Problem 68 - Fundamentals of Physics

Question

A wire lying along a \(y\) axis from \(y=0\) to \(y=0.250 \mathrm{~m}\) carries a current of \(2.00 \mathrm{~mA}\) in the negative direction of the axis. The wire fully lies in a nonuniform magnetic field that is given by \(\vec{B}=(0.300 \mathrm{~T} / \mathrm{m}) y \hat{\mathrm{i}}+(0.400 \mathrm{~T} / \mathrm{m}) y \hat{\mathrm{j}} .\) In unit-vector notation, what is the magnetic force on the wire?

Step-by-Step Solution

Answer

The magnetic force on the wire is \( \vec{F} = 1.875 \times 10^{-5} \hat{k} \text{ N} \).

Step 1: Identify the given parameters

We have a wire carrying a current of \( I = 2.00 \text{ mA} = 2.00 \times 10^{-3} \text{ A} \) along the negative \( y \)-axis from \( y = 0 \) to \( y = 0.250 \text{ m} \). The magnetic field is given by \( \vec{B}=(0.300 \text{ T/m}) y \hat{\mathrm{i}}+(0.400 \text{ T/m}) y \hat{\mathrm{j}} \). The goal is to find the magnetic force on the wire in unit-vector notation.

Step 2: Recall the formula for magnetic force

The magnetic force on a current-carrying wire in a magnetic field is given by the formula \( \vec{F} = I \int \vec{dl} \times \vec{B} \), where \( \vec{dl} \) is a differential length element of the wire.

Step 3: Define the differential length element

Since the wire lies along the \( y \)-axis, the differential length element \( \vec{dl} \) can be defined as \( \vec{dl} = dy \hat{\mathrm{j}} \).

Step 4: Compute the cross product

The cross product \( \vec{dl} \times \vec{B} \) is \( dy \hat{\mathrm{j}} \times \, [(0.300 y \hat{\mathrm{i}}) + (0.400 y \hat{\mathrm{j}})] \). This simplifies to \( (0.300 y \hat{k}) dy \), as \( \hat{\mathrm{j}} \times \hat{\mathrm{i}} = \hat{k} \) and \( \hat{\mathrm{j}} \times \hat{\mathrm{j}} = 0 \).

Step 5: Integrate the cross product over the wire length

Integrate \( (0.300 y \hat{k}) dy \) from \( y = 0 \) to \( y = 0.250 \). This gives \( \int_0^{0.250} 0.300 y \hat{k} \, dy = [0.300 \frac{y^2}{2} \hat{k}]_0^{0.250} = 0.300 \times \frac{(0.250)^2}{2} \hat{k} = 0.009375 \hat{k} \).

Step 6: Calculate the magnetic force on the wire

Substitute the integrated value back into the force equation: \( \vec{F} = I \cdot 0.009375 \hat{k} = (2.00 \times 10^{-3}) \cdot 0.009375 \hat{k} = 1.875 \times 10^{-5} \hat{k} \text{ N} \).

Step 7: Express the magnetic force in unit-vector notation

The magnetic force on the wire is \( \vec{F} = 1.875 \times 10^{-5} \hat{k} \text{ N} \), which is directed in the \( \hat{k} \) direction.

Key Concepts

Cross Product

In physics, the cross product is a fundamental operation used to determine the magnetic force on a wire. This operation involves two vectors—here, the length element of the wire (\( \vec{dl} \)) and the magnetic field vector (\( \vec{B} \)). To find the direction of the force on a current-carrying wire, we apply the right-hand rule to this cross product.
The formula for force is \( \vec{F} = I \int \vec{dl} \times \vec{B} \), where:

• \( I \) is the current.
• \( \vec{dl} \) is the infinitesimal vector along the wire.
• \( \vec{B} \) is the magnetic field.

Here, the solution showed how the cross product helps in finding the z-direction component of magnetic force. A crucial part to remember is that the cross product's result is always perpendicular to both vectors being multiplied.
In this case, \( \hat{\mathrm{j}} \times \hat{\mathrm{i}} \) results in \( \hat{k} \), introducing the third dimension—out of the original x-y plane.

Nonuniform Magnetic Field

A nonuniform magnetic field means that the field varies with position. In this exercise, the magnetic field \( \vec{B} = (0.300 \mathrm{~T/m}) y \hat{\mathrm{i}} + (0.400 \mathrm{~T/m}) y \hat{\mathrm{j}} \) changes with the y-coordinate, dependent on the position along the wire.
The significance of a nonuniform field is that the force computation requires evaluating the force contribution at each point separately. This is different from a uniform field, where one could simply apply a straightforward multiplication. Because \( \vec{B} \) is a function of y here, it means that different segments of the wire experience different magnetic forces.
Understanding the impact of nonuniformity ensures that the force calculations are accurate, accounting for all parts of the wire placed in varying magnetic intensities.

Current-Carrying Wire

A current-carrying wire produces a magnetic force when placed in a magnetic field. The direction of the current directly affects the resultant magnetic force.
In this problem, the wire carries a current of \( 2.00 \text{ mA} \) along the negative y-axis. This flow of charges interacts with the surrounding magnetic field, contributing to the magnetic force.
The orientation and magnitude of the current play a crucial role in determining the resultant force vector. Be aware that reversing current direction will generally reverse the direction of the magnetic force, showing the vector nature of these quantities.

Integration of Forces

Integration is used in determining the net magnetic force when dealing with varying conditions, such as a nonuniform magnetic field. In the exercise, the integration process computes the cumulative effect of magnetic forces over the wire's length.
In this case, \( \int_0^{0.250} 0.300 y \hat{k} \, dy \) captures how different segments of the wire, starting from \( y=0 \) to \( y=0.250 \), contribute to the total force.

• The integrand \( 0.300 y \hat{k} \) reflects the cross product's impact on the force direction.
• The limits of integration ( 0 to 0.250 m ) correspond to the wire's physical boundaries.

Through integration, one essentially sums up the infinitesimal contributions of force along the wire, simplifying it into a singular force vector that acts upon the entire wire.