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Chapter 28: Problem 31

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 Andr茅-Marie Amp猫re, 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

The current in the windings of a toroidal solenoid is \(2.400 \mathrm{~A}\). There are 500 turns, and the mean radius is \(25.00 \mathrm{~cm} .\) The toroidal solenoid is filled with a magnetic material. The magnetic field inside the windings is found to be 1.940 T. Calculate (a) the relative permeability and (b) the magnetic susceptibility of the material that fills the toroid.

A \(-4.80 \mu \mathrm{C}\) charge is moving at a constant speed of \(6.80 \times 10^{5} \mathrm{~m} / \mathrm{s}\) in the \(+x\) -direction relative to a reference frame. At the instant when the point charge is at the origin, what is the magneticfield vector it produces at the following points: (a) \(x=0.500 \mathrm{~m}, y=0\) \(z=0 ;\) (b) \(x=0, y=0.500 \mathrm{~m}, z=0 ;\) (c) \(x=0.500 \mathrm{~m}, y=0.500 \mathrm{~m}\) \(z=0 ;(\mathrm{d}) x=0, y=0, z=0.500 \mathrm{~m} ?\)

A solid conductor with radius \(a\) is supported by insulating disks on the axis of a conducting tube with inner radius \(b\) and outer radius \(c\) (Fig. E28.39). The central conductor and tube carry equal currents \(I\) in opposite directions. The currents are distributed uniformly over the cross sections of each conductor. Derive an expression for the magnitude of the magnetic field (a) at points outside the central, solid conductor but inside the tube \((ac)\)

The Magnetic Field from a Lightning Bolt. Lightning bolts can carry currents up to approximately 20 kA. We can model such a current as the equivalent of a very long, straight wire. (a) If you were unfortunate enough to be \(5.0 \mathrm{~m}\) away from such a lightning bolt, how large a magnetic field would you experience? (b) How does this field compare to one you would experience by being \(5.0 \mathrm{~cm}\) from a long, straight household current of \(10 \mathrm{~A}\) ?

We can estimate the strength of the magnetic field of a refrigerator magnet in the following way: Imagine the magnet as a collection of current-loop magnetic dipoles. (a) Derive the force between two current loops with radius \(R\) and current \(I\) separated by distance \(d \ll R\). Very close to the wire its magnetic field is about the same as for an infinitely long wire, and Eq.( 28.11 ) can be used. (b) Using Eq. ( 28.17 ), express the current \(I\) in terms of the magnetic field at the middle of the loop, and express the radius \(R\) in terms of the area of the loop. In this way, derive an expression for the force \(F\) between two identical current loops separated by a small distance \(d\) in terms of their mutual area \(A\) and center magnetic field \(B\). (c) Rearrange your result to obtain an expression for the magnetic field of a dipole with area \(A\) in terms of the force \(F\) from an identical dipole separated by a small distance \(d\). (d) Now notice that the force it takes to separate one magnet from your refrigerator is nearly the same as the force it takes to separate two magnets stuck together. Estimate that force \(F\). (e) Estimate the area of a refrigerator magnet. (f) Assume that when these magnets are stuck together or to the refrigerator, they are separated by an effective distance \(d=25 \mu \mathrm{m}\). Use the formula derived above to estimate the magnetic field strength of the magnet.
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