The radius of a piece of Nichrome wire is 0.314 mm. (A) Calculate the resistance

Question

The radius of a piece of Nichrome wire is 0.314 mm. (A) Calculate the resistance per unit length of this wire. (B) If a potential difference of 10 V is maintained across a 1.0-m length of the Nichrome wire, what is the current in the wire? SOLVE IT (A) Calculate the resistance per unit length of this wire. Conceptualize Nichrome has a resistivity two orders of magnitude larger than the best conductors. Therefore, we expect it to have some special practical applications that the best conductors may not have. Categorize We model the wire as a cylinder so that a simple geometric analysis can be applied to find the resistance. Analyze Use the equation and the resistivity of Nichrome to find the resistance per unit length: R = A = r2 = 1.50 10-6 ·m (3.14 10-4 m)2 = 4.84 Correct: Your answer is correct. /m (B) If a potential difference of 10 V is maintained across a 1.0-m length of the Nichrome wire, what is the current in the wire? Analyze Find the current: I = V R = V (4.84 /m) = 10 V (4.84 /m)(1.0 m) = 2.07 Correct: Your answer is correct. A Finalize A copper wire of the same radius would have a resistance per unit length of only 0.055 /m. A 1.0-m length of copper wire of the same radius would carry the same current with an applied potential difference of only 0.114 V. Because of its high resistivity and resistance to oxidation, Nichrome is often used for heating elements in toasters, irons, and electric heaters. MASTER IT HINTS: GETTING STARTED | I'M STUCK! For the wire in the example, calculate the following assuming that the wire carries a current of 4.9 A. (a) the current density J = 1.58E7 Correct: Your answer is correct. A/m2 (b) the electric field in the wire E = 27. Incorrect: Your answer is incorrect. N/C

Explanation / Answer

Data from example:

= 1.50 x 10-6 -m

r = 3.14 x 10-4 m

Given Data:

wire carries a current = 4.9 A

Current density J = I/A

J = 4.9A / (*(3.14 x 10-4m)2 )

J = 4.9A / 3.095914 x 10-7m2

J = 15827313 A/m2

J = 1.58 x 107 A/m2

(b)

= E/J

E = *J

where = resistivity, E = electric field, and J = current density

E = 1.50 x 10-6 -m *1.58 x 107 A/m2

E = 2.37409695 -m * 10A/m2

E = 23.7409695 N/C

E = 23.7 N/C

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