Problem 84 - Fundamentals of Physics

Question

A wire lying along an \(x\) axis from \(x=0\) to \(x=1.00 \mathrm{~m}\) carries a current of \(3.00 \mathrm{~A}\) in the positive \(x\) direction. The wire is immersed in a nonuniform magnetic field that is given by \(\vec{B}=\) \(\left(4.00 \mathrm{~T} / \mathrm{m}^{2}\right) x^{2 \hat{i}}-\left(0.600 \mathrm{~T} / \mathrm{m}^{2}\right) x^{2} \mathrm{j} .\) In unit-vector notation, what is the magnetic force on the wire?

Step-by-Step Solution

Answer

The magnetic force on the wire is \(-0.60 \hat{k} \, \mathrm{N}\).

Step 1: Understand the Formula

The magnetic force on a wire carrying current in a magnetic field is given by the formula \( \vec{F} = I \int \vec{dl} \times \vec{B} \). Here, \( I = 3.00 \, \mathrm{A} \) is the current, \( \vec{dl} \) is the differential vector length along the wire, and \( \vec{B} \) is the magnetic field.

Step 2: Set Up the Equation

Each small section of the wire can be expressed as a differential length vector, \( \vec{dl} = dx \, \hat{i} \), because the wire is along the x-axis. So the force on an infinitesimal section can be given by \( d\vec{F} = I (\vec{dl} \times \vec{B}) = I(dx \, \hat{i} \times \vec{B}) \).

Step 3: Calculate the Cross Product

We need to find the cross product \( \hat{i} \times \vec{B} = \hat{i} \times [(4.00 \, \mathrm{T}/\mathrm{m}^2)x^2 \hat{i} - (0.600 \, \mathrm{T}/\mathrm{m}^2) x^2 \hat{j}] \). Use the cross product identities: \( \hat{i} \times \hat{i} = 0 \) and \( \hat{i} \times \hat{j} = \hat{k} \). Hence, \( \hat{i} \times \vec{B} = -0.600 \, x^2 \hat{k} \).

Step 4: Integrate to Find Total Force

Substitute back into the equation: \( d\vec{F} = -3.00 \, \mathrm{A} \cdot 0.600 \, x^2 dx \, \hat{k} = -1.80 \, x^2 dx \, \hat{k} \). Integrate this from \( x=0 \) to \( x=1.00 \ \), the length of the wire. The integral is \( \int_0^1 -1.80 \, x^2 dx \hat{k} = -1.80 \left[ \frac{x^3}{3}\right]_0^1 \hat{k} = -1.80 \times \frac{1}{3} \hat{k} = -0.60 \hat{k} \).

Step 5: Finalize the Answer

Thus, the magnetic force on the wire is given by \( \vec{F} = -0.60 \hat{k} \, \mathrm{N} \). Since all calculations are along the x and defined components, the answer is confirmed in terms of the unit vector notation.

Key Concepts

current-carrying wire

A current-carrying wire is a fundamental concept where an electric current flows through a wire. In this scenario, the wire exists along a specific path, like the x-axis, and carries a steady current, denoted as \( I \). For example, in our case, the wire carries a current of 3.00 A along the positive x-direction. This steady flow of charge creates a situation where the wire can interact with a magnetic field present in the environment.
This is significant because the flow of current generates its own magnetic field, which can then interact with external magnetic fields. This interaction is precisely what results in the magnetic force exerted on the wire. It is crucial to consider that the force is dependent not only on the current but also on the orientation and length of the wire in the magnetic field.

magnetic field

The magnetic field is a vector field surrounding magnetic materials and currents, represented by the symbol \( \vec{B} \). In this exercise, the nonuniform magnetic field is given by the expression \( \vec{B} = (4.00 \mathrm{~T/m}^{2}) x^{2} \hat{i} - (0.600 \mathrm{~T/m}^{2}) x^{2} \hat{j} \).
This means that the magnetic field changes with the position along the wire. Components in the x-direction \((x^2 \hat{i})\) and y-direction \((x^2 \hat{j})\) together describe its magnitude and direction. The nonuniform nature implies variations in strength and direction at different points along the wire.

• This field acts on the wire, impacting the electrons traveling along it.
• The interaction between the current and magnetic field is what leads to force exertion on the wire, calculated through vector operations such as the cross product.

cross product

The cross product is a mathematical operation used to determine the vector perpendicular to two given vectors. In the context of magnetic forces, it helps in evaluating the direction and magnitude of the force experienced by a current-carrying wire in a magnetic field.
In our scenario, we compute the cross product as \( \hat{i} \times \vec{B} \). This simplifies calculating the interaction force between the current element and the magnetic field:

• The identities \( \hat{i} \times \hat{i} = 0 \) and \( \hat{i} \times \hat{j} = \hat{k} \) are employed.
• It results in the force direction being along the \( \hat{k} \) axis (or z-axis) because the force is perpendicular to both the current’s direction and the field’s direction.

This specific operation allows us to find that the magnetic force is \(-0.600 \, x^2 \hat{k}\), acting in the z-direction, at every segment along the wire.

vector calculus

Vector calculus involves using calculus methods on vector fields, and is essential in solving problems related to magnetic forces on wires. In this exercise, a basic operation involves integrating the force over the length of the wire to find the total force.
Since the force on an infinitesimally small section of wire is represented as \( d\vec{F} = -1.80 \, x^2 \, dx \, \hat{k} \), this force needs to be summed (integrated) over the entire length of the wire from 0 to 1.00 m.
The integral, \( \int_0^1 -1.80 \, x^2 \, dx \), is evaluated to find the total exerted force.

• Integration aggregates the small forces across the wire, giving the total magnetic force on the wire.
• This results in \( -0.60 \, \hat{k} \, \text{N} \), showing the magnitude and direction of the force acting along the z-axis.

Vector calculus thus enables the transition from micro-scale forces to understanding the overall force exerted on the wire.