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Chapter 4: Q21E (page 842)

A current-carrying gold wire has diameter 0.84 mm. The electric field in the wire is 0.49 V/m. What are (a) the current carried by the wire; (b) the potential difference between two points in the wire 6.4 mapart; (c) the resistance of a 6.4 mlength of this wire?

Short Answer

Expert verified

(a) The current carried by the wire, l = 11.125 A

(b) The potential difference between two points in the wire, V = 3.136 V

(c) The resistance of the wire,R=0.282惟

Step by step solution

01

Determine the area of the wire

A=蟺4d2A=蟺4脳0.84脳10-32A=5.54脳10-7m2

02

Determine the formula to find current carried by the wire

We know,

E=pJ

And

J=lA

Therefore,

l=EAp

03

Determine the current carried by the wire

(a) Use the formula found above,

l=Eapl=0.49V/m5.54脳10-7m22.44脳10-8惟ml=11.125A

Therefore, the current carried by the wire is 11.125 A

04

Determine the potential difference between two points in the wire

(b) Use the formula

V=EL

Substitute the given values

V=0.49V/m6.4mV=3.136V

Thus, the potential difference between two points in the wire 6.4m apart is 3.136 V

05

Determine the resistance of the wire

Use the formula of resistance

R=蚁LA

Substitute the values

R=2.44脳10-8惟m6.4m5.54脳10-7m2R=0.282惟

Therefore, the resistance of a 6.4 M length of this wire is 0.282惟

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