A current-carrying gold wire has a diameter of 0.84 mm. The electric field in th

Question

A current-carrying gold wire has a diameter of 0.84 mm. The electric field in the wire is 0.48 V/m. What is the current carried by the wire? Express your answer using two significant figures. Submit My Answers Give Up What is the potential difference between two points in the wire 6.3 m apart? Express your answer using two significant figures. Submit My Answers Give Up What is the resistance of a 6.3-m length of this wire? Express your answer using two significant figures. Submit My Answers Give Up

Explanation / Answer

a)

p of AU = 2 .44 * 10 -8

A = pi * r^2

= pi * [0.84 * 10-3 / 2 ]^2

I = [0.48 * pi(0.84 / 2 )^2] / 2 .44 * 10 -8

= 10.9 A

----------------------------------------------------

b)

V = I R

R = p L / A

V = 10.9 *2.44 * 10-8*6.3 / pi(0.84 / 2 )^2

= 3. 0 V

--------------------------------------------------

c)

R = V / I

= 3 / 10.9

= 0.26 ohm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.