Asilver wire with resistivity 1.59 × 10-8 ω-m carries a current density of 4.0 a/mm2, what is the magnitude of the electric field inside the wire? 10 a) 0.064 v/m b) 2.5 v/m c) 0.10 v/m d) 0.040 v/m

Electric field, E = 0.064 V/m

Explanation:

It is given that,

Resistivity of silver wire,

Current density of the wire,

We need to find the magnitude of the electric field inside the wire. The relationship between electric field and the current density is given by :

E = 0.0636 V/m

or

E = 0.064 V/m

So, the magnitude of electric field inside the wire is 0.064 V/m. Hence, this is the required solution.

a

explanation:

no diagram was provided, but i'm guessing it's the one i've attached to my answer.

here, a mechanical force turns the rod.   this generates an electric current in the coil, which powers the bulb.