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If the magnetic field due to a long, straight, current-carrying wire has a magnitude \(B\) at a distance \(R\) from the wire's center, how far away must you be (in terms of \(R\) ) for the magnetic field to decrease to \(B / 3 ?\)

Short Answer

Expert verified
You must be at a distance of 3R for the magnetic field to decrease to B/3.

Step by step solution

01

Understanding the Magnetic Field Formula

The magnetic field due to a long, straight, current-carrying wire can be calculated using Ampère's law. The formula is given by \( B = \frac{\mu_0 I}{2 \pi r} \), where \( B \) is the magnetic field, \( \mu_0 \) is the permeability of free space, \( I \) is the current, and \( r \) is the distance from the wire.
02

Identifying the Given Values

In the problem, it's stated that the magnetic field \( B \) at a distance \( R \) from the wire needs to change to \( \frac{B}{3} \). The new magnetic field \( B' = \frac{B}{3} \) when the distance is \( r' \).
03

Setting Up the Equation for New Distance

From the formula \( B = \frac{\mu_0 I}{2 \pi r} \), we express \( \frac{B}{3} \) for the new distance \( r' \):\[ \frac{B}{3} = \frac{\mu_0 I}{2 \pi r'} \].
04

Relating the Two Equations

We can equate the expressions for \( B \) and \( B' \ \frac{\mu_0 I}{2 \pi R} = B \) and \( \frac{\mu_0 I}{2 \pi r'} = \frac{B}{3} \).So, \( \frac{\mu_0 I}{2 \pi R} = 3 \cdot \frac{\mu_0 I}{2 \pi r'} \).
05

Solving for the New Distance

By simplifying, we see that \( R = 3r' \), which leads to:\( r' = 3R \). This means that the distance where the magnetic field measures \( \frac{B}{3} \) is \( 3R \).

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Key Concepts

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

Ampère's Law
Ampère's law is a fundamental principle in electromagnetism. It's a practical way to determine the relationship between electric currents and magnetic fields. According to this law, the magnetic field around a closed loop is proportional to the electric current flowing through the loop. This relationship is expressed mathematically as:
\[ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I \]
Where:
  • \( \oint \mathbf{B} \cdot d\mathbf{l} \) represents the line integral of the magnetic field \( \mathbf{B} \) around a closed path.
  • \( \mu_0 \) is the permeability of free space.
  • \( I \) is the current enclosed by the path.
Ampère's law allows us to calculate the magnetic field generated by a current-carrying wire, as seen in the given problem. It shows that the magnetic field's strength decreases with distance from the wire, emphasizing the inverse relationship between distance and field strength.
Current-Carrying Wire
A current-carrying wire is a common source of magnetic fields in physics. When an electric current flows through a wire, it generates a magnetic field that encircles the wire. The direction of this magnetic field is determined by the right-hand rule:
  • Point your thumb in the direction of the current.
  • Your fingers will curl in the direction of the magnetic field lines.
The magnetic field is strongest close to the wire and weakens as you move away. This concept is essential in explaining why at a distance 3R from a wire, the field is reduced to one-third of its strength at distance R. Understanding this interaction helps in designing electrical devices and managing electromagnetic fields in technology.
Permeability of Free Space
The permeability of free space, denoted as \( \mu_0 \), is a physical constant essential in electromagnetism. It quantifies the ability of a vacuum to support magnetic field lines. Its value is approximately \( 4\pi \times 10^{-7} \text{ T} ext{m/A} \) (Tesla meter per Ampere), which is fundamental in the equations that describe magnetic fields:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
Here, \( \mu_0 \) helps determine how effective a magnetic field is generated by a current in free space. It is a cornerstone in calculating the magnetic field effects around current-carrying conductors. Understanding \( \mu_0 \) is crucial for accurately solving problems related to magnetic interactions, like the one given, where it helps link the magnetic field to the current and the geometry of the system.
Magnetic Field Formula
The magnetic field formula for a long, straight, current-carrying wire is given by:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
This formula helps calculate the magnetic field \( B \) at a distance \( r \) from the wire. Several key components determine the field's strength:
  • \( \mu_0 \): Permeability of free space, which reflects how easily a magnetic field can pass through the vacuum.
  • \( I \): The current flowing through the wire, which is directly proportional to the strength of the magnetic field.
  • \( r \): The distance from the wire, which is inversely proportional to the magnetic field strength.
Knowing how to manipulate this formula is crucial for understanding various practical applications, from designing circuit boards to magnetic resonance imaging (MRI) in medical fields. It illustrates the foundational relationship between current, distance, and magnetic field strength.

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Most popular questions from this chapter

A circular coil of 50 loops and diameter \(20.0 \mathrm{~cm}\) is lying flat on a tabletop and carries a clockwise current of 2.50 A. A magnetic field of \(0.450 \mathrm{~T},\) directed to the north and at an angle of \(45.0^{\circ}\) from the vertical down through the coil and into the tabletop, is turned on. (a) What is the torque on the coil, and (b) which side of the coil (north or south) will tend to rise from the tabletop?

A beam of protons is accelerated through a potential difference of \(0.745 \mathrm{kV}\) and then enters a uniform magnetic field traveling perpendicular to the field. (a) What magnitude of field is needed to bend these protons in a circular are of diameter \(1.75 \mathrm{~m} ?\) (b) What magnetic field would be needed to produce a path with the same diameter if the particles were electrons having the same speed as the protons?

A solenoid that is \(35 \mathrm{~cm}\) long and contains 450 circular coils \(2.0 \mathrm{~cm}\) in diameter carries a 1.75 A current. (a) What is the magnetic field at the center of the solenoid, \(1.0 \mathrm{~cm}\) from the coils? (b) Suppose we now stretch out the coils to make a very long wire carrying the same current as before. What is the magnetic field \(1.0 \mathrm{~cm}\) from the wire's center? Is it the same as you found in part (a)? Why or why not?

A particle having a mass of \(0.195 \mathrm{~g}\) carries a charge of \(-2.50 \times 10^{-8} \mathrm{C} .\) The particle is given an initial horizontal northward velocity of \(4.00 \times 10^{4} \mathrm{~m} / \mathrm{s}\). What are the magnitude and direction of the minimum magnetic field that will balance the earth's gravitational pull on the particle?

(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 orientations of the vectors \(\overrightarrow{\boldsymbol{v}}, \overrightarrow{\boldsymbol{E}},\) and \(\overrightarrow{\boldsymbol{B}}\). (c) When the electric field is removed, what is the radius of the electron orbit? What is the period of the orbit?

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