Problem 74 - University Physics with Modern Physics

Question

A wire 25.0 \(\mathrm{cm}\) long lies along the \(z\) -axis and carries a current of 9.00 \(\mathrm{A}\) in the \(+z\) -direction. The magnetic field is uniform and has components \(B_{x}=-0.242 \mathrm{T}, B_{y}=-0.985 \mathrm{T}\) , and \(B_{z}=\) \(-0.336 \mathrm{T} .\) (a) Find the components of the magnetic force on the wire. (b) What is the magnitude of the net magnetic force on the wire?

Step-by-Step Solution

Answer

\(F_x = 2.22\, \mathrm{N},\, F_y = 0.54\, \mathrm{N},\, F_z = 0\, \mathrm{N}"; magnitude = 2.28 \mathrm{N}."

Step 1: Identify the relevant formula

To find the magnetic force on the wire, we'll use the formula for the magnetic force on a current-carrying conductor: \[ \vec{F} = I \vec{L} \times \vec{B} \]where \( I \) is the current, \( \vec{L} \) is the vector representing the length of the wire, and \( \vec{B} \) is the magnetic field.

Step 2: Define the vectors

Since the wire is along the "+z" direction, the length vector \( \vec{L} \) can be defined as:\[ \vec{L} = 0\hat{i} + 0\hat{j} + 0.25\hat{k} \ \mathrm{m} \]The magnetic field vector \( \vec{B} \) is:\[ \vec{B} = -0.242 \hat{i} - 0.985 \hat{j} - 0.336 \hat{k} \]

Step 3: Calculate the cross product \(\vec{L} \times \vec{B}\)

The cross product \(\vec{L} \times \vec{B}\) is calculated as follows:\[\vec{L} \times \vec{B} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \0 & 0 & 0.25 \-0.242 & -0.985 & -0.336\end{vmatrix}\]Calculating the determinants, we find:\[\vec{L} \times \vec{B} = \left[(0)(-0.336) - (0.25)(-0.985)\right] \hat{i} - \left[(0)(-0.336) - (0.25)(-0.242)\right] \hat{j} + [0 - 0] \hat{k}\] \[= 0.24625 \hat{i} + 0.0605 \hat{j} \]

Step 4: Multiply by the current to get \(\vec{F}\)

Using the current \( I = 9.00\, \mathrm{A} \), multiply the result of the cross product by the current to find the magnetic force:\[\vec{F} = 9.00 \times (0.24625 \hat{i} + 0.0605 \hat{j})\]\[\vec{F} = 2.21625 \hat{i} + 0.5445 \hat{j}\] \mathrm{N}Thus, the force components on the wire are:\(F_x = 2.21625 \), \(F_y = 0.5445 \), and \(F_z = 0\) \(\mathrm{N}\).

Step 5: Calculate the magnitude of the net force

The magnitude of the net magnetic force is given by the magnitude of \(\vec{F}\):\[|\vec{F}| = \sqrt{(F_x)^2 + (F_y)^2 + (F_z)^2}\]\[= \sqrt{(2.21625)^2 + (0.5445)^2 + 0^2}\]\[= \sqrt{4.91041 + 0.29646}\]\[= \sqrt{5.20687}\]\[= 2.28293 \mathrm{N}\]

Key Concepts

Cross Product

The cross product is a mathematical operation crucial for finding the magnetic force on a current-carrying wire. This operation is important because it calculates the force by multiplying two vectors: the current direction and the magnetic field. When calculating the cross product, we often use the determinant method, especially for three-dimensional vector components. This is represented as a matrix where the top row holds the unit vectors \( \hat{i}, \hat{j}, \hat{k} \). The subsequent rows contain the components of the length vector \( \vec{L} \) and the magnetic field vector \( \vec{B} \). By solving this determinant, we can find a new vector that gives the direction and magnitude of the force.
This operation captures the integral vector nature of the magnetic force, ensuring that the result

• tells us the direction of the force, which is perpendicular to both the current and magnetic field directions
• has a magnitude that depends on the sine of the angle between these vectors

Understanding the cross product allows students to effectively apply the right-hand rule and predict the direction of magnetic forces.

Magnetic Field Components

Components of a magnetic field refer to its individual parts along the x, y, and z axes. In this exercise, the magnetic field is represented as having components \( B_x = -0.242 \), \( B_y = -0.985 \), and \( B_z = -0.336 \) Tesla.
These components give us critical information. They tell us the magnetic field's structure and allow us to predict how it interacts with the current. A uniform magnetic field means these components remain constant across the region considered, helping simplify calculations. Each component affects the resulting force based on vector operations. Understanding these interactions is vital, as they influence the force experienced by the wire.
When solving problems, always:

• Be mindful of the sign of each component, as it affects the calculation direction.
• Recognize that these components cumulatively contribute to the force the wire experiences.

Vector Calculations

Vector calculations are a cornerstone of physics problems involving electricity and magnetism. They help solve problems where quantities have both magnitude and direction. In this particular exercise, we used vector calculations to determine the magnetic force on a current-carrying wire. To tackle problems with vectors:

• Represent vectors using their components based on the given axes.
• Apply vector operations such as addition, subtraction, and importantly, the cross product to find the needed vector quantities.

These methods ensure precise calculations of forces and movements in a multi-dimensional space. Vectors not only provide a powerful way to quantify forces but also help visualize physical phenomena. Through careful set-up of vectors and rigorous calculations, students can solve complex problems reliably and confidently.
For students, mastering vector calculations can greatly enhance their understanding and ability to navigate physics problems effectively.