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Chapter 28: Problem 16

A very long, straight horizontal wire carries a current such that \(8.20 \times 10^{18}\) 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 magnitude of the magnetic field produced by the wire at a point \(4.00 \) cm directly above it is \( 6.58 \times 10^{-6} \) T and its direction is towards the south.

Step by step solution

01

Convert Number of Electrons to Current

The current \( I \) in a conductor is defined as the charge \( Q \) flowing through it per unit of time \( t \). Given that the charge of a single electron is approximately \( e = -1.602 \times 10^{-19} \) C, we can express the current produced by the electrons as \( I = n \times e \), where \( n \) is the number of electrons per second. Substituting \( n = 8.20 \times 10^{18} \) s$^{-1}$ and \( e = -1.602 \times 10^{-19} \) C, we find that \( I = 8.20 \times 10^{18} \times -1.602 \times 10^{-19} = -1.31 \) A. Please note that the direction of the current (west to east) is opposite to the direction of electron flow.
02

Apply Ampere's law

We can now substitute the current \( I = 1.31 \) A, the permeability of free space \( \mu_{0} = 4\pi \times 10^{-7} \) Tm/A, and the distance \( r = 4.00 \times 10^{-2} \) m into the formula for Ampere's law such that \( B = \frac{\mu_{0}I}{2 \pi r} = \frac{4 \pi \times 10^{-7} \times 1.31}{2 \pi \times 4.00 \times 10^{-2}} = 6.58 \times 10^{-6} \) T.
03

Determine the Direction of Magnetic Field

The direction of the magnetic field can be found using the right-hand rule. If we point the thumb of our right hand in the direction of conventional current flow (west to east), then the fingers will curl in the direction of the magnetic field. Therefore, the magnetic field circulates around the wire in a counterclockwise direction. Above the wire (looking down), the magnetic field points south.

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Key Concepts

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

Ampere's Law
Ampere's Law is a fundamental principle in electromagnetism that relates the magnetic field around a current-carrying wire to the current itself. It is expressed mathematically as: \[B = \frac{\mu_{0} I}{2\pi r}\]where:
  • \(B\) is the magnetic field strength,
  • \(\mu_{0}\) is the permeability of free space (\(4\pi \times 10^{-7}\) Tm/A),
  • \(I\) is the electric current,
  • \(r\) is the radial distance from the wire.
This formula allows us to calculate the magnitude of the magnetic field at a certain distance from a current-carrying wire.
In the provided exercise, after calculating the current to be 1.31 A, Ampere's Law is applied to find the magnetic field at 4 cm above the wire.
By inserting these values into the formula, we see that the magnetic field generated by the wire is approximately \(6.58 \times 10^{-6}\) T. This highlights how current and distance affect the magnetic field strength around a wire.
right-hand rule
The right-hand rule is a mnemonic that helps us determine the direction of the magnetic field generated by a current. It is particularly useful when dealing with the circular magnetic field lines around a straight current-carrying wire.

How to use the right-hand rule:
  • Point your thumb in the direction of the conventional current flow (not electron flow), which in this scenario is from west to east along the wire.
  • Let your fingers naturally curl around the wire.
Your fingers now point in the direction of the magnetic field. In the exercise's context, if you apply this rule above the wire, your fingers will curl in a counterclockwise direction when viewed from above.
This means at the point above the wire (from our perspective looking down), the magnetic field points south. The right-hand rule is a quick and effective way to visualize and understand the orientation of magnetic fields around current-carrying wires.
current direction
Understanding the current direction is paramount in electromagnetism since it affects both the calculation and direction of the resulting magnetic field.

Current is defined as the flow of positive charge, meaning conventional current flows in the opposite direction to the flow of electrons. In the exercise, while electrons move from east to west, the conventional current direction is considered from west to east.
With this knowledge, you can apply formulas like Ampere's Law correctly and use the right-hand rule to ensure the magnetic field direction is determined correctly.
It's essential to distinguish between the two concepts:
  • Electrons' movement: Actual flow of negative charges, in this case from east to west.
  • Conventional current direction: For calculations, consider the flow of positive charge, west to east.
This understanding helps students avoid confusion when analyzing problems involving electric fields and currents.

