91Ó°ÊÓ

Question: A circuit is constructed from two batteries and two wires, as shown in Figure 18.104. Each battery has an emf of $\mathbf{1}\mathbf{.}\mathbf{3}\mathbf{}\mathit{V}$. Each wire is$\mathbf{26}\mathbf{}\mathit{c}\mathit{m}$long and has a diameter of $\mathbf{7}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathit{m}$. The wires are made of a metal that has$\mathbf{7}\mathbf{×}{\mathbf{10}}^{\mathbf{28}}$mobile electrons per cubic meter; the electron mobility is $\mathbf{5}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{}\mathbf{(}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}\mathbf{/}\mathbf{(}\mathbf{V}\mathbf{/}\mathbf{m}\mathbf{)}$. A steady current runs through the circuit. The locations marked by $\mathbf{×}$and labeled by a letter are in the interior of the wire. (a) Which of these statements about the electric field in the interior of the wires, at the locations marked by $\mathbf{×}\mathbf{\text{'}}\mathit{s}$, are true? List all that apply. (1) The magnitude of the electric field at location G is larger than the magnitude of the electric field at location F. (2) At every marked location the magnitude of the electric field is the same. (3) At location B the electric field points to the left. (b) Write a correct energy conservation (round-trip potential difference) equation for this circuit, along a round-trip path starting at the negative end of battery 1 and traveling counterclockwise through the circuit (that is, traveling to the left through the battery, and continuing on around the circuit in the same direction). (c) What is the magnitude of the electric field at location B? (d) How many electrons per second enter the positive end of battery 2? (e)If the cross-sectional area of both wires were increased by a factor of 2, what would be the magnitude of the electric field at location B? (f) Which of the diagrams in Figure 18.105 best shows the approximate distribution of excess charge on the surface of the circuit?

Step by step solution

01

Write the given data from the question.

Emf of the battery, $V=1.3\text{V}$

Length of wire, $L=26\text{cm}$

Diameter of wire,$d=7×{10}^{-4}\text{m}$

Number of mobile electrons,role="math" localid="1668583532869" $N=7×{10}^{28}1/{\mathrm{m}}^3$

Electron mobility,$\mathrm{μ}=5×{10}^{-5}(\mathrm{m}/\mathrm{s})/(\mathrm{V}/\mathrm{m})$

02

Determine the equation to find out the correct statement.

The electric field is defined as the ratio of the voltage and length of the wire.

The expression to calculate the electric field is given as follows.

$\mathit{E}\mathbf{=}\frac{\mathbf{V}}{\mathbf{L}}$ …… (i)

Here, is$\mathbf{L}$ the length of the wire.

03

Find out the correct statement.

The circuit is connected in the series connection, and the current in the series is always the same. The direction of the electric field produce inside the wire is always towards the positive terminal and away from the negative terminal.

From equation (i), it is clear that the electric field depends on the voltage and length of the wire. Therefore, the electric field in the battery's wire is the same at all locations.

Hence the correct statement is (2): At every marked location, the magnitude of the electric field is the same.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!