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Chapter 7: Problem 82

A long, straight wire of length \(2 L\) on the \(y\) -axis carries a current \(I\). According to the Biot-Savart Law, the magnitude of the magnetic field due to the current at a point \((a, 0)\) is given by $$B(a)=\frac{\mu_{0} I}{4 \pi} \int_{-L}^{L} \frac{\sin \theta}{r^{2}} d y$$ where \(\mu_{0}\) is a physical constant, \(a>0,\) and \(\theta, r,\) and \(y\) are related as shown in the figure. a. Show that the magnitude of the magnetic field at \((a, 0)\) is $$B(a)=\frac{\mu_{0} I L}{2 \pi a \sqrt{a^{2}+L^{2}}}$$ b. What is the magnitude of the magnetic field at \((a, 0)\) due to an infinitely long wire \((L \rightarrow \infty) ?\)

Short Answer

Expert verified
Question: Calculate the magnitude of the magnetic field at point (a, 0) due to a current-carrying wire of infinite length. Answer: The magnitude of the magnetic field at point (a, 0) due to an infinitely long wire is given by the expression $$B(a) = \frac{\mu_{0} I}{2 \pi a}$$, where $\mu_{0}$ is the permeability of free space, I is the current flowing through the wire, and a is the perpendicular distance from the wire to the point (a, 0).

Step by step solution

01

Determine the relation between 胃, r, and y

To find the relationship between 胃, r, and y, observe the following triangles: Triangle A, formed by points (0,0), (0,y), and (a,y); Triangle B, formed by points (0,0), (a,0), and (a,y). From Triangle A, we get $$\tan \theta = \frac{a}{y}$$ and from Triangle B, we observe that r is the hypotenuse. Therefore, $$r^{2} = a^{2} + (y - 0)^{2}$$ We can also rewrite \(\tan \theta\) in terms of sin and cos: $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$ Substitute the expressions we have derived to find the relationship.
02

Substitute the relationship in the integral formula

Rewrite $$\tan \theta = \frac{a}{y}$$ as $$\sin \theta = a \cdot \frac{\cos{\theta}}{y}$$ and substitute this into the integral formula $$B(a)=\frac{\mu_{0} I}{4 \pi} \int_{-L}^{L} \frac{\sin \theta}{r^{2}} d y$$ We get $$B(a) = \frac{\mu_{0} I}{4 \pi} \int_{-L}^{L} \frac{a \cdot \frac{\cos \theta}{y}}{(a^{2} + y^{2})} dy$$
03

Integrate

Now, integrate the expression we got in step 2: $$B(a) = \frac{\mu_{0} I a \cos{\theta}}{4 \pi} \int_{-L}^{L} \frac{d y}{y^{2}+a^{2}}$$ Let $$u = \frac{y}{a}$$, so $$\frac{d u}{d y} = \frac{1}{a}$$ and $$d y = a d u$$. The limits will change to $$u_{min}=-\frac{L}{a}$$ and $$u_{max}=\frac{L}{a}$$. Substituting these, we get $$B(a) = \frac{\mu_{0} I a \cos{\theta}}{4 \pi} \int_{-\frac{L}{a}}^{\frac{L}{a}} \frac{a \: d u}{u^2+ 1}$$ which simplifies to $$B(a) = \frac{\mu_{0} I \cos{\theta}}{4 \pi} \int_{-\frac{L}{a}}^{\frac{L}{a}} \frac{d u}{u^2+ 1}$$ Evaluate the integral: $$B(a) = \frac{\mu_{0} I \cos{\theta}}{4 \pi} [ \arctan{u} ]_{-\frac{L}{a}}^{\frac{L}{a}}$$ Now substitute back for u: $$B(a) = \frac{\mu_{0} I \cos{\theta}}{4 \pi} [ \arctan{(\frac{y}{a})}]_{-L}^{L}$$
04

Evaluate the integral and simplify the expression.

