Answer:

0.45 mm

Explanation:

The complete question is

a certain fuse “blows” if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?

A) 0.45 mm

B) 0.63 mm

C.) 0.68 mm

D) 0.91 mm

Current in the fuse is 1.0 A

Current density of the fuse when it melts is 620 A/cm^2

Area of the wire in the fuse = I/ρ

Where I is the current through the fuse

ρ is the current density of the fuse

Area = 1/620 = 1.613 x 10^-3 cm^2

We know that 10000 cm^2 = 1 m^2, therefore,

1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2

Recall that this area of this wire is gotten as

A = [tex]\frac{\pi d^{2} }{4}[/tex]

where d is the diameter of the wire

1.613 x 10^-7 = [tex]\frac{3.142* d^{2} }{4}[/tex]

6.448 x 10^-7 = 3.142 x [tex]d^{2}[/tex]

[tex]d^{2}[/tex] =[tex]\sqrt{ 2.05*10^-7}[/tex]

d = 4.5 x 10^-4 m = 0.45 mm