Answer:
The diameter of the wire is 1.596 x 10⁻⁵ m.
Explanation:
Given;
length of the wire, L = 4.0 m
emf of the battery, V = 1.5 V
current through the wire, I = 4 mA = 0.004 A
The resistance of the wire is given by;
R = V / I
R = 1.5 / 0.004
R = 375 ohms
The resistance of the wire in terms of resistivity is given by;
[tex]R = \frac{\rho L}{A}\\\\ A = \frac{\rho L}{R}\\\\[/tex]
Where;
A is area of the wire
ρ is the resistivity of gold = 2.44 x 10⁻⁸ ohm. m
[tex]A = \frac{2.44*10^{-8}* 4}{375}\\\\A = 2.603*10^{-10} \ m^2[/tex]
The diameter of the wire is given by;
[tex]A = \frac{\pi d^2}{4}\\\\ \pi d^2 = 4A\\\\d = \sqrt{\frac{4A}{\pi} }\\\\ d = \sqrt{\frac{4(2.602*10^{-10})}{\pi} }\\\\d = 1.596 *10^{-5} \ m[/tex]
Therefore, the diameter of the wire is 1.596 x 10⁻⁵ m.
