to design systems where precise control over magnetic field strength is necessary, such as in electromagnets or MRI machines. Understanding this relationship is pivotal for students exploring electromagnetism and its applications.

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) How large a current would a very long, straight wire have to carry so that the magnetic field 2.00 cm from the wire is equal to 1.00 G (comparable to the earth's northward-pointing magnetic field)? (b) If the wire is horizontal with the current running from east to west, at what locations would the magnetic field of the wire point in the same direction as the horizontal component of the earth's magnetic field? (c) Repeat part (b) except the wire is vertical with the current going upward.

A square wire loop 10.0 cm on each side carries a clockwise current of 8.00 A. Find the magnitude and direction of the magnetic field at its center due to the four 1.20-mm wire segments at the midpoint of each side.

Two long, parallel transmission lines, 40.0 cm apart, carry 25.0-A and 75.0-A currents. Find all locations where the net magnetic field of the two wires is zero if these currents are in (a) the same direction and (b) the opposite direction.

Certain bacteria (such as \(Aquaspirillum\) \(magnetotacticum\)) tend to swim toward the earth's geographic north pole because they contain tiny particles, called magnetosomes, that are sensitive to a magnetic field. If a transmission line carrying 100 A is laid underwater, at what range of distances would the magnetic field from this line be great enough to interfere with the migration of these bacteria? (Assume that a field less than 5\(\%\) of the earth's field would have little effect on the bacteria. Take the earth's field to be 5.0 \(\times\) 10\(^{-5}\) T, and ignore the effects of the seawater.)

The current in the windings of a toroidal solenoid is 2.400 A. There are 500 turns, and the mean radius is 25.00 cm. The toroidal solenoid is filled with a magnetic material. The magnetic field inside the windings is found to be 1.940 T. Calculate (a) the relative permeability and (b) the magnetic susceptibility of the material that fills the toroid.

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.