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Chapter 28: Problem 16

A very long, straight horizontal wire carries a current such that \(3.50 \times 10^{18}\) electrons per second pass any given point going from west to east. What are the magnitude and direction of the magnetic field this wire produces at a point 4.00 \(\mathrm{cm}\) directly above it?

Short Answer

Expert verified
The magnetic field is \(2.8 \times 10^{-6}\) T pointing north.

Step by step solution

01

Understand the Problem

We are tasked with finding the magnetic field produced by a long, straight current-carrying wire at a point above it. Given the number of electrons passing a point per second, we'll first convert this to current, then use it to find the magnetic field.
02

Convert Electron Flow to Current

The number of electrons per second is given as \(3.50 \times 10^{18}\). Since each electron has a charge of \(1.6 \times 10^{-19}\) C, the current \(I\) is calculated by \(I = 3.50 \times 10^{18} \times 1.6 \times 10^{-19} = 0.56\) A, flowing west to east.
03

Use Biot-Savart Law for Magnitude

For a long, straight wire, the magnetic field at a distance \(r\) is \(B = \frac{\mu_0 I}{2 \pi r}\). With \(\mu_0 = 4\pi \times 10^{-7}\, \text{T}\cdot\text{m/A}\) and \(r = 0.04\) m, substitute these into the equation to get \(B = \frac{(4\pi \times 10^{-7}) \times 0.56}{2 \pi \times 0.04} = 2.8 \times 10^{-6}\) T.
04

Determine Direction Using Right-Hand Rule

By the right-hand rule, point your thumb in the direction of the current (west to east). Your fingers curl in the direction of the magnetic field. At a point above the wire, this results in the magnetic field pointing north.

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Key Concepts

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

Biot-Savart Law
The Biot-Savart Law is essential for understanding how magnetic fields originate from moving charges or electric currents. This law gives us the mathematical way to calculate the magnetic field produced at a specific point in space by a current-carrying conductor. It's particularly useful when dealing with geometry and configurations where Ampere鈥檚 Law becomes challenging to apply.

In essence, the Biot-Savart Law states that the magnetic field, denoted as \( B \), at a point is proportional to the current \( I \) and depends inversely on the distance \( r \) from the current element. For an infinitely straight conductor, the calculation simplifies to the formula:
\[B = \frac{\mu_0 I}{2\pi r}\]
where \(\mu_0\) is the permeability of free space, valued at \(4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}\). This formula is useful for finding the magnetic field at any point around a long, straight wire, like the example from the exercise.
  • The direct proportionality to current means stronger currents produce stronger fields.
  • The inversely proportional relationship with distance reminds us that moving further from the wire decreases the magnetic field strength.
Right-Hand Rule
The Right-Hand Rule is a handy mnemonic for understanding the direction of the magnetic field relative to the current in a wire. It simplifies the potentially confusing task of determining magnetic field directions in three dimensions.

To apply the right-hand rule, you use your right hand to mimic the orientation of the current and the resultant magnetic field.
  • Extend your thumb in the direction of the current flow. In the scenario from the exercise, this means pointing your thumb from west to east.
  • Your fingers then naturally curl around in the direction the magnetic field lines circle the wire.
  • Above the wire, your fingers point north, indicating that the magnetic field at that point wraps around the wire upwards towards the north.

The right-hand rule is intuitive once practiced, making it a powerful tool for visualizing and verifying field directions without calculations.
Current in a Wire
Current represents the flow of electric charge, and in our exercise, this flow is quantified by the number of electrons per second. Electron flow presents its behavior slightly differently from the conventional current since in reality, electrons move in the opposite direction of the conventional current.

In the given exercise, we compute the current \( I \) by multiplying the given electron flow rate by the charge per electron \( e \). This transformation to current (measured in amperes, A) becomes:
\[I = (3.50 \times 10^{18} \text{ electrons/s}) \times (1.6 \times 10^{-19} \text{ C/electron}) = 0.56 \text{ A}\]
This transformation helps us understand how much charge is moving past a certain point in the wire per second, highlighting the nature of current as a rate of charge flow.
  • With current established, the problem can then utilize this value to determine the magnetic field produced.
  • Understanding that the current flows west to east guides how we interpret and apply the right-hand rule for magnetic field direction.

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Most popular questions from this chapter

A wooden ring whose mean diameter is 14.0 \(\mathrm{cm}\) is wound with a closely spaced toroidal winding of 600 turns. Compute the magnitude of the magnetic field at the center of the cross section of the windings when the current in the windings is 0.650 \(\mathrm{A}\) .

Two identical circular, wire loops 40.0 \(\mathrm{cm}\) in diameter each carry a current of 1.50 \(\mathrm{A}\) in the same direction. These loops are parallel to each other and are 25.0 \(\mathrm{cm}\) apart. Line \(a b\) is normal to the plane of the loops and passes through their centers. A proton is fired at 2400 \(\mathrm{km} / \mathrm{sperpendicular}\) to line \(a b\) from a point midway between the centers of the loops. Find the magnitude and direction of the magnetic force these loops exert on the proton just after it is fired.

Fields within the Atom. In the Bohr model of the hydrogen atom, the electron moves in a circular orbit of radius \(5.3 \times 10^{-11} \mathrm{m}\) with a speed of \(2.2 \times 10^{6} \mathrm{m} / \mathrm{s}\) . If we are viewing the atom in such a way that the electron's orbit is in the plane of the paper with the electron moving clockwise, find the magnitude and direction of the electric and magnetic fields that the electron produces at the location of the nucleus (treated as a point).

A Charged Dielectric Disk. A thin disk of dielectric material with radius \(a\) has a total charge \(+Q\) distributed uniformly over its surface. It rotates \(n\) times per second about an axis perpendicular to the surface of the disk and passing through its center. Find the magnetic field at the center of the disk. (Hint: Divide the disk into concentric rings of infinitesimal width.)

A long solenoid with 60 turns of wire per centimeter carries a current of 0.15 \(\mathrm{A}\) . The wire that makes up the solenoid is wrapped around a solid core of silicon steel \(\left(K_{\mathrm{m}}=5200\right) .\) (The wire of the solenoid is jacketed with an insulator so that none of the current flows into the core.) (a) For a point inside the core, find the magnitudes of (i) the magnetic field \(\overrightarrow{\boldsymbol{B}}_{0}\) due to the solenoid current; (ii) the magnetization \(\vec{M} ;\) (iii) the total magnetic field \(\overrightarrow{\boldsymbol{B}}\) . (b) In a sketch of the solenoid and core, show the directions of the vectors \(\overrightarrow{\boldsymbol{B}}, \overrightarrow{\boldsymbol{B}}_{0}\) , and \(\overrightarrow{\boldsymbol{M}}\) inside the core.
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