Evaluate the arctan term: $$B(a) = \frac{\mu_{0} I \cos{\theta}}{4\pi} \left[\arctan{\left(\frac{L}{a}\right)} - \arctan{\left(\frac{-L}{a}\right)}\right]$$ From properties of arctan, we can simplify further: $$B(a) = \frac{\mu_{0} I \cos{\theta}}{4\pi} \left[2\arctan{\left(\frac{L}{a}\right)}\right]$$ The final expression becomes: $$B(a) = \frac{\mu_{0} I \cos{\theta}}{2\pi} \arctan{\left(\frac{L}{a}\right)}$$ Using the previous expression for tan胃 and the relation between sin胃 and cos胃, we can simplify the expression to: $$B(a) = \frac{\mu_{0} I L}{2 \pi a \sqrt{a^2 + L^2}}$$
05

Find the magnetic field due to an infinitely long wire

To find the magnitude of the magnetic field at (a, 0) due to an infinitely long wire, take the limit of the expression found in step 4 as L approaches infinity: $$B(a) = \lim_{L\to\infty} \frac{\mu_{0} I L}{2 \pi a \sqrt{a^2 + L^2}}$$ Simplify this expression by dividing both numerator and denominator by L: $$B(a) = \lim_{L\to\infty} \frac{\mu_{0} I}{2 \pi a \frac{\sqrt{a^2 + L^2}}{L}}$$ Now, taking the limit as L approaches infinity, the term inside the square root converges to 1: $$B(a) = \frac{\mu_{0} I}{2 \pi a}$$ So the magnitude of the magnetic field at (a, 0) due to an infinitely long wire is $$\frac{\mu_{0} I}{2 \pi a}$$

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Key Concepts

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

Understanding Magnetic Fields
Magnetic fields are invisible forces that exist around electric currents and magnetic materials. They are a fundamental part of electromagnetism, a major branch of physics. When an electric current flows through a wire, it creates a magnetic field around it. This field influences other charged particles and magnetic materials nearby.

In the context of the Biot-Savart Law, the magnetic field's magnitude at a specific point is determined by several factors, including the current's strength, the point's position relative to the wire, and the geometry of the current path.

The Biot-Savart Law is an essential tool in physics for calculating the magnetic field generated by a current. It helps us understand how fields behave in various situations, from simple circuits to complex superconductors.
Applying Integration Techniques
Integration techniques are crucial when calculating the magnetic field in the context of the Biot-Savart Law. Since we deal with continuous variables and fields, precise calculation often requires integrating functions over specified limits.

In our exercise, we used integration to determine the magnitude of the magnetic field at a point (a, 0) on the y-axis. The process began with substituting the trigonometric identities into the integral.

Next, we handled the integral using a change of variable technique called substitution. By substituting variables, such as setting \(u = \frac{y}{a}\), the integral became more manageable and allowed us to evaluate the field's magnitude accurately.

Evaluating integrals is a pivotal part of finding quantitative answers in physics, helping bridge the gap between theory and practical measurements.
Basics of Electric Current
Electric current is the flow of electric charge, mainly carried through conductors by creating movement of electrons. In our exercise, this current takes the symbol \(I\) and travels along a straight wire on the y-axis. When calculating magnetic fields, the current's magnitude and direction play essential roles.

The current influences the resulting magnetic field's strength. The Biot-Savart Law shows that a larger current generates a stronger magnetic field at any given point.

Additionally, the configuration of the wire, long in this exercise, affects how the magnetic field spreads around the wire. Understanding these characteristics of electric current is crucial for grasping more advanced topics in electromagnetism.
Importance of Trigonometric Identities
Trigonometric identities help simplify expressions involving angles, especially when integrating functions. In this exercise, they were crucial to relate angles \(\theta\), distances \(r\), and positions \(y\) about the wire.

By understanding these identities, such as \(\tan \theta = \frac{a}{y}\), we can effectively substitute in terms of sine or cosine when needed. This substitution helps simplify the integral we needed to solve for the magnetic field.

Utilizing trigonometric identities ensures that complex expressions can be tackled more efficiently, making them a valuable tool in mathematical physics. They reduce complex problems to manageable tasks, which is essential when calculating electromagnetic fields.

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Most popular questions from this chapter

Find the volume of the following solids. The region bounded by \(y=1 /(x+2), y=0, x=0,\) and \(x=3\) is revolved about the line \(x=-1\)

When is the volume finite? Let \(R\) be the region bounded by the graph of \(f(x)=x^{-p}\) and the \(x\) -axis, for \(x \geq 1.\) a. Let \(S\) be the solid generated when \(R\) is revolved about the \(x\) -axis. For what values of \(p\) is the volume of \(S\) finite? b. Let \(S\) be the solid generated when \(R\) is revolved about the \(y\) -axis. For what values of \(p\) is the volume of \(S\) finite?

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