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Chapter 7: Problem 1

Current in a bottle \(*\) An ocean current flows at a speed of 2 knots (approximately \(1 \mathrm{~m} / \mathrm{s}\) ) in a region where the vertical component of the earth's magnetic field is \(0.35\) gauss. The conductivity of seawater in that region is 4 (ohm-m) \(^{-1}\). On the assumption that there is no other horizontal component of \(\mathbf{E}\) than the motional term \(\mathbf{v} \times \mathbf{B}\), find the density \(J\) of the horizontal electric current. If you were to carry a bottle of seawater through the earth's field at this speed, would such a current be flowing in it?

Short Answer

Expert verified
The current density is \( J = \sigma \mathbf{E} = 4 ohm-m^{-1} \times 3.5\times10^{-5} T = 1.4\times10^{-4} A/m^{2}\). Yes, technically, a small current would be flowing in a bottle of seawater moving at the same speed as the ocean current through the Earth's magnetic field, because of the motional term for electric field.

Step by step solution

01

Convert Units

Start by converting the magnetic field from gauss to tesla since the SI unit of magnetic field is the Tesla (T). 1 Gauss is equal to \(1\times 10^{-4}\) Tesla, so \(0.35 Gauss = 0.35\times10^{-4} T\). The speed of the ocean current is given as approximately 1 m/s.
02

Calculate the Electric Field

Use the formula for the motional term for electric field \( \mathbf{E} = \mathbf{v} \times \mathbf{B} \). Plug in the values given for the speed of the ocean current (v) and the magnetic field (B) in the equation to get \( \mathbf{E} = 1m/s \times 0.35\times10^{-4} T\).
03

Calculate the Current Density

The Current density (J) can be found using the relation \( J = \sigma \mathbf{E} \) where \(\sigma\) is the conductivity of the seawater. Substitute the given conductivity (4 ohm-m\(^{-1}\)) and the calculated electric field into the equation to find J.
04

Analyze the Current in a Bottle of Seawater

If a bottle of seawater is moving at the same speed as the ocean current through the Earth's magnetic field, indeed a current would be flowing in it, as the seawater would still experience the same electric field. In reality though, the current is tiny and hard to measure, as the conductivity of seawater is relatively low.

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Key Concepts

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

Motional EMF
Electromagnetic induction is a fascinating phenomenon where a change in magnetic field induces an electromotive force (EMF).
One popular concept of this principle is the Motional EMF.
When a conductor moves through a magnetic field, electrons present in the conductor experience a force that leads to an induced voltage across the conductor. This is essentially the motional EMF.
  • The formula for motional EMF is derived from the Lorentz force: \[\mathbf{E} = \mathbf{v} \times \mathbf{B}\]
  • In this exercise, the ocean current acts like a moving conductor while the Earth's magnetic field provides the necessary magnetic field for induction.
  • The speed (\(\mathbf{v}\)) and the magnetic field (\(\mathbf{B}\)) are perpendicular to each other, maximizing the induced EMF.
The concept emphasizes the interaction between motion and magnetic fields, fundamental to understanding electromagnetic devices like generators.
Current Density
Current density is a measure of electric current flowing per unit area of a conductor.
In the context of this exercise, it's about deriving the current density (\(J\)) in a moving body of seawater due to the motional EMF generated.
  • To calculate current density, use the equation:\[J = \sigma \mathbf{E}\]where \(\sigma\) is the conductivity of the seawater.
  • The value of the electric field (\(\mathbf{E}\)) is obtained from the motional EMF formula.
  • With seawater conductivity given as 4 (ohm-m\(^{-1}\)), the equation allows us to find the actual current density.
Although seawater's conductivity impacts the current magnitude, the exercise highlights how internal properties of materials influence electric phenomena. Using conductivity to find \(J\) provides deeper insights into how materials manage electric currents.
Magnetic Field Conversion
The Earth's magnetic field affects how we calculate motional EMF and consequent electric phenomena.
Before implementing equations like \(\mathbf{E} = \mathbf{v} \times \mathbf{B}\), unit conversion of the magnetic field is necessary.
  • We often measure magnetic fields in Gauss in certain contexts but require Tesla in the SI system.
  • Conversion is straightforward: 1 Gauss equals \(1 \times 10^{-4}\) Tesla.
  • In this scenario, converting from 0.35 Gauss to Tesla by multiplying with \(10^{-4}\) simplifies subsequent calculations.
Understanding magnetic field conversion ensures proper unit consistency, which is crucial for accurate computations and better comprehension of magnetic effects in physics.
Seawater Conductivity
Seawater conductivity is a central aspect of studies involving electric currents in marine environments.
The conductivity of seawater plays a pivotal role in determining how easily electric currents can flow.
  • Conductivity in this exercise is measured as 4 (ohm-m\(^{-1}\)).
  • This relatively low conductivity means induced currents due to motional EMF, such as in our exercise, are small and often difficult to detect.
  • The presence of salt and ions in seawater assists conductivity, albeit at lower levels compared to metals.
This property of seawater aligns with the exercise's question regarding whether current would be detectable in a bottle of seawater. Understanding the nuances of seawater conductivity helps clarify why detecting current in an individual bottle moving at ocean current speed through the Earth's magnetic field is challenging.

