Integral Arising From A Problem In Magnetostatics

Introduction

In the realm of magnetostatics, the calculation of magnetic fields is a crucial aspect of understanding the behavior of electric currents. One of the fundamental problems in this field is the determination of the magnetic field of a stationary current loop. This problem has far-reaching implications in various areas of physics, including electromagnetism and quantum mechanics. In this article, we will delve into the intricacies of this problem and explore a specific integral that arises from it.

The Problem

The magnetic field of a stationary current loop can be calculated using the Biot-Savart law, which states that the magnetic field at a point due to a small element of a current-carrying wire is proportional to the current and inversely proportional to the square of the distance between the element and the point. However, when dealing with a closed loop, the calculation becomes more complex, and the integral arises.

The Integral

The integral in question is given by:

$$I=\int_0^{2\pi}\frac{\alpha\cos{\theta}}{\big(1-\cos{(\theta-\phi)}\big)}d\theta$$

where $\alpha$ is a constant, $\theta$ is the angle between the current loop and the point of interest, and $\phi$ is the angle between the current loop and the axis of the loop.

Breaking Down the Integral

To tackle this integral, we need to break it down into manageable parts. The first step is to recognize that the denominator can be expanded using the trigonometric identity:

$$\cos{(\theta-\phi)} = \cos{\theta}\cos{\phi} + \sin{\theta}\sin{\phi}$$

Substituting this into the integral, we get:

$$I=\int_0^{2\pi}\frac{\alpha\cos{\theta}}{\big(1-\cos{\theta}\cos{\phi} - \sin{\theta}\sin{\phi}\big)}d\theta$$

Using Trigonometric Identities

To simplify the integral further, we can use the trigonometric identity:

$$\cos{\theta}\cos{\phi} + \sin{\theta}\sin{\phi} = \cos{(\theta-\phi)}$$

Substituting this into the integral, we get:

$$I=\int_0^{2\pi}\frac{\alpha\cos{\theta}}{\big(1-\cos{(\theta-\phi)}\big)}d\theta$$

Evaluating the Integral

To evaluate the integral, we can use the substitution:

$$u = \theta - \phi$$

This gives us:

$$du = d\theta$$

Substituting this into the integral, we get:

$$I=\int_{-\phi}^{2\pi-\phi}\frac{\alpha\cos{(u+\phi)}}{\big(1-\cos{u}\big)}du$$

Using the Substitution

To simplify the integral further, we can use the substitution:

$$v = \cos{u}$$

This gives us:

$$dv = -\sin{u}du$$

Substituting this into the integral, we get:

$$I=-\int_{\cos{(-\phi)}}^{\cos{(2\pi-\phi)}}\frac{\alpha\sqrt{1-v^2}}{\big(1-v\big)}dv$$

Evaluating the Integral

To evaluate the integral, we can use the substitution:

$$w = \frac{1-v}{\sqrt{1-v^2}}$$

This gives us:

$$dw = \frac{v}{\sqrt{1-v^2}}dv$$

Substituting this into the integral, we get:

$$I=-\int_{\frac{1-\cos{(-\phi)}}{\sqrt{1-\cos^2{(-\phi)}}}}^{\frac{1-\cos{(2\pi-\phi)}}{\sqrt{1-\cos^2{(2\pi-\phi)}}}}\alpha w^2dw$$

Simplifying the Integral

To simplify the integral further, we can use the fact that:

$$\cos{(-\phi)} = \cos{\phi}$$

and

$$\cos{(2\pi-\phi)} = \cos{(\phi+2\pi)} = \cos{\phi}$$

Substituting this into the integral, we get:

$$I=-\int_{\frac{1-\cos{\phi}}{\sqrt{1-\cos^2{\phi}}}}^{\frac{1-\cos{\phi}}{\sqrt{1-\cos^2{\phi}}}}\alpha w^2dw$$

Evaluating the Integral

To evaluate the integral, we can use the fact that:

$$\int w^2dw = \frac{w^3}{3}$$

Substituting this into the integral, we get:

$$I=-\alpha\left[\frac{\left(\frac{1-\cos{\phi}}{\sqrt{1-\cos^2{\phi}}}\right)^3}{3}\right]$$

Simplifying the Result

To simplify the result, we can use the fact that:

$$\left(\frac{1-\cos{\phi}}{\sqrt{1-\cos^2{\phi}}}\right)^3 = \left(\frac{1-\cos{\phi}}{\sin{\phi}}\right)^3$$

Substituting this into the result, we get:

$$I=-\alpha\left[\frac{\left(\frac{1-\cos{\phi}}{\sin{\phi}}\right)^3}{3}\right]$$

Conclusion

In conclusion, the integral in question can be evaluated using a combination of trigonometric identities and substitutions. The final result is a function of the constant $\alpha$ and the angle $\phi$. This result has far-reaching implications in various areas of physics, including electromagnetism and quantum mechanics.

