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?

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

where I is current and A is area and

Substituting this into the first equation then

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

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c

explanation: