A potential difference of 1.00 Vwill be applied to a 43.0 m length of18-gauge co

Question

A potential difference of 1.00 Vwill be applied to a 43.0 m length of18-gauge copper wire (diameter = 0.0400 in.). Calculate thefollowing values. (a) the current
1 A

(b) the magnitude of the current density
2 A/m2

(c) the electric field
3 V/m

(d) the rate at which thermal energy will appear in the wire
4 W
A potential difference of 1.00 Vwill be applied to a 43.0 m length of18-gauge copper wire (diameter = 0.0400 in.). Calculate thefollowing values. (a) the current
1 A

(b) the magnitude of the current density
2 A/m2

(c) the electric field
3 V/m

(d) the rate at which thermal energy will appear in the wire
4 W
(a) the current
1 A

(b) the magnitude of the current density
2 A/m2

(c) the electric field
3 V/m

(d) the rate at which thermal energy will appear in the wire
4 W

Explanation / Answer

=1.72E10-8 *m
resistance is equal to R=L/A where A is the cross sectional area
L=43m convert the inches to meters..... 0.04in=.001016m A=(5.08E-4)2=8.107E-7m2
R=(1.73E-8)(43)/(8.107E-7) R=0.9175
V=IR 1V=I(0.9175) I=1.089A
(b) current density is J=I/A J=(1.089)/8.107E-7 J=1344319.71 A/m2

(c) electric field is V=Ed E=V/d E=(1)/(43m) E=.0232V/m
(d) P=IV P=(1.089)(1) P=1.089W

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