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The magnetic field around the head has been measured to be approximately \(3.0 \times 10^{-8} \mathrm{G}\). Although the currents that cause this field are quite complicated, we can get a rough estimate of their size by modeling them as a single circular current loop \(16 \mathrm{~cm}\) (the width of a typical head) in diameter. What is the current needed to produce such a field at the center of the loop?

Short Answer

Expert verified
The current needed to produce a magnetic field of \(3.0 \times 10^{-8} \, G\) at the center of the circular loop with a diameter of \(16 \, cm\) is approximately \(1.2 \times 10^{-6}\) A.

Step by step solution

01

Understand Ampere's Law

Ampere's law, in its simplified form for the magnetic field \(B\) at the center of a circular loop, states that \(B = \mu_0 (I / 2R)\), where \(\mu_0\) is the magnetic permeability of free space (\(4\pi \times 10^{-7} \, T \cdot m/A\)), \(I\) is the current and \(R\) is the radius of the loop.
02

Convert the Provided Data into Appropriate Units

The magnetic field was given in \(G\), but we need it in \(T\) (Tesla). So convert \(3.0 \times 10^{-8}\) G to T, knowing that \(1 \, T = 10^4 \, G\). So, \(B = 3.0 \times 10^{-8} \, G = 3.0 \times 10^{-8} \times 10^{-4} \, T = 3 \times 10^{-12} \, T\). Also, convert the diameter of the circular loop into radius in meters: \(R = 16\, cm / 2 = 0.08 \, m\).
03

Apply Ampere's Law to Calculate the Required Current

Rearrange Ampere's law to solve for \(I\), getting \(I = 2R \cdot B / \mu_0\). Substitute the values for \(B\), \(R\), and \(\mu_0\) into this equation to find \(I = 2 \cdot 0.08 \, m \cdot 3 \times 10^{-12} \, T / (4\pi \times 10^{-7} \, T \cdot m/A) = 1.2 \times 10^{-6} \, A\). Note that Ampere (A) is the SI unit for electric current.

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Key Concepts

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

Magnetic Field
Imagine a space around a magnet or a conductor with an electric current where magnetic forces can be detected, and you've got the idea of a magnetic field. This invisible field can be characterized by both its strength and direction. In our exercise, the magnetic field at the center of a current-carrying loop is being analyzed.

To visualize this, think of the magnetic field lines as closed loops emanating from the north pole of a magnet, looping around to the south pole. These fields are not just theoretical; they affect compass needles and can even influence the paths of charged particles. Essential devices like MRI machines rely on powerful magnetic fields to function. The strength of a magnetic field is measured in Teslas (T) in the SI system, and in this particular exercise, we converted it from Gauss (G), a smaller unit. The formula linking magnetic field strength, current, and loop radius provided by Ampere's Law is a powerful tool in calculating these electric phenomena.
Magnetic Permeability
Magnetic permeability is a core concept when discussing magnetic fields and currents. It represents the ability of a material to support the formation of a magnetic field within itself, essentially measuring how well a material can become magnetized.

It's symbolized by the Greek letter \(\mu\) and often is referred to in terms of \(\mu_0\), the magnetic permeability of free space, which is a constant. This value provides the backbone for calculations involving Ampere's Law, as seen in our exercise. When considering different materials, such as air, iron, or vacuum, magnetic permeability helps determine how strong the resulting magnetic field will be for a given current. A higher permeability indicates that the material can support a stronger magnetic field. Understanding this concept is essential for fields like electromagnetism, electrical engineering, and it is directly applied when designing electric motors, inductors, and transformers.
Electric Current
Let's dial into electric current, which is essentially the flow of electric charge carriers, usually electrons or ions. The movement of these charge carriers is what allows us to power our homes, devices, and so much more. Electric current flows from areas of high electronic potential (voltage) to lower potential, similar to how water flows downhill.

In our exercise, we're dealing with a steady current that flows through the circular loop, generating the magnetic field around it. The standard unit of electric current in the International System of Units (SI) is the Ampere (A), named after French physicist André-Marie Ampère, one of the main discoverers of electromagnetism. Ampere's Law, an integral part of calculating magnetic fields due to currents, allows us to use the current, along with the loop's radius and magnetic permeability, to work out the necessary current to produce a specific magnetic field, as we did in the solution steps. Understanding electric current is crucial for modern electronics, electrical engineering, and essentially all technology that is powered electrically.

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Most popular questions from this chapter

You use a teslameter (a Hall-effect device) to measure the magnitude of the magnetic field at various distances from a long, straight, thick cylindrical copper cable that is carrying a large constant current. To exclude the earth's magnetic field from the measurement, you first set the meter to zero. You then measure the magnetic field \(B\) at distances \(x\) from the surface of the cable and obtain these data: $$ \begin{array}{l|lllll} x(\mathrm{~cm}) & 2.0 & 4.0 & 6.0 & 8.0 & 10.0 \\ \hline B(\mathrm{mT}) & 0.406 & 0.250 & 0.181 & 0.141 & 0.116 \end{array} $$ (a) You think you remember from your physics course that the magnetic field of a wire is inversely proportional to the distance from the wire. Therefore, you expect that the quantity \(B x\) from your data will be constant. Calculate \(B x\) for each data point in the table. Is \(B x\) constant for this set of measurements? Explain. (b) Graph the data as \(x\) versus \(1 / B\). Explain why such a plot lies close to a straight line. (c) Use the graph in part (b) to calculate the current \(I\) in the cable and the radius \(R\) of the cable.

A wire of length \(20.0 \mathrm{~cm}\) lies along the \(x\) -axis with the center of the wire at the origin. The wire carries current \(I=8.00 \mathrm{~A}\) in the \(-x\) -direction. (a) What is the magnitude \(B\) of the magnetic field of the wire at the point \(y=5.00 \mathrm{~cm}\) on the \(y\) -axis? (b) What is the percent difference between the answer in (a) and the value you obtain if you assume the wire is infinitely long and use Eq. ( 28.9 ) to calculate \(B\) ?

A solenoid is designed to produce a magnetic field of \(0.0270 \mathrm{~T}\) at its center. It has radius \(1.40 \mathrm{~cm}\) and length \(40.0 \mathrm{~cm},\) and the wire can carry a maximum current of 12.0 A. (a) What minimum number of turns per unit length must the solenoid have? (b) What total length of wire is required?

In the Bohr model of the hydrogen atom, the electron moves in a circular orbit of radius \(5.3 \times 10^{-11} \mathrm{~m}\) with a speed of \(2.2 \times 10^{6} \mathrm{~m} / \mathrm{s}\). If we are viewing the atom in such a way that the electron's orbit is in the plane of the paper with the electron moving clockwise, find the magnitude and direction of the electric and magnetic fields that the electron produces at the location of the nucleus (treated as a point).

Long, straight conductors with square cross sections and each carrying current \(I\) are laid side by side to form an infinite current sheet (Fig. \(\mathbf{P 2 8 . 6 9}\) ). The conductors lie in the \(x y\) -plane, are parallel to the \(y\) -axis, and carry current in the \(+y\) -direction. There are \(n\) conductors per unit length measured along the \(x\) -axis. (a) What are the magnitude and direction of the magnetic field a distance \(a\) below the current sheet? (b) What are the magnitude and direction of the magnetic field a distance \(a\) above the current sheet?

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