/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 76 Two long, parallel wires separat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 28: Problem 76

Two long, parallel wires separated by a distance, \(d\), carry currents in opposite directions. If the left-hand wire carries a current \(i / 2,\) and the right-hand wire carries a current \(i\), determine where the magnetic field is zero.

Short Answer

Expert verified
Answer: The position where the magnetic field is zero is $\frac{d}{3}$ from the left-hand wire and $\frac{2d}{3}$ from the right-hand wire, where $d$ is the distance between the two wires.

Step by step solution

01

Determine the magnetic field produced by each wire

First, let's determine the magnetic field produced by each wire using the Biot-Savart Law. The magnetic field at a distance \(x\) from a wire carrying current \(i\) is given by: \(B = \frac{\mu_0 i}{2\pi x}\) where \(\mu_0\) is the permeability of free space, equal to \(4\pi \times 10^{-7} \, Tm/A\).
02

Set up an equation for the position where the magnetic fields cancel out

We want to find the position where the magnetic fields produced by the two wires cancel out. This means that \(B_1 = B_2\), where \(B_1\) is the magnetic field due to the left-hand wire and \(B_2\) is the magnetic field due to the right-hand wire. Let \(x_1\) be the distance from the left-hand wire to the position where the magnetic fields cancel, and \(x_2\) be the distance from the right-hand wire to the same position. The magnetic fields at this position are: \(B_1 = \frac{\mu_0 (i/2)}{2\pi x_1}\) \(B_2 = \frac{\mu_0 i}{2\pi x_2}\) Since we want to find the position where \(B_1 = B_2\), we can set the expressions for \(B_1\) and \(B_2\) equal to each other: \(\frac{\mu_0 (i/2)}{2\pi x_1} = \frac{\mu_0 i}{2\pi x_2}\)
03

Solve for the ratio between \(x_1\) and \(x_2\)

Now we need to solve the equation for the ratio between \(x_1\) and \(x_2\). We can first cancel the terms that appear in both sides of the equation: \(\frac{i/2}{x_1} = \frac{i}{x_2}\) Next, let's make \(x_2\) the subject of the equation: \(x_2 = \frac{2x_1}{(i/2)}\) We can further simplify it: \(x_2 = 2x_1\) This tells us that the position at which the magnetic fields cancel out is twice as far from the right-hand wire as it is from the left-hand wire. To determine the actual position along the wire, we can use the fact that \(x_1 + x_2 = d\): \(d = x_1 + 2x_1 = 3x_1\) Thus, \(x_1 = \frac{d}{3}\) and \(x_2 = \frac{2d}{3}\).
04

Conclusion

The position where the magnetic field is zero is \(\frac{d}{3}\) from the left-hand wire and \(\frac{2d}{3}\) from the right-hand wire.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Biot-Savart Law
The Biot-Savart Law is a critical principle in electromagnetism, specifically concerning the magnetic effects of electric currents. It provides a mathematical model for calculating the magnetic field produced at a point in space due to a current-carrying conductor. According to this law, the magnetic field \( B \) at a point is directly proportional to the current \( i \) and inversely proportional to the distance \( x \) from the conductor.

For a straight, infinite wire, the equation is expressed as:
\[ B = \frac{\mu_0 i}{2\pi x} \]
where \( \mu_0 \) is the permeability of free space, representing how much resistance the vacuum offers against the formation of the magnetic field. This law is indispensable when dealing with problems involving magnetic fields around currents, as evident in our textbook problem where it helps determine the magnetic field produced by two parallel wires carrying electric current.
Magnetic Fields Cancellation
Magnetic fields cancellation is a phenomenon where the magnetic fields from different sources combine to produce a net field of zero at certain points in space. This occurs when the magnetic fields are equal in magnitude but opposite in direction. In our exercise, we explored this concept by looking for a point where the magnetic fields produced by two currents in opposite directions cancel each other out.

Using the Biot-Savart Law, we find that at the point of cancellation, the field contributions from both wires must satisfy the condition:\[ B_1 = B_2 \]
By setting up and solving the equations, we determined the specific locations relative to the wires where this cancellation occurs. Understanding magnetic field cancellation is crucial for designing magnetic shielding and in applications that require neutral magnetic environments, such as in certain medical or scientific instrumentation.
Permeability of free space
The permeability of free space, denoted as \( \mu_0 \) and also known as the magnetic constant, is a fundamental physical constant that represents the measure of the magnetic permeability of the classical vacuum. Essentially, it's a measure of how much magnetic field is produced in a vacuum by an electric current or a magnetic moment. The value of \( \mu_0 \) is roughly \(4\pi \times 10^{-7} \, Tm/A\) (tesla meter per ampere).

In the equations derived from the Biot-Savart Law, \( \mu_0 \) appears as a multiplier, showing how it influences the strength of the magnetic field around a current-carrying wire. It's noteworthy that \( \mu_0 \) not only plays a role in this law but also in other electromagnetic phenomena, including the inductance of coils and the operation of electromagnetic waves.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You are standing at a spot where the magnetic field of the Earth is horizontal, points due northward, and has magnitude \(40.0 \mu \mathrm{T}\). Directly above your head, at a height of \(12.0 \mathrm{~m},\) a long, horizontal cable carries a steady \(\mathrm{DC}\) current of 500.0 A due northward. Calculate the angle \(\theta\) by which your magnetic compass needle is deflected from true magnetic north by the effect of the cable. Don't forget the sign of \(\theta-\) is the deflection eastward or westward?

A hairpin configuration is formed of two semiinfinite straight wires that are \(2.00 \mathrm{~cm}\) apart and joined by a semicircular piece of wire (whose radius must be \(1.00 \mathrm{~cm}\) and whose center is at the origin of \(x y z\) -coordinates). The top straight wire is along the line \(y=1.00 \mathrm{~cm},\) and the bottom straight wire is along the line \(y=-1.00 \mathrm{~cm} ;\) these two wires are in the left side \((x<0)\) of the \(x y\) -plane. The current in the hairpin is \(3.00 \mathrm{~A},\) and it is directed toward the right in the top wire, clockwise around the semicircle, and to the left in the bottom wire. Find the magnetic field at the origin of the coordinate system.

In a solenoid in which the wires are wound such that each loop touches the adjacent ones, which of the following will increase the magnetic field inside the magnet? a) making the radius of the loops smaller b) increasing the radius of the wire c) increasing the radius of the solenoid d) decreasing the radius of the wire e) immersion of the solenoid in gasoline

A square loop, with sides of length \(L\), carries current i. Find the magnitude of the magnetic field from the loop at the center of the loop, as a function of \(i\) and \(L\).

A loop of wire of radius \(R=25.0 \mathrm{~cm}\) has a smaller loop of radius \(r=0.900 \mathrm{~cm}\) at its center such that the planes of the two loops are perpendicular to each other. When a current of \(14.0 \mathrm{~A}\) is passed through both loops, the smaller loop experiences a torque due to the magnetic field produced by the larger loop. Determine this torque assuming that the smaller loop is sufficiently small so that the magnetic field due to the larger loop is same across the entire surface.
See all solutions

Recommended explanations on Physics Textbooks

Famous Physicists

Read Explanation

Fields in Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Space Physics

Read Explanation

Magnetism

Read Explanation

Conservation of Energy and Momentum

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.