A circuit is constructed from two batteries and two wires, as shown in the diagr

Question

A circuit is constructed from two batteries and two wires, as shown in the diagram below. Battery l Battery Each battery has an emf of 1.3 volts. Each wire is 23 cm long, and has a diameter of 0.0007 meters. The wires are made of a metal which has 7e+28 mobile electrons per cubic meter; the electron mobility is 3e-05 (m/s)/(V/m). A steady current runs through the circuit. The locations marked by "x" and labeled by a letter are in the interior of the wire. Which of these statements about the electric field in the interior of the wires, at the locations marked by "x"s, are true? Check all that apply. The magnitude of the electric field at location D is larger than the magnitude of the electric field at location C. At location B the electric field points to the left n At every marked location the magnitude of the electric field is the same Which of the following equations is 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)? a +1.3 E 0.23 m) 1.3 0.23 m) 0 1.3 +1.3 V E 0.23 m) 0 1.3 V E 0.23 m) 1.3 V E 0.23 m) 0 +1.3 V E 0.23 m) 1.3 V E 0.23 m) 0 1.3 V E 0.23 m) 1.3 V 0.23 m) 0 1.3 V E 0.23 m) 0 What is the magnitude of the electric field at location C? V/m How many electrons per second enter the positive end of battery #2? electrons/s If the cross-sectional area of both wires were increased by a factor of 3, what would the be magnitude of the electric field at location C? new value of Ec V/m which of the diagrams below best shows the approximate distribution of excess charge on the surface of the circuit? C

Explanation / Answer

What is the magnitude of the electric field at location C?

E = i/L = 1.3V/0.23m =5.62 V/M

#2)

How many electrons per second enter the positive end of battery ?

You have a mobile electron per cubic meter of 7E28. Multiply this by the volume of the wire in order to find the

number of electrons.

7E28 * (0.00035^2 * pi) = 2.69E22 e

Multiply this number by the electron mobility and the electric field to find the e/s.

2.69E22 * 2E-5 * 5.62 = 3.023E18 e/s

If the cross-sectional area of both wires were increased by a factor of 3, what would the be magnitude of the electric

field at location C?

The electric field only depends on the length of the wire, not the radius,

so

it will still be 5.62 v/m.

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