/*! 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 20 Each of the given formulas arise... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 20

Each of the given formulas arises in the technical or scientific area of study shown. Solve for the indicated letter. \(B=\frac{\mu_{0} I}{2 \pi R},\) for \(R \quad\) (magnetic field)

Short Answer

Expert verified
\( R = \frac{\mu_{0} I}{2 \pi B} \)

Step by step solution

01

Identify the equation

We start with the given equation for the magnetic field: \( B=\frac{\mu_{0} I}{2 \pi R} \). We need to solve this equation for \( R \), which means we need to express \( R \) in terms of the other variables in the equation.
02

Isolate the denominator

Multiply both sides of the equation by \( 2 \pi R \) to get rid of the fraction: \[ B \cdot 2 \pi R = \mu_{0} I \] This step isolates \( 2 \pi R \) on one side of the equation.
03

Solve for R

Now, solve for \( R \) by dividing both sides of the equation by \( 2\pi B \): \[ R = \frac{\mu_{0} I}{2 \pi B} \] This gives us the value of \( R \) in terms of the other variables.

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.

Solving Equations
Solving equations often feels like a puzzle, where you need to find the value of the unknown variable. In our example, we need to solve for \( R \) in the formula for the magnetic field: \[ B=\frac{\mu_{0} I}{2 \pi R} \].
To solve for \( R \), the initial step is to eliminate any fractions, which makes handling equations simpler.
  • Multiply both sides by \( 2 \pi R \). This cancels out the fraction and nicely rearranges the equation to: \[ B \cdot 2 \pi R = \mu_{0} I \]
  • Next, to isolate \( R \), just divide both sides by \( 2 \pi B \): \[ R = \frac{\mu_{0} I}{2 \pi B} \]

This step-by-step breakdown helps you understand how each manipulation moves you closer to finding the variable you want.
Algebra
Algebra is the language of mathematics that helps us methodically unwrap problems. It involves rearranging terms to make complicated expressions more manageable. Consider the formula for magnetic field we have: \[ B=\frac{\mu_{0} I}{2 \pi R} \].
Here, algebraic manipulation lightly guides you through finding \( R \):
  • Transforming the formula by multiplying both sides by \( 2 \pi R \) is a critical algebraic technique. This isolates terms, making it easier to redefine the equation.
  • Division follows as a crucial algebraic move, efficiently isolating the variable \( R \).

Through these steps, you engage with algebra as a practical tool that enables you to adapt mathematical expressions towards solution-focused approaches.
Physics
Physics often combines diverse formulas to describe how the physical world operates. In this context, the magnetic field formula expresses the relationship between magnetic fields and their dependent variables. The formula given, \[ B=\frac{\mu_{0} I}{2 \pi R} \], reveals how the
  • Magnetic Constant (\( \mu_{0} \)): Influences magnetic fields and is considered a constant of nature.
  • Electric Current (\( I \)): Drives the magnetic force through the wire.
  • Radius (\( R \)): Represents the distance from the wire, inversely affecting the magnitude of the field.

By understanding and manipulating these variables, physics allows us to predict and explain the behavior of electric currents and magnetic forces in the real world. This understanding is crucial for innovations and applications in technology and everyday gadgets.

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

Solve the given problems. A tsunami is a very high ocean tidal wave (or series of waves) often caused by an earthquake. An Alaskan tsunami in 1958 measured over \(500 \mathrm{m}\) high; an Asian tsunami in 2004 killed over 230,000 people; a tsunami in Japan in 2011 killed over 10,000 people. An equation that approximates the speed \(v\) (in \(\mathrm{m} / \mathrm{s}\) ) of a tsunami is \(\left.v=\sqrt{g d}, \text { where } g=9.8 \mathrm{m} / \mathrm{s}^{2} \text { and } d \text { is the average depth (in } \mathrm{m}\right)\) of the ocean floor. Find \(v\) (in \(\mathrm{km} / \mathrm{h}\) ) for \(d=3500 \mathrm{m}\) (valid for many parts of the Indian Ocean and Pacific Ocean).

Answer the given questions. Refer to Appendix \(B\) for units of measurement and their symbols. A technician records 4.4 s as the time for a robot arm to swing from the extreme left to the extreme right, 2.72 s as the time for the return swing, and 1.68 s as the difference in these times. What is wrong with this conclusion?

Solve the given problems. Refer to Appendix B for units of measurement and their symbols. A faulty gauge on a fire engine pump caused the apparent pressure in the hose to change every few seconds. The pressures (in \(\mathrm{Ib} / \mathrm{in}\). \(^{2}\) above and below the set pressure were recorded as: +7,-2,-9,-6 What was the change between (a) the first two readings, (b) between the middle two readings, and (c) the last two readings?

An athlete who was jogging and wearing a Fitbit found that she burned 250 calories in 20 minutes. At that rate, how long will it take her to burn 400 calories? Assume all numbers are exact.

Perform the indicated multiplications. By multiplication, show that \((x+y)^{3}\) is not equal to \(x^{3}+y^{3}\).
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Applied Mathematics

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.