A circuit consists of a battery, whose emf is K, and five Nichrome wires, three thick and two thin as shown in Figure 19.78. The thicknesses of the wires have been exaggerated in order to give you room to draw inside the wires. The internal resistance of the battery is negligible compared to the resistance of the wires. The voltmeter is not attached until part (e) of the problem. (a) Draw and label appropriately the electric field at the locations marked × inside the wires, paying attention to appropriate relative magnitudes of the vectors that you draw. (b) Show the approximate distribution of charges for this circuit. Make the important aspects of the charge distribution very clear in your drawing, supplementing your diagram if necessary with very brief written descriptions on the diagram. Make sure that parts (a) and (b) of this problem are consistent with each other. (c) Assume that you know the mobile-electron density n and the electron mobility u at room temperature for Nichrome. The lengths (L1,L2,L3)and diameters (d1,d2)of the wires are given on the diagram. Calculate accurately the number of electrons that leave the negative end of the battery every second. Assume that no part of the circuit gets very hot. Express your result in terms of the given quantities (K,L1,L2,L3,d1,d2,nandu). Explain your work and identify the principles you are using. (d) In the case that d2d1, what is the approximate number of electrons that leave the negative end of every second? (e) A voltmeter is attached to the circuit with its + lead connected to location B (halfway along the leftmost thick wire) and its - lead connected to location C (halfway along the leftmost thin wire). In the case that d2d1, what is the approximate voltage shown on the voltmeter, including sign? Express your result in terms of the given quantities (K,L1,L2,L3,d1,d2,nandu).

Short Answer

Expert verified

(a) The direction of the electric field is anticlockwise in the circuit, and the field is smaller in the thick wire than the field in the thin wire.

Step by step solution

01

Given data

Given data can be listed below as:

A circuit is given which has a battery, with emf value K and five Nichrome wires out of 5, three are thick, and two are thin.

02

Concept

The number of charge carriers moving in a conductor can be written as,

ne=nAuEt (1)

Here n is the charge density of the charge carrier, A is the cross-sectional area of the conductor, E is the electric field in the conductor, u is the mobility of the charge carriers, t is the time.

03

(a) Drawing and labeling the electric field at the locations marked × inside the wires

The charge carriers are electrons; in this case, thus for thick wire, we can write equation 1 for t = 1 s as,

n1=nA1uE1 (2)

Here A1is the cross-sectional area of the thick wire, E1is the electric field in the thick wire.

The charge carriers are electrons; in this case, thus for thin wire, we can write equation 1 for t = 1 s as,

n2=nA2uE2 (3)

Here A2 is the cross-sectional area of the thin wire, E2 is the electric field in the thin wire.

The number of electrons that are passing through the circuit is constant; thus, the number of electrons flowing through the thick and thin wire will be the same.

From equations 2 and 3, we can write,

n1=n2

Substitute the values in the above expression, and we get,

nA1uE1=nA2uE2A1E1=A2E2

From the above expression, we can conclude that the thick wire has a larger cross-sectional area; thus, the electric field will be small, and for thin wire, as the cross-sectional area is small, the electric field will be large, and the electric fields will be uniform throughout the wires.

The direction of the electric field can be drawn as,

Thus, the direction of the electric field is anticlockwise in the circuit, and the field is smaller in the thick wire than the field in the thin wire.

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