A 4.0 m length of gold wire is connected to a 1.5 V battery, and a current of 4.0 mA flows through it. What is the diameter of the wire? (The resistivity of gold is 2.44 × 10-8 Ω·m.) A. 9.0 μm B. 8.5 μm C. 17 μm D. 48 μm

The diameter of the wire is 18μm and the closest answer is option C.

The resistance of a wire is proportional to its length and inversely proportional to its area.

Given that a  4.0 m length of gold wire is connected to a 1.5 V battery, and a current of 4.0 mA flows through it. Then,

R = (ρL)/A

where

ρ = resistivity = 2.44 × 10-8 Ω·m

L = length

A = Area = D/2

D = diameter of the wire

From Ohm's law, V = IR

make resistance R the subject of the formula

R = V/I

R = 1.5/4 x

R = 375 Ω

Substitute all the parameters into the formula

R = (ρL)/A

375 = (2.44 × x 4)/A

A = (9.76 × )/ 375

A = 2.603 x

but

A =  

2.603 x =  

/4 = 2.603 x

= 1.04 x

= (1.04 x )/

D =

D = 1.8 x m

D = 18 μm

Therefore, the diameter of the wire is 18μm and the closest answer is option C.

Learn more about Resistivity here: link

Explanation:

Given that,

Length of gold wire, l = 4 m

Voltage of battery, V = 1.5 V

Current, I = 4 mA

The resistivity of gold,

Resistance in terms of resistivity is given by :

Also, V = IR

So,

A is area of wire,

, r is radius, r = d/2 (diameter=d)

Out of four option, near option is (C) 17 μm.

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