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Most popular questions from this chapter

Maximum emffor a thin loop *** A long straight stationary wire is parallel to the \(y\) axis and passes through the point \(z=h\) on the \(z\) axis. A current \(I\) flows in this wire, returning by a remote conductor whose field we may neglect. Lying in the \(x y\) plane is a thin rectangular loop with two of its sides, of length \(\ell\), parallel to the long wire. The length \(b\) of the other two sides is very small. The loop slides with constant speed \(v\) in the \(\hat{x}\) direction. Find the magnitude of the electromotive force induced in the loop at the moment the center of the loop has position \(x .\) For what values of \(x\) does this emf have a local maximum or minimum? (Work in the approximation where \(b \ll x\), so that you can approximate the relevant difference in \(B\) fields by a derivative.)

Ring in a solenoid ** An infinite solenoid with radius \(b\) has \(n\) turns per unit length. The current varies in time according to \(I(t)=I_{0} \cos \omega t\) (with positive defined as shown in Fig. 7.36). A ring with radius \(r

Critical frequency of a dynamo *** A dynamo like the one in Exercise \(7.47\) has a certain critical speed \(\omega_{0}\). If the disk revolves with an angular velocity less than \(\omega_{0}\), nothng happens. Only when that speed is attained is the induced \(\mathcal{E}\) arge enough to make the current large enough to make the mag- netic field large enough to induce an \(\mathcal{E}\) of that magnitude. The critial speed can depend only on the size and shape of the conductors, ic dimension expressing the size of the dynamo, such as the radius f the disk in our example. a) Show by a dimensional argument that \(\omega_{0}\) must be given by a relation of this form: \(\omega_{0}=K / \mu_{0} \sigma d^{2}\), where \(K\) is some dimensionless numerical factor that depends only on the arrangement and relative size of the various parts of the dynamo. (b) Demonstrate this result again by using physical reasoning that relates the various quantities in the problem \((R, \mathcal{E}, E, I, B,\), etc.). You can ignore all numerical factors in your calculations and absorb them into the constant \(K\). Additional comments: for a dynamo of modest size made owever, with \(d\) measured in hundreds of kilometers rather than neters, the critical speed is very much smaller. The earth's magnetic field is almost certainly produced by a nonferromagnetic dynamo involving motions in the fluid metallic core. That fluid happens to be molten iron, but it is not even slightly ferromagnetic because it is too hot. (That will be explained in Chapter 11.) We don't know how the conducting fluid moves, or what configuration of electric currents and magnetic fields its motion generates in the core. The magnetic field we observe at the earth's surface is the external field of the dynamo in the core. The direction of the earth's field a million years ago is preserved in the magnetization of rocks that solidified at that time. That magnetic record shows that the field has reversed its direction nearly 200 times in the last 100 million years. Although a reversal cannot have been instantaneous (see Exercise 7.46), it was a relatively sudden event on the geological time scale. The immense value of paleomagnetism as an indelible record of our planet's history is well explained in Chapter 18 of Press and Siever (1978).

Mutual-inductance symmetry \(* *\) In Section \(7.7\) we made use of the vector potential to prove that \(M_{12}=M_{21} .\) We can give a second proof, this time in the spirit of Exercise 3.64. Imagine increasing the currents in two circuits gradually from zero to the final values of \(I_{1 \mathrm{f}}\) and \(I_{2 \mathrm{f}}(\) "'f" for "final"). Due to the induced emfs, some external agency has to supply power to increase (or maintain) the currents. The final currents can be brought about in many different ways. Two possible ways are of particular interest. (a) Keep \(I_{2}\) at zero while raising \(I_{1}\) gradually from zero to \(I_{1 \mathrm{f}}\). Then raise \(I_{2}\) from zero to \(I_{2 \mathrm{f}}\) while holding \(I_{1}\) constant at \(I_{1 \mathrm{f}}\). (b) Carry out a similar program with the roles of 1 and 2 exchanged, that is, raise \(I_{2}\) from zero to \(I_{2} \mathrm{f}\) first, and so on.

Sliding loop A long straight stationary wire is parallel to the \(y\) axis and passes through the point \(z=h\) on the \(z\) axis. A current \(I\) flows in this wire, returning by a remote conductor whose field we may neglect. Lying in the \(x y\) plane is a square loop with two of its sides, of length \(b\), parallel to the long wire. This loop slides with constant speed \(v\) in the \(\hat{\mathbf{x}}\) direction. Find the magnitude of the electromotive force induced in the loop at the moment when the center of the loop crosses the \(y\) axis.
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