All of the wires in the circuit shown in the diagram are made of the same materi
ID: 1599328 • Letter: A
Question
All of the wires in the circuit shown in the diagram are made of the same material, but one wire has a smaller radius than the other wires. Which of the following statements are true of this circuit in the steady state? Check all that apply. The magnitude of the electric field is the same at each location labeled by a letter. The electric field at location F points up. The electric field at G is larger in magnitude than the electric field at C. The drift speed of electrons passing location D is greater than the drift speed of electrons passing location G. The number of electrons passing location B each second is the same as the number of electrons passing location D each second. The radius of the thin wire is 0.18 mm, and the radius of the thick wire is 0.45 mm. There are 4e+28 mobile electrons per cubic meter of this material, and the electron mobility is 0.0006 (m/s)/V/m). If 4e+ 18 electrons pass location D each second, how many electrons pass location A each second? 6.55 electrons per second What is the magnitude of the electric field at location A? |E^rightarrow| = v/mExplanation / Answer
The cross section area of the wire is not same at all the locations labelled by a letter.so the electric field is not uniform in the wire.so first option is false.
The direction of the electric field in the wire is from positive terminal to the neagtive terminal of the battery.so second option is false.
The cross area of the wire at point G and C is same.So the electric field at these points is same.So third option is false.
As the cross section area of the wire is less than that of point G the drift speed of the electron passing at loction D is greater than that of point G.so fourth option is false.
Current flow is defined as the amount of charge passes through any cross section per unit time .Here current flow is constant so number of electrons passing through point B and point D per unit time is same.So fifth option is true.
THE NUMBER OF ELECTRONS PASSING THROUGH ANY POINT IS SAME. SO NUMBER OF ELECTRONS PASSING TROUGH B IS 4*10^18 electrons per second
Current flow in the wire is
I=Q/T=Ne/t
4*10^18 X 1.6*10^-19 C/1s
=0.64 A
Cross section area of the wire at location A is
=pi *(0.45*10^-3 m)^2
6.362*10^-7 m2
Now magnitude of elecric field at location A is
EA =I/UneA
=> 0.64A/(0.0006m/s/V/m)(4*10^28)(1.6*10^-19 C)(6.362*10^-7m2)
=0.262 v/m