A wire with cross-sectional area \(A\) carries a current \(I\) Show that the electric field strength \(E\) in the wire is proportional to the current per unit area \((I / A)\) and identify the constant of proportionality. [Hint: Assume a length \(L\) of wire. How is the potential difference across the wire related to the electric field in the wire? (Which is uniform?) Use \(V=I R\) and the connection between resistance and resistivity.]

Short Answer

Expert verified
Question: Identify the constant of proportionality between the electric field strength, \(E\), in a wire and the current per unit area, \(\frac{I}{A}\), and explain how it is derived. Answer: The constant of proportionality between the electric field strength, \(E\) and the current per unit area, \(\frac{I}{A}\), is the resistivity, \(\rho\). It is derived by relating the potential difference, resistance, and electric field strength formulas and solving for \(E\). The resulting equation is \(E = \frac{I}{A} \cdot \rho\), which shows that \(E \propto \frac{I}{A}\) with the constant of proportionality \(\rho\).

Step by step solution

01

Write the formula for potential difference

The formula for potential difference is given by: \(V = IR\)
02

Write the formula for resistance

The formula for resistance in terms of resistivity, length, and cross-sectional area is: \(R = \rho \frac{L}{A}\)
03

Substitute the resistance formula into the potential difference formula

Using the resistance formula, we can rewrite the potential difference formula as: \(V = I (\rho \frac{L}{A})\)
04

Write the formula for electric field strength

The formula for electric field strength in terms of potential difference is: \(V = EL\)
05

Equate the potential difference formulas

Equate the potential difference formulas from Steps 3 and 4: \(E L = I (\rho \frac{L}{A})\)
06

Solve for electric field strength

We will now solve for the electric field strength \(E\): \(E = \frac{I (\rho \frac{L}{A})}{L}\) Then, cancel the terms \(L\): \(E = \frac{I}{A} \cdot \rho\)
07

Identify the constant of proportionality

From the above formula, we can see that the electric field strength \(E\) is proportional to \(\frac{I}{A}\) with the constant of proportionality being the resistivity \(\rho\): \(E \propto \frac{I}{A}\), with the constant of proportionality \(\rho\).

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