Finding diameter of a wire based on the current and current density.

[mkienbau]

You need to design a 0.9 A fuse that "blows" if the current exceeds 0.9 A. The fuse material in your stockroom melts at a current density of 540 A/cm2. What diameter wire of this material will do the job?

540x10^-2=5.40 A/m^2

5.40= .9/x

Solving for x gives the answer: .166666666666667

Then:

.166666666667=pi(r^2)

sqrt(.166666666667/pi)=r

The result of that answer multiplied by 2 gives 4.607m. The answer asks for it in mm so I converted to 460.7mm and it says its wrong. Where did I go wrong, I'm just using the current density formula: J=I/A

Where J is the current density
I is the current
A is the area

 

[OlderDan]

You need to design a 0.9 A fuse that "blows" if the current exceeds 0.9 A. The fuse material in your stockroom melts at a current density of 540 A/cm2. What diameter wire of this material will do the job?

540x10^-2=5.40 A/m^2

5.40= .9/x

Solving for x gives the answer: .166666666666667

Then:

.166666666667=pi(r^2)

sqrt(.166666666667/pi)=r

The result of that answer multiplied by 2 gives 4.607m. The answer asks for it in mm so I converted to 460.7mm and it says its wrong. Where did I go wrong, I'm just using the current density formula: J=I/A

Where J is the current density
I is the current
A is the area

Do what you did again, only this time keep track of your units every step along the way. Unit consistency will do wonders for you.