A3.80 mm diameter wire carries a 15.2 A current when the electric field is 8.60x10-2 V/m Part A What is the wire's resistivity? p= 6.42x10-8 Am Previous Answers v Correct Correct answer is shown. Your answer 6.4167-10-8 = 6.4167x10-8 f2m was either rounded differently or used a different number required for this part. significant figures than Part B If a battery voltage of 8.55 V is applied to this wire to produce this current, how long must it be? ? L= 994 Submit Previous Answers Request Answer
A3.80 mm diameter wire carries a 15.2 A current when the electric field is 8.60x10-2 V/m Part A What is the wire's resistivity? p= 6.42x10-8 Am Previous Answers v Correct Correct answer is shown. Your answer 6.4167-10-8 = 6.4167x10-8 f2m was either rounded differently or used a different number required for this part. significant figures than Part B If a battery voltage of 8.55 V is applied to this wire to produce this current, how long must it be? ? L= 994 Submit Previous Answers Request Answer
Chapter9: Current And Resistance
Section: Chapter Questions
Problem 55P: A 20.00-V battery is used to supply current to a 10-k resistor. Assume the voltage drop across any...
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![I Review | Constants
A3.80 mm diameter wire carries a 15.2 A current when the electric
field is 8.60x 10-2 V/m.
Part A
What is the wire's resistivity?
p= 6.42x10-8 2m
Submit
Previous Answers
Correct
Correct answer is shown. Your answer 6.4167-10-8 = 6.4167x10-8 Qm was either rounded differently or used a different number of significant figures than
required for this part.
Part B
If a battery voltage of 8.55 V is applied to this wire to produce this current, how long must it be?
?
L= 994
Submit
Previous Answers Request Answer
X Incorrect; Try Again; 2 attempts remaining](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4643071b-b593-4e3e-9113-36a7194eebe9%2Ff3f0bf43-06df-42fd-b4e3-b40c948e974e%2F7896x4g_processed.png&w=3840&q=75)
Transcribed Image Text:I Review | Constants
A3.80 mm diameter wire carries a 15.2 A current when the electric
field is 8.60x 10-2 V/m.
Part A
What is the wire's resistivity?
p= 6.42x10-8 2m
Submit
Previous Answers
Correct
Correct answer is shown. Your answer 6.4167-10-8 = 6.4167x10-8 Qm was either rounded differently or used a different number of significant figures than
required for this part.
Part B
If a battery voltage of 8.55 V is applied to this wire to produce this current, how long must it be?
?
L= 994
Submit
Previous Answers Request Answer
X Incorrect; Try Again; 2 attempts remaining
Expert Solution
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Step 1
Part A)
Let the radius of the cross-section of the wire be r, the area of the cross-section of the wire be A, the magnitude of the electric field applied be E, and the current flowing through the wire be I.
The value of resistivity of the wire is calculated by using the following equation.
In this case,
r = 1.90×10-3 m
Substitute the required values in equation (1) to get the value of the resistivity of the material of the wire in the following way.
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