References

• [1] Jackson, J. D. (1999). Classical Electrodynamics. John Wiley & Sons.
• [2] Landau, L. D., & Lifshitz, E. M. (1975). The Classical Theory of Fields. Pergamon Press.
• [3] Griffiths, D. J. (2013). Introduction to Electrodynamics. Pearson Education.

Appendix

The following is a list of the trigonometric identities used in this article:

• $\cos{(\theta-\phi)} = \cos{\theta}\cos{\phi} + \sin{\theta}\sin{\phi}$
• $\cos{\theta}\cos{\phi} + \sin{\theta}\sin{\phi} = \cos{(\theta-\phi)}$
• $\cos{u} = v$
• $\sin{u} = \sqrt{1-v^2}$
• $\frac{1-v}{\sqrt{1-v^2}} = w$
• $\frac{v}{\sqrt{1-v^2}} = dw$

Introduction

In our previous article, we explored the intricacies of the magnetic field of a stationary current loop and the integral that arises from it. In this article, we will address some of the most frequently asked questions about this topic.

Q: What is the Biot-Savart law?

A: The Biot-Savart law is a fundamental principle in electromagnetism that describes the magnetic field generated by a current-carrying wire. It states that the magnetic field at a point due to a small element of a current-carrying wire is proportional to the current and inversely proportional to the square of the distance between the element and the point.

Q: How does the Biot-Savart law relate to the integral in question?

A: The Biot-Savart law is used to calculate the magnetic field of a current loop, which is a closed loop of wire. The integral in question arises from the calculation of the magnetic field at a point due to a small element of the loop.

Q: What is the significance of the angle $\phi$ in the integral?

A: The angle $\phi$ represents the angle between the current loop and the axis of the loop. It is an important parameter in the calculation of the magnetic field, as it affects the orientation of the loop and the resulting magnetic field.

Q: How does the integral relate to the magnetic field of a current loop?

A: The integral in question is used to calculate the magnetic field of a current loop at a point due to a small element of the loop. The result of the integral is a function of the constant $\alpha$ and the angle $\phi$, which can be used to determine the magnetic field at a point.

Q: What are some of the applications of the integral in question?

A: The integral in question has far-reaching implications in various areas of physics, including electromagnetism and quantum mechanics. Some of the applications of the integral include:

• Calculating the magnetic field of a current loop
• Determining the force on a current-carrying wire due to a magnetic field
• Understanding the behavior of electric currents in various materials

Q: How can the integral be evaluated?

A: The integral in question can be evaluated using a combination of trigonometric identities and substitutions. The final result is a function of the constant $\alpha$ and the angle $\phi$.

Q: What are some of the challenges associated with evaluating the integral?

A: One of the challenges associated with evaluating the integral is the complexity of the trigonometric identities and substitutions required. Additionally, the integral involves a combination of algebraic and trigonometric manipulations, which can be time-consuming and error-prone.

Q: What are some of the resources available for learning more about the integral in question?

A: Some of the resources available for learning more about the integral in question include:

• Textbooks on electromagnetism and quantum mechanics
• Online resources and tutorials
• Research papers and articles on the topic

Conclusion

In conclusion, the integral in question is a fundamental aspect of magnetostatics and has far-reaching implications in various areas of physics. By understanding the intricacies of the integral and its applications, we can gain a deeper appreciation for the behavior of electric currents and the magnetic field.

References

• [1] Jackson, J. D. (1999). Classical Electrodynamics. John Wiley & Sons.
• [2] Landau, L. D., & Lifshitz, E. M. (1975). The Classical Theory of Fields. Pergamon Press.
• [3] Griffiths, D. J. (2013). Introduction to Electrodynamics. Pearson Education.

Appendix

The following is a list of the resources mentioned in this article:

• Textbooks on electromagnetism and quantum mechanics
• Online resources and tutorials
• Research papers and articles on the topic

These resources can be used to learn more about the integral in question and its applications in various areas of physics.