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A long wire stretches along the \(x\) axis and carries a 3.0-A current to the right \((+x)\). The wire is in a uniform magnetic field \(\overrightarrow{\mathbf{B}}=(0.20 \hat{\mathbf{i}}-0.36 \hat{\mathbf{j}}+0.25 \hat{\mathbf{k}})\) T. Determine the components of the force on the wire per \(\mathrm{cm}\) of length.

Short Answer

Expert verified
The force components per cm on the wire are: \( F_x = 0 \) N/cm, \( F_y = -0.0075 \) N/cm, and \( F_z = 0.0108 \) N/cm.

Step by step solution

01

Understand the Force on a Current-Carrying Wire

The force on a segment of a wire carrying a current in a magnetic field is given by the equation \( \overrightarrow{\mathbf{F}} = I \cdot \left( \overrightarrow{\mathbf{L}} \times \overrightarrow{\mathbf{B}} \right) \), where \( I \) is the current, \( \overrightarrow{\mathbf{L}} \) is the vector length of the wire, and \( \overrightarrow{\mathbf{B}} \) is the magnetic field.
02

Identify the Components

The problem states that the wire stretches along the \( x \) axis, so the length vector \( \overrightarrow{\mathbf{L}} \) can be represented as \( L \hat{\mathbf{i}} \). For a 1 cm length, \( L = 0.01 \) m because we need to calculate force per cm.
03

Set-up the Cross Product

The force, per cm of the wire, can be found using the cross product \( \overrightarrow{\mathbf{L}} \times \overrightarrow{\mathbf{B}} = (0.01 \hat{\mathbf{i}}) \times (0.20 \hat{\mathbf{i}} - 0.36 \hat{\mathbf{j}} + 0.25 \hat{\mathbf{k}}) \). Only terms involving different unit vectors are non-zero, hence we get \( (0.01 \cdot 0.20) \hat{\mathbf{i}} \times \hat{\mathbf{i}} \rightarrow 0 \). The same calcualtion for all unit pairs.
04

Perform the Cross Product

Calculating only non-zero terms, we find \((0.01 \hat{\mathbf{i}}) \times (-0.36 \hat{\mathbf{j}}) = 0.01 \cdot 0.36 \hat{\mathbf{k}} = 0.0036 \hat{\mathbf{k}} \) m. Similarly, \((0.01 \hat{\mathbf{i}}) \times (0.25 \hat{\mathbf{k}}) = -0.0025 \hat{\mathbf{j}} \) m.
05

Calculate the Force Components

Since the current \( I = 3.0 \) A, plug into \( \overrightarrow{\mathbf{F}} = I \cdot \left( \overrightarrow{\mathbf{L}} \times \overrightarrow{\mathbf{B}} \right) \).1. \( F_j = 3.0 \times (-0.0025) = -0.0075 \) N2. \( F_k = 3.0 \times (0.0036) = 0.0108 \) N.3. \( F_i = 0 \) N, as shown before.
06

Conclude the Components of the Force per cm of Wire

The components of the force per cm on the wire are:- \( F_x = 0 \) N/cm- \( F_y = -0.0075 \) N/cm- \( F_z = 0.0108 \) N/cm.

