Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In electromagnetic theory, the magnetic potential \(u\) at a point on the axis of a circular coil is given by $$ u=A r \int_{a}^{\infty} \frac{d x}{\left(r^{2}+x^{2}\right)^{3 / 2}} $$ where \(A, r\), and \(a\) are constants. Evaluate \(u\).

Short Answer

Expert verified
\( u = \frac{A}{r} \left( 1 - \frac{a}{\sqrt{a^2 + r^2}} \right) \).

Step by step solution

01

Identify the Integral Form

The given expression for magnetic potential is \( u = A r \int_{a}^{\infty} \frac{d x}{\left(r^{2}+x^{2}\right)^{3 / 2}} \). This is an improper integral from \( a \) to infinity.
02

Perform the Substitution

To simplify the integral, use the substitution \( x = r \tan{\theta} \) where \( dx = r \sec^2{\theta} \, d\theta \). This leads to \( r^2 + x^2 = r^2 \sec^2{\theta} \).
03

Substitute into the Integral

Substitute the expressions from the substitution \( x = r \tan{\theta} \) into the integral: \[ \int \frac{r \sec^2{\theta} \, d\theta}{(r^2 \sec^2{\theta})^{3/2}} \] and simplify to \[ \int \frac{r \, d\theta}{r^3 \sec^3{\theta}} = \int \frac{d\theta}{r^2 \sec{\theta}} = \int \frac{\cos{\theta}}{r^2} \, d\theta \].
04

Evaluate the Simplified Integral

The integral simplifies to \( \int \frac{\cos{\theta}}{r^2} \, d\theta = \frac{1}{r^2} \int \cos{\theta} \, d\theta \). The antiderivative of \( \cos{\theta} \) is \( \sin{\theta} \).
05

Determine the Limits of Integration in Terms of \( \theta \)

Convert the original limits of \( x \) from \( a \to \infty \) to \( \theta \): - As \( x = a \), \( \theta = \tan^{-1}{\frac{a}{r}} \). - As \( x \to \infty \), \( \theta \to \frac{\pi}{2} \).
06

Compute the Final Evaluated Integral

Substitute the angular bounds to find the value of \( u \): \[ u = \frac{A r}{r^2} \left[ \sin{\theta} \right]_{\theta = \tan^{-1}{(a/r)}}^{\pi/2} = \frac{A}{r} \left( 1 - \frac{a}{\sqrt{a^2 + r^2}} \right) \].

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 Vaia!

Key Concepts

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

Improper Integral
Improper integrals are a type of integral where the limits of integration approach infinity or involve a point of discontinuity. In the context of electromagnetic theory, calculating the magnetic potential at a point involves solving such an integral. Here, the integral is given from a finite point \( a \) to infinity, which makes it an improper integral.
To handle this, one often applies limits to substitute the infinity point with a suitable variable and solve the integral within a finite range. This makes the solution of the integral mathematically feasible.
Improper integrals are crucial in physics and engineering fields as they allow the estimation of various physical quantities that extend over infinite domains. Properties of convergence, where the integral results in a finite number, determine the validity of the solution for physical applications.
Trigonometric Substitution
Trigonometric substitution is a powerful technique used to simplify integrals involving radicals. It transforms complex expressions into trigonometric functions that are often easier to handle. In this exercise, the substitution \( x = r \tan{\theta} \) simplifies the term \( r^2 + x^2 \) to a form using \( r^2 \sec^2{\theta} \).
This method also changes the differential \( dx \) to \( r \sec^2{\theta} \ d\theta \). Consequently, the integral becomes easier to integrate as we've converted it into trigonometric functions whose antiderivatives are well-known.
Trigonometric substitution is very useful for solving integrals that involve square roots in the denominator. It's an essential tool in calculus for dealing with integrals where direct integration isn't straightforward. It also serves other mathematical areas, like solving certain differential equations.
Limits of Integration
The limits of integration denote the range over which the integration occurs. Understanding and correctly transforming these limits are integral to solving problems accurately. In this example, the original integration limits were from \( a \) to infinity in terms of \( x \).
Upon performing trigonometric substitution, these limits change into angles. For \( x = a \), the new limit becomes \( \theta = \tan^{-1}{\frac{a}{r}} \). As \( x \) increases towards infinity, the angular limit approaches \( \frac{\pi}{2} \).
This conversion of limits is a pivotal step in evaluating the integral with respect to the new variable \( \theta \). It ensures that the integral remains within a feasible range within the new framework, facilitating easier calculation and solution of the expression.

One App. One Place for Learning.

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

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free