20 Nov 2019

The circuit shown above consists of a single battery, whose emf is1.4 V, and three wires made of the samematerial, but havingdifferent cross-sectional areas. Each thick wirehas cross-sectionalarea 1.4e-6 m2,and is 21 cm long. The thin wirehascross-sectional area 6.4e-8 m2,and is 7.5 cm long. Inthis metal, theelectron mobility is 6e-4 (m/s)/(V/m), andthere are8e+28 mobile electrons/m3.
Which of the following statements about the circuit in the steadystateare true?

The magnitude of the electric field atlocations D andA is thesame.The magnitude of the electric fieldatlocations A and C is thesame.The electroncurrent at location Dis the same as the electron currentat location A.At location B the electricfieldpoints toward the top of the page.

The symbol EF represents themagnitude of theelectric field at location F, and thesymbol EDrepresents themagnitude of the electric field at location D.Which of thefollowing equations is a correct energy conservation (loop)equationfor this circuit, following a path that starts at the negative endof the battery and goes counter-clockwise?
0 = +1.4V - EF*0.21m -ED*0.075m -EF*0.21m 0 = -1.4V +EF*0.21m +ED*0.075m +EF*0.21m 0 = -1.4V - EF*0.21m -ED*0.075m -EF*0.21m 0 = 1.4V +EF*0.21m +ED*0.075m +EF*0.21m 1.4V =EF*0.21m 1.4V =ED*0.075m

The symbol iF represents the electroncurrent atlocation F, etc. Which of the followingequations is a correctcharge conservation (node) equation for thiscircuit?
2*iF =iDiF =iD iF= 2*iDiF +iD= 0

Use the appropriate equation(s), plus the equation relatingelectroncurrent to electric field, to solve for the factor thatgoes in theblank below:
EF = 4 *ED
Use the appropriate equation(s) to calculate the magnitude ofED
ED = 5V/m
Use the appropriate equati

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Jamar Ferry
Jamar FerryLv2
9 Apr 2019

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