g 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 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;

Where;

A is area of the wire

ρ is the resistivity of gold = 2.44 x 10⁻⁸ ohm. m

The diameter of the wire is given by;

Therefore, the diameter of the wire is 1.596 x 10⁻⁵ m.

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j