Problem 60 - Calculus - AP Edition

Question

A long high-voltage power line is 18 feet above the ground. The electric current in the line generates a magnetic field whose magnitude \(F\) (in microtesla) is given by \(F=36 / L\) where \(L\) is the (perpendicular) distance to the line in feet. Suppose that a new regulation requires the power line to be raised everywhere by \(0.9 \mathrm{ft} .\) Use a differential to estimate the decrease in the value of \(F\) at a point on the ground directly beneath the line.

Step-by-Step Solution

Answer

The magnetic field decreases by approximately 0.1 microteslas.

Step 1: Understanding the Problem

We need to estimate the change in the magnetic field strength when the power line is raised by 0.9 feet. The original distance from the ground to the line is 18 feet.

Step 2: Setting Up the Formula

The formula for the magnetic field strength is given by \( F = \frac{36}{L} \), where \( L = 18 \) feet initially. We need to find the change in \( F \) when \( L \) increases by 0.9 feet.

Step 3: Finding the Differential

Use the differential formula \( dF \approx \frac{dF}{dL} \times dL \). First, find the derivative of \( F \) with respect to \( L \), which is \( \frac{dF}{dL} = -\frac{36}{L^2} \).

Step 4: Substitute Values into the Derivative

Substitute \( L = 18 \) feet into \( \frac{dF}{dL} = -\frac{36}{L^2} \) to find \( \frac{dF}{dL} = -\frac{36}{18^2} = -\frac{36}{324} = -\frac{1}{9} \).

Step 5: Calculate the Estimated Change in \( F \)

Using the differential form, \( dF \approx \frac{dF}{dL} \times dL \), where \( dL = 0.9 \) feet, find \( dF \approx -\frac{1}{9} \times 0.9 = -0.1 \).

Step 6: Interpret the Result

The decrease in the magnitude of the magnetic field \( F \) is approximately \(-0.1\) microteslas when the height of the power line is increased by 0.9 feet.

Key Concepts

Magnetic Field

A magnetic field is a vector field that represents the magnetic influence of electric currents and magnetic materials. It is present around magnets and electric currents, and it generates a force on other nearby magnets and currents. In this context, the magnetic field is generated by the electric current flowing through the power line. The strength of this magnetic field at a given distance can be represented with the formula:
\[ F = \frac{36}{L} \]Where \(F\) is the magnitude of the magnetic field in microtesla, and \(L\) is the perpendicular distance from the line in feet.
The magnetic field decreases as you move further away from the source. This is because the force exerted by the magnetic field is inversely proportional to the distance from the source. Hence, as \(L\) increases, \(F\) decreases.

Electric Current

Electric currents are streams of charged particles, such as electrons or ions, moving through an electrical conductor or space. They play a crucial role in generating magnetic fields. When electric current flows through a conductor, like a power line, it creates a magnetic field around it.
This is the principle at work in power lines, which transmit electricity over long distances efficiently. The magnetic field produced by the electric current provides essential data for understanding how much electromagnetic interference (EMI) might be affecting nearby devices or structures.
The magnitude of the magnetic field depends significantly on the amount of current flowing and the distance from the conductor. Thus, managing electric current is critical in reducing harmful magnetic interference.

Differential Approximation

Differential approximation is a powerful tool in calculus used to estimate small changes in a function's value. It allows us to approximate how a small change in one variable affects another variable, which can simplify complex real-world problems like the one given.
In the context of this problem, differential approximation helps us estimate the change in the magnetic field \( F \) as the height of the power line changes by \(0.9\) feet. We start by finding the derivative of \( F \) with respect to \( L \), which tells us the rate at which \( F \) changes:
\[ \frac{dF}{dL} = -\frac{36}{L^2} \]This derivative helps in setting up the differential:

• \( dF \approx \frac{dF}{dL} \times dL \)

Substituting the given values, \( L = 18 \) and \( dL = 0.9 \), gives:
\[ dF \approx -\frac{1}{9} \times 0.9 = -0.1 \]This result shows a decrease of approximately \(0.1\) microteslas in the field's strength, emphasizing the practical utility of differential approximation in estimating small changes.