A 3.0 m length of wire is made by welding the end of a 110 cm long silver wire t

Question

A 3.0 m length of wire is made by welding the end of a 110 cm long silver wire to the end of a 190 cm long copper wire . Each piece of wire is 0.80 mm in diameter. The wire is at room temperature, so the resistivities are as given in the table below. A potential difference of 7.0 V is maintained between the ends of the 3.0 m composite wire

A 3.0 m length of wire is made by welding the end of a 110 cm long silver wire to the end of a 190 cm long copper wire . Each piece of wire is 0.80 mm in diameter. The wire is at room temperature, so the resistivities are as given in the table below. A potential difference of 7.0 V is maintained between the ends of the 3.0 m composite wire (a) What is the current in the copper section? A (b) What is the current in the silver section? A (c) What is the magnitude of Ein the copper? V/m (d) What is the magnitude of Ein the silver? V/m (e) What is the potential difference between the ends of the sliver section of wire? V

Explanation / Answer

Area

A=pi*d2/4 =pi*(0.8*10-3)2/4

A=5.03*10-7 m2

Resistance of silver section

Rs=psLs/A=(1.47*10-8)*1.1/(5.03*10-7) =0.032 ohms

Resistance of Copper section

Rc=pcLc/A=(1.72*10-8)*1.9/(5.03*10-7) =0.065 ohms

Total Resistance of wire is

R=Rs+Rc=0.065+0.032=0.097 ohms

Total Current in the wire is

I=V/R =7/0.097 =72.16 A

a)

Current in copper section

Ic=72.16 A

b)

Current in Silver section

Is=72.16 A

c)

Current density in wire is

J=I/A =72.16/(5.03*10-7)=1.4356*108 A/m2

Electric field in copper sectionis

Ec=Jpc=(1.4356*108)(1.72*10-8)=2.47 V/m

d)

Electric field in silver section is

Ec=Jpc=(1.4356*108)(1.47*10-8)=2.11 V/m

e)

Potential difference between the ends of wire in silver sectionis

Vs=IRs=72.16*0.032

Vs=2.31 Volts

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