Answer: A 2.0-mm length of wire is made by welding the end

Chapter 25, Problem 58P

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QUESTION:

A 2.0-mm length of wire is made by welding the end of a 120-cm-long silver wire to the end of an 80-cm-long copper wire. Each piece of wire is 0.60 mm in diameter. The wire is at room temperature, so the resistivities are as given in Table 25.1. A potential difference of 5.0 V is maintained between the ends of the 2.0-m composite wire. (a) What is the current in the copper section? (b) What is the current in the silver section? (c) What is the magnitude of $\vec{E}$ in the copper? (d) What is the magnitude of $\vec{E}$ in the silver? (e) What is the potential difference between the ends of the silver section of wire?

Equation Transcription:

$\vec{E}$

Text Transcription:

Vector E

Questions & Answers

QUESTION:

Equation Transcription:

$\vec{E}$

Text Transcription:

Vector E

ANSWER:

Solution 58P

Step 1 of 8:

In the given problem, a L=2m length of wire is made by welding a silver wire of length ${L}_{si}$ =120 cm=1.2 m with voltage ${V}_{si}$ and resistance ${R}_{si}$ . To the end of this wire, a copper wire of length ${L}_{cu}$ =80 cm=0.8 m with voltage ${V}_{cu}$ and resistance ${R}_{cu}$ . Also the diameter of both the wires is ${d}_{si}$ = ${d}_{cu}$ =0.6 mm=0.6 $\times{}{10}^{-3}m$ . This total wire is kept across the potential difference of V= 5 volt.

In part (a) and (b) using ohm’s law, we need to find the current through the copper and silver wire ${I}_{cu}=?$ and ${I}_{si}=?$ . Also in part (c) and (d) using relation between voltage and electric field, we need to find the magnitude of electric field through the copper and silver wire ${E}_{cu}=?$ and ${E}_{si}=?$ . Also in part (e) we need to find the potential difference between the ends of the silver section of wire, ${V}_{si}=?$