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

Step by step solution

Determine the area of the wire

$A = \frac{π}{4} d^{2} A = \frac{π}{4} \times {(0.84 \times 10^{- 3})}^{2} A = 5.54 \times 10^{- 7} m^{2}$

Determine the formula to find current carried by the wire

We know,

$E = p J$

And

$J = \frac{l}{A}$

Therefore,

$l = \frac{E A}{p}$

Determine the current carried by the wire

(a) Use the formula found above,

$l = \frac{E a}{p} l = \frac{(0.49 V / m) (5.54 \times 10^{- 7} m^{2})}{2.44 \times 10^{- 8} Ω m} l = 11.125 A$

Therefore, the current carried by the wire is 11.125 A

Determine the potential difference between two points in the wire

(b) Use the formula

$V = E L$

Substitute the given values

$V = (0.49 V / m) (6.4 m) V = 3.136 V$

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

Determine the resistance of the wire

Use the formula of resistance

$R = \frac{ρ L}{A}$

Substitute the values

$R = \frac{(2.44 \times 10^{- 8} Ω m) (6.4 m)}{5.54 \times 10^{- 7} m^{2}} R = 0.282 Ω$

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

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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

Step by step solution

Determine the area of the wire

Determine the formula to find current carried by the wire

Determine the current carried by the wire

Determine the potential difference between two points in the wire

Determine the resistance of the wire

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