91Ó°ÊÓ

Navigation of electric fish. Certain fish, such as the Nile fish (Gnathonemus), concentrate charges in their head and tail, thereby producing an electric field in the water around them. (See Figure \(19.67 . )\) This field creates a potential difference of a few volts between the head and tail, which in turn causes current to flow in the conducting seawater. As the fish swims, it passes near objects that have resistivities different from that of seawater, which in turn causes the current to vary. Cells in the skin of the fish are sensitive to this current and can detect changes in it. The changes in the current allow the fish to navigate. (In the next chapter, we shall investigate how the fish might detect this current.) Since the electric field is weak far from the fish, we shall consider only the field running directly from the head to the tail. We can model the seawater through which that field passes as a conducting tube of area 1.0 \(\mathrm{cm}^{2}\) and having a potential difference of 3.0 \(\mathrm{V}\) across its ends. The length of a Nile fish is about \(20 \mathrm{cm},\) and the resistivity of seawater is 0.13\(\Omega \cdot \mathrm{m} .\) (a) How large is the current through the tube of seawater? (b) Suppose the fish swims next to an object that is 10 \(\mathrm{cm}\) long and 1.0 \(\mathrm{cm}^{2}\) in cross-sectional area and has half the resistivity of seawater. This object replaces the seawater for half the length of the tube. What is the current through the tube now? How large is the change in the current that the fish must detect? (Hint: How are this object and the remaining water in the tube connected, in series or in parallel?)

Step by step solution

01

Understand the Problem

We need to calculate the current through a conducting tube of seawater, which functions as a resistor. We then need to calculate the change in current when the fish swims past an object with different resistivity, which also acts as a resistor.

02

Apply Ohm's Law to Seawater

Use Ohm's Law, which states that current \( I \) is given by \( I = \frac{V}{R} \). The resistance \( R \) can be calculated using the formula \( R = \rho \frac{L}{A} \), where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area.

03

Calculate Resistance of Seawater

The resistance of seawater \( R_{sea} \) is:\[R_{sea} = 0.13 \times \frac{0.2}{0.0001} = 260 \ \Omega.\]

04

Calculate Current in Seawater

Using Ohm's Law:\[I_{sea} = \frac{3.0}{260} \approx 0.0115 \ A.\]

05

Determine Resistance of the Object

The object has half the resistivity of seawater, so \( \rho = 0.065 \ \Omega \cdot \mathrm{m} \). The resistance of the object \( R_{obj} \) is:\[R_{obj} = 0.065 \times \frac{0.1}{0.0001} = 65 \ \Omega.\]

06

Model the New Circuit

The circuit now consists of the original seawater resistance and the resistance of the object. Since the object replaces seawater for half the length, they are in series. The total resistance \( R_{total} \) is the sum:\[R_{total} = 65 + 130 = 195 \ \Omega.\]

07

Calculate New Current

Using Ohm's Law for the new configuration:\[I_{new} = \frac{3.0}{195} \approx 0.0154 \ A.\]

08

Determine Change in Current

The change in current \( \Delta I \) is the difference between the original and new currents:\[\Delta I = 0.0115 - 0.0154 = -0.0039 \ A.\]

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Key Concepts

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

Ohm's Law

Ohm's Law is a fundamental principle in electricity that links the current flowing through a conductor to the voltage across it and its resistance. Formally, Ohm's Law is stated as \( I = \frac{V}{R} \), where \( I \) is the current in amperes, \( V \) is the voltage in volts, and \( R \) is the resistance in ohms. This law is vital for analyzing electrical circuits.

In the context of the electric fish, the seawater between the fish's head and tail acts as a resistor. The fish creates a potential difference of about 3 volts across this 'water resistor', and the subsequent current can be calculated using Ohm's Law. Understanding this helps explain how the fish emits an electric field to navigate by detecting changes in the current as it passes by objects with different resistivities.

Electric Field

An electric field is an invisible field around charged particles that exerts force on other charges within the field. The electric field produced by the fish assists it in navigation. In this case, the fish creates an electric field from its head to its tail.

This electric field leads to a potential difference much like the voltage applied across a resistor. For electric fish, the electric field isn't uniform throughout the water; it diminishes with distance. However, for simplicity, we consider the direct line from head to tail, where the electric field is strongest, to calculate the potential difference that drives the current.

Resistivity

Resistivity is a property of materials that quantifies how strongly a given material opposes the flow of electric current. It is represented by \( \rho \) and is measured in ohm-meters (\( \Omega \cdot m \)).

The resistivity of seawater is a known factor which influences how electricity flows around the fish. For seawater, it's given as 0.13 \( \Omega \cdot m \). When the fish swims near objects with different resistivities, the current flowing through the water changes, which the fish can detect. For example, an object with half the resistivity of seawater modifies the current, allowing the fish to sense its presence and navigate efficiently.

Current Calculation

Current is defined as the flow of electric charge and is measured in amperes (A). To calculate the current in the seawater around the fish, we first determine the resistance using the formula \( R = \rho \frac{L}{A} \), where \( L \) is the water length and \( A \) is its cross-sectional area.

Let's recap how we calculate it in this exercise:

• The length \( L \) of the water is 0.2 m, and the area \( A \) is 0.0001 m².
• The resistance of seawater \( R \) is computed as 260 ohms.
• Applying Ohm's Law, the current \( I \) through the water is \( \frac{3.0}{260} \approx 0.0115 \) A.

When the fish swims by an object with different resistivity, this resistance and the resultant current are recalculated. The object with lower resistivity is essentially a new resistor in series, altering the total resistance and therefore the current in the circuit. This sensitivity to current changes aids the fish in navigating its environment.