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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?

Short Answer

Expert verified
The magnetic field is \(4 \times 10^{-5}\, \text{T}\), slightly less than Earth's magnetic field.

Step by step solution

01

Identify the Formula

To find the magnetic field created by a long straight wire, we use Ampere's Law in the form of the formula for the magnetic field around a straight conductor: \( B = \frac{\mu_0 I}{2\pi r} \), where \( \mu_0 \) (the permeability of free space) is \(4\pi \times 10^{-7}\, \text{T m/A} \), \( I \) is the current, and \( r \) is the distance from the wire.
02

Substitute the Values

Substitute the given values into the formula: \( I = 10\, \text{A} \) and \( r = 0.05\, \text{m} \). The formula becomes \( B = \frac{(4\pi \times 10^{-7}\, \text{T m/A}) \times 10\, \text{A}}{2\pi \times 0.05\, \text{m}} \).
03

Simplify the Formula

Carry out the multiplications and divisions:\[ B = \frac{4\pi \times 10^{-6}\, \text{T}}{2 \pi \times 0.05} \]This simplifies to:\[ B = \frac{4 \times 10^{-6}}{0.1} \]\[ B = 4 \times 10^{-5}\, \text{T} \]
04

Compare with Earth's Magnetic Field

The Earth's magnetic field is about \(5 \times 10^{-5}\, \text{T} \). Comparing the two, the magnetic field from the current is \(4 \times 10^{-5}\, \text{T} \), which is slightly less than the Earth's magnetic field.

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Key Concepts

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

Magnetic Field
A magnetic field is an invisible force that exerts influence on particles that are charged and in motion, like electrons in a wire. This field surrounds electric currents, magnets, and moving charges. Magnetic fields are often depicted by field lines, which indicate the field's direction and strength.

In everyday terms, the Earth itself has a magnetic field, which is why compass needles point north. Similarly, electrical devices, like a long straight wire carrying current, create their magnetic fields, though not as strong as the Earth's.

Using Ampere's Law, we can determine the strength of the magnetic field around a long straight wire using the formula: \[ B = \frac{\mu_0 I}{2\pi r} \]where:
  • \( B \) is the magnetic field in Tesla
  • \( \mu_0 \) is the permeability of free space
  • \( I \) is the current in Amperes
  • \( r \) is the distance from the wire in meters
This formula shows that the field strength decreases as we move further away from the wire.
Current
Current, denoted by \( I \), is the rate of flow of electric charge. It is typically measured in Amperes (A). When electrons move through a conductor like a wire, they create an electric current.

In the context of a household circuit, a current of 10 A is not uncommon. This current flows through wires and can create a magnetic field around them. The magnitude of this current directly affects the strength of the magnetic field created by the wire.
  • A larger current means stronger magnetic fields.
  • If the current changes direction, the magnetic field direction changes too.
Understanding how current influences the magnetic field is key when analyzing wire circuits and for practical applications such as designing electrical systems or magnetic devices.
Permeability of Free Space
The permeability of free space, represented by \( \mu_0 \), is a fundamental physical constant that expresses how much resistance the vacuum of space offers to the formation of a magnetic field. Its value is approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \).

This constant is crucial when calculating magnetic fields using Ampere's Law since it helps us understand how efficiently a magnetic field can be established in a given environment. In simpler terms, think of \( \mu_0 \) as the magnetic field's permission slip to exist in space.

Why is it important? Well, it affects the calculation of magnetic field strength in any scenario involving a vacuum, which is the perfect model for many real-life situations like electronic circuits and electromagnetic calculations.
  • It is used in the formula \( B = \frac{\mu_0 I}{2\pi r} \) to calculate magnetic fields.
  • The value tells us how easily magnetic field lines form in space; the smaller the resistance, the easier it is for the field to form.
Therefore, understanding \( \mu_0 \) is essential for anyone studying magnetism or working with electromagnetic applications.

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Most popular questions from this chapter

\(\bullet\) As a new electrical technician, you are designing a large solenoid to produce a uniform 0.150 T magnetic field near its center. You have enough wire for 4000 circular turns, and the solenoid must be 1.40 m long and 2.00 \(\mathrm{cm}\) in diameter. What current will you need to produce the necessary field?

Magnetic force on a lightning bolt. Currents during lightning strikes can be up to \(50,000\) A (or more!). We can model such a strike as a \(50,000\) A vertical current perpendicular to the earth's magnetic field, which is about \(\frac{1}{2}\) gauss. What is the force on each meter of this current due to the earth's magnetic field?

A horizontal rod 0.200 \(\mathrm{m}\) long carries a current through a uniform horizontal magnetic field of magnitude 0.067 T that points perpendicular to the rod. If the magnetic force on this rod is measured to be \(0.13 \mathrm{N},\) what is the current flowing through the rod?

\(\bullet\) In a cloud chamber experiment, a proton enters a uniform 0.250 T magnetic field directed perpendicular to its motion. You measure the proton's path on a photograph and find that it follows a circular arc of radius 6.13 \(\mathrm{cm} .\) How fast was the proton moving?

\(\bullet\) Singly ionized (onne electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is 155 \(\mathrm{V} / \mathrm{m}\) and the magnetic field is 0.0315 \(\mathrm{T}\) . The ions next enter a uniform magnetic field of magnitude 0.0175 \(\mathrm{T}\) that is oriented perpendicular to their velocity. (a) How fast are the ions moving when they emerge from the velocity selector? (b) If the radius of the path of the ions in the second magnetic field is \(17.5 \mathrm{cm},\) what is their mass?

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