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Most popular questions from this chapter

A long, straight, solid cylinder, oriented with its axis in the \(z\) -direction, carries a current whose current density is \(\overrightarrow{\boldsymbol{J}}\). The current density, although symmetric about the cylinder axis, is not constant but varies according to the relationship $$ \begin{array}{rlr} \overrightarrow{\boldsymbol{J}} & =\frac{2 I_{0}}{\pi a^{2}}\left[1-\left(\frac{r}{a}\right)^{2}\right] \hat{k} & \text { for } r \leq a \\ & =\mathbf{0} \quad & \text { for } r \geq a \end{array} $$ where \(a\) is the radius of the cylinder, \(r\) is the radial distance from the cylinder axis, and \(I_{0}\) is a constant having units of amperes. (a) Show that \(I_{0}\) is the total current passing through the entire cross section of the wire. (b) Using Ampere's law, derive an expression for the magnitude of the magnetic field \(\vec{B}\) in the region \(r \geq a\). (c) Obtain an expression for the current \(I\) contained in a circular cross section of radius \(r \leq a\) and centered at the cylinder axis. (d) Using Ampere's law, derive an expression for the magnitude of the magnetic field \(\vec{B}\) in the region \(r \leq a\). How do your results in parts (b) and (d) compare for \(r=a ?\)

A \(-4.80 \mu \mathrm{C}\) charge is moving at a constant speed of \(6.80 \times 10^{5} \mathrm{~m} / \mathrm{s}\) in the \(+x\) -direction relative to a reference frame. At the instant when the point charge is at the origin, what is the magneticfield vector it produces at the following points: (a) \(x=0.500 \mathrm{~m}, y=0\) \(z=0 ;\) (b) \(x=0, y=0.500 \mathrm{~m}, z=0 ;\) (c) \(x=0.500 \mathrm{~m}, y=0.500 \mathrm{~m}\) \(z=0 ;(\mathrm{d}) x=0, y=0, z=0.500 \mathrm{~m} ?\)

Currents in dc transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such lines near their homes could pose health dangers. For a line that has current \(150 \mathrm{~A}\) and a height of \(8.0 \mathrm{~m}\) above the ground, what magnetic field does the line produce at ground level? Express your answer in teslas and as a percentage of the earth's magnetic field, which is \(0.50 \mathrm{G}\). Is this value cause for worry?

A closely wound, circular coil with radius \(2.40 \mathrm{~cm}\) has 800 turns. (a) What must the current in the coil be if the magnetic field at the center of the coil is \(0.0770 \mathrm{~T}\) ? (b) At what distance \(x\) from the center of the coil, on the axis of the coil, is the magnetic field half its value at the center?

A plasma is a gas of ionized (charged) particles. When plasma is in motion, magnetic effects "squeeze" its volume, inducing inward pressure known as a pinch. Consider a cylindrical tube of plasma with radius \(R\) and length \(L\) moving with velocity \(\overrightarrow{\boldsymbol{v}}\) along its axis. If there are \(n\) ions per unit volume and each ion has charge \(q\), we can determine the pressure felt by the walls of the cylinder. (a) What is the volume charge density \(\rho\) in terms of \(n\) and \(q ?\) (b) The thickness of the cylinder "surface" is \(n^{-1 / 3}\). What is the surface charge density \(\sigma\) in terms of \(n\) and \(q ?\) (c) The current density inside the cylinder is \(\vec{J}=\rho \overrightarrow{\boldsymbol{v}}\). Use this result along with Ampere's law to determine the magnetic field on the surface of the cylinder. Denote the circumferential unit vector as \(\hat{\phi}\). (d) The width of a differential strip of surface current is \(R d \phi .\) What is the differential current \(d I_{\text {surface }}\) that flows along this strip? (e) What differential force is felt by this strip due to the magnetic field generated by the volume current? (f) Integrate to determine the total force on the walls of the cylinder; then divide by the wall area to obtain the pressure in terms of \(n, q, R\), and \(v\). (g) If a plasma cylinder with radius \(2.0 \mathrm{~cm}\) has a charge density of \(8.0 \times 10^{16}\) ions \(/ \mathrm{cm}^{3},\) where each ion has a charge of \(e=1.6 \times 10^{-19} \mathrm{C}\) and is moving axially with a speed of \(20.0 \mathrm{~m} / \mathrm{s},\) what is its pinch pressure?
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