Question

A "gauge 8" jumper cable has a diameter d of 0.326 centimeters. The cable carries a current I of 30.0 amperes. The electric field E in the cable is 0.062 newtons per coulomb. part a: What electric field E′ would have been required to create a current of 30.0 amperes in a copper "gauge 10" wire with diameter d equal to 0.259 centimeters?

Answer 1

Answer:

0.0979 N/c

Explanation:

Electric field, E is given as a product of resistivity and current density

E=jP where P is resistivity and j is current density

But the current density is given as

$j=\frac {I}{A}$ where I is current and A is area and $A=\pi r^{2}$

Substituting this into the first equation then $E=P\times \frac {I}{\pi r^{2}}$

Given diameter of 0.259 cm= 0.00259 m and the radius will be half of it which is 0.001295 m

$E=1.72\times 10^{-8}\times \frac {30}{\pi \times 0.001295^{2}}=9.79\times 10^{-2} N/c=0.0979 N/c$