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A long Iron slab of width w and height h emerges from a furnace, as shown in Figure 19.79. Because the end of the slab near the furnace is hot and the other end Is cold, the electron mobility increases significantly with the distance x. The electron mobility is u=u0+kxwhere u0is the mobility of the iron at the hot end of the slab. There are n iron atoms per cubic meter, and each atom contributes one electron to the sea of the mobile electron (we can neglect the small thermal expansion of the iron). A steady state conventional current runs through the slab from the hot end towards cold end, and an ammeter (not shown) measures the current to have a magnitude I in amperes. A voltmeter is connected to two locations a distance d apart, as shown. (a) Show the electric field inside the slab at two locations marked with ×. Pay attention to the relative magnitudes of the two vectors that you draw. (b) Explain why the magnitude of the electric field is different at these two locations. (c) At a distance x from the left voltmeter connection, what is the magnitude of the electric field in terms x and the given quantities w,h,d,u0,k,l, and n ( and fundamental constants)? (d) What is the sign of potential difference displayed on the voltmeter? Explain briefly. (e) In terms of the given quantitiesw,h,d,u0,k,l, and n and ( and fundamental constants), what is the magnitude of the voltmeter reading? Check your work. (f) What is the resistance of this length of the iron slab?

Short Answer

Expert verified

(f) The resistance of this length of the iron slab is 1newhklogu0+kdu0.

Step by step solution

01

Given data

The given data can be listed below as,

  • A long Iron slab with w width and h height has a hot and cold end.
  • n is the number of iron atoms per cubic meter which is given.
  • The electron mobility is given with the help of the equation u=u0+kx, in whichu0 is given as the mobility of the iron at the hot end of the slab, and the mobility increases as we move away from the left end with respect to distance x.
  • The steady-state current that is flowing in the slab is given as I. A voltmeter is there, which is connected to the circuit to measure the potential drop between two locations that are at a finite distance d.
02

Concept

Ohms's law provides us with the relation between the flowing current I in the conductor and the voltage V between two ends of the conductor when we know the resistance R of the conductor.

The formula is given as,

R=VI

03

(f) Calculation of the resistance

The magnitude of the voltage across this length is Inewhklogu0+kdu0. (refer to sid: 875865-19-61 P-e)

Substitute the value in the above equation, and we get,

R=Inewhklogu0+kdu0I=1newhklogu0+kdu0

Thus, the resistance of this length of the iron slab is 1newhklogu0+kdu0.

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Most popular questions from this chapter

Question: in circuit 1 (Figure 19.72), an uncharged capacitor is connected in series with two batteries and one light bulb. Circuit 2 (Figure 19.72) contains two light bulbs identical to the bulb in the circuit; in all other respects, it is identical to circuit 1. In circuit 1, the light bulb stays lit for 25 s. The following questions refer to these circuits. You should draw diagrams representing the fields and charges in each circuit at the times mentioned, in order to answer the questions.

(a)One microsecond after connecting both circuits, which of the following are true? Chose all that apply: (1) the net electric field at location B in circuit 1 is larger than the net electric field at location B in circuit 2. (2) At location A in 1, electrons flow to the left. (3) At location A in circuit 1, the electric fields due to charges on the surface of the wires and batteries points to the right. (4) in circuit 1 the potential difference across the capacitor plates is equal to the emf of the batteries. (5) The current in circuit 1 is larger than the current in circuit 2.

(b)Two seconds after connecting both circuits, which of the following are true? Choose all that apply: (1) there is more charge on the plates of capacitor 1 than there is on the plates of capacitor 2. (2) there is negative charge on the right plate of the capacitor in circuit 1. (3) At location B in circuit 2 the net electric field points to the right. (4) At location B in circuit 2 the fringe field of the capacitor points to the right. (5) At location A in circuit 1 the fringe field of the capacitor points to the left.

(c)Which of the graphs in Figure 19.73 represents the amount of charge on the positive plate of the capacitor in circuit 1 as a function of time?

(d)Which of the graphs in Figure 19.73 represents the current in circuit 1 as a function of time?

Consider two capacitors whose only difference is that the plates of capacitor number 2 are closer together than those of capacitor number 1 (Figure 19.56). Neither, capacitors has an insulating layer between the plates. They are placed in two different circuits having similar batteries and bulbs in series with the capacitor.

Show that in the first fraction of a second, the current stays nearly constant (decreases less rapidly) in the circuit with capacitor number 2. Explain your reasoning in detail.

Hint: Show charges on metal plates, and consider the electric fields they produce in the nearby wires. Remember that the fringe field near a plate outside a circular capacitor is approximately-

(QAεo)(s2R)

More extensive analysis shows that this trend holds true for the entire charging process: the capacitor with the narrower gap ends up with more charge on the plates.

A circuit consists of two batteries (with negligible internal resistance), five ohmic resistors (Figure 19.88). The connecting wires that have negligible resistance. The letters A through are shown to make it possible to refer to specific parts of the circuit.

(a) Write all the equations necessary to solve for the unknown currents I1, I2, I3, I4 and I5, whose directions are indicated on the circuit diagram. Do not solve the equations but do explain very clearly what your equations are based on and to what they refer.

Assume that a computer program has solved your equations in terms of known battery voltages and known resistances so that the currents I1, I2,I3 ,I4and I5are are known. (b) In terms of known quantities calculate VD-VAand check that your sign makes sense. (c) In terms of known quantities, calculate the power produced in battery number 2.

A circuit consists of two batteries (with negligible resistance), six ohmic resistors and connecting wires that have negligible resistance. The resistance R1is 10Ω, R2 is 20Ω, R3 is 30Ω, R4is 12Ω, R5is 15Ω and R6 is 20Ω. Unknown currents I1,I2 ,I3 ,I4 , I5 and I6 have their directions marked on the circuit diagram in figure 19.87.

(a) Write down a set of equations that could be solved for the six unknown currents. Make sure you can explain how to you got these equations. (b) When a correct set of equations is solved the currents are as follows (to the nearest miiampeares) I1=0.4394A, I2=0.3312A, I3=0.0065A, I4=0.1082A, I5=0.3247Aand I6=0.4329A. Check your equations by substituting in these numbers. (c) Suppose that you connect the negative lead of a voltmeter to location C. What does the voltmeter read, including both magnitude and sign? (d) What does the power output of the 5 V battery? (e) Resistor is made of a very thin metal wire that is 3 mm long, with a diameter of 0.1 mm. What is the electric field inside the metal resistor.

Two resistor each with resistance of 4×106Ω are connected in series to a 60 V power supply whose internal resistance is negligible. You connect the voltmeter across one of these resistors and this voltmeter has an internal resistance of 1×106Ω. What is the reading on the voltmeter?

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