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Key Concepts

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

Current-Carrying Wire
When talking about a current-carrying wire, imagine it as a pathway through which electrons flow. These electrons create an electric current, symbolized as \( I \), where its unit is measured in amperes (A). In simpler terms, a current-carrying wire is much like a garden hose, but instead of water, it carries an invisible flow of electrical juice called current.
For instance, a wire with a 3.0 A current, like the one mentioned in the original exercise, means that 3.0 coulombs of charge are flowing through the wire per second. This current can interact with other elements, such as a magnetic field, to produce some interesting phenomena.
Key elements to remember about current-carrying wires:
  • Current direction is important. It dictates how the wire interacts with external fields.
  • Current is measured in amperes (A).
  • The wire's length along a particular axis can be represented as a vector (e.g., along the \( x \) axis, it is \( L \hat{\mathbf{i}} \)).
Cross Product
The cross product is a mathematical operation that takes two vectors and combines them to produce a third vector. It’s particularly significant in physics when determining forces, like the magnetic force on a current-carrying wire. This operation tells us how two vectors interact in a three-dimensional space.
In the formula \( \overrightarrow{\mathbf{F}} = I \cdot \left( \overrightarrow{\mathbf{L}} \times \overrightarrow{\mathbf{B}} \right) \), the cross product \( \overrightarrow{\mathbf{L}} \times \overrightarrow{\mathbf{B}} \) produces a new vector perpendicular to both \( \overrightarrow{\mathbf{L}} \) and \( \overrightarrow{\mathbf{B}} \). This new vector is crucial for calculating the magnetic force.
Considerations when dealing with cross products:
  • The cross product of two parallel vectors is zero.
  • The right-hand rule helps determine the direction of the resultant vector.
  • Unit vector combinations, such as \( \hat{\mathbf{i}} \times \hat{\mathbf{j}} = \hat{\mathbf{k}} \), are helpful references for solving cross products.
Magnetic Field
A magnetic field is a region of space where magnetic forces can be observed. You can think of it as an invisible force field surrounding a magnet. In this field, other magnets or magnetic materials, including current-carrying wires, experience a force.
We represent a magnetic field with a vector \( \overrightarrow{\mathbf{B}} \), which showcases both its magnitude and direction. In the exercise, the magnetic field \( \overrightarrow{\mathbf{B}} = (0.20 \hat{\mathbf{i}} - 0.36 \hat{\mathbf{j}} + 0.25 \hat{\mathbf{k}}) \), suggests a field pointing in a 3D space with specific influences along the \( \hat{\mathbf{i}}, \hat{\mathbf{j}}, \) and \( \hat{\mathbf{k}} \) directions.
Key points to note about magnetic fields:
  • They exist around magnets and moving electric charges.
  • The strength and direction of the magnetic field are described by the vector \( \overrightarrow{\mathbf{B}} \).
  • Magnetic fields exert forces on current-carrying wires, which can be calculated using the cross product.

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Most popular questions from this chapter

Zeeman effect. In the Bohr model of the hydrogen atom, the electron is held in its circular orbit of radius \(r\) about its proton nucleus by electrostatic attraction. If the atoms are placed in a weak magnetic field \(\overrightarrow{\mathbf{B}}\), the rotation frequency of electrons rotating in a plane perpendicular to \(\overrightarrow{\mathbf{B}}\) is changed by an amount $$ \Delta f=\pm \frac{e B}{4 \pi m} $$ where \(e\) and \(m\) are the charge and mass of an electron. (a) Derive this result, assuming the force due to \(\overrightarrow{\mathbf{B}}\) is much less than that due to electrostatic attraction of the nucleus. (b) What does the \(\pm\) sign indicate?

A 3.40-g bullet moves with a speed of \(155 \mathrm{~m} / \mathrm{s}\) perpendicular to the Earth's magnetic field of \(5.00 \times 10^{-5} \mathrm{~T}\). If the bullet possesses a net charge of \(18.5 \times 10^{-9} \mathrm{C},\) by what distance will it be deflected from its path due to the Earth's magnetic field after it has traveled \(1.00 \mathrm{~km} ?\)

A proton and an electron have the same kinetic energy upon entering a region of constant magnetic field. What is the ratio of the radii of their circular paths?

The force on a wire is a maximum of \(7.50 \times 10-2 \mathrm{~N}\) when placed betw een the pole faces of a magnet. The current flows horizontally to the right and the magnetic field is vertical. The wire is observed to "jump" toward the observer when the current is turned on. (a) What type of magnetic pole is the top pole face? (b) If the pole faces have a diameter of \(10.0 \mathrm{~cm},\) estimate the current in the wire if the field is \(0.220 \mathrm{~T}\) (c) If the wire is tipped so that it makes an angle of \(10.0^{\circ}\) with the horizontal, what force will it now feel?

(I) If the restoring spring of a galvanometer weakens by 15\(\%\) over the years, what current will give full-scale deflection if it originally required 46\(\mu \mathrm{A}\) ?

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