Question

A cylindrical wire of radius 2 mm carries a current of 3.0 A. The potential difference between points on the wire that are 44 m apart is 3.8 V.
Required:
a. What is the electric field in the wire?
b. What is the resistivity of the material of which the wire is made?

Answer 1

Answer:

a. E = 86.36 x 10⁻³ V/m = 86.36 mV/m

b. ρ = 3.6 x 10⁻⁷ Ωm

Explanation:

a.

The electric field in terms of the voltage is given by the following formula:

E = V/d

where,

E = Electric Field in the Wire = ?

V = Potential Difference = 3.8 V

d = distance between the points = 44 m

Therefore,

E = 3.8 V/44 m

E = 86.36 x 10⁻³ V/m = 86.36 mV/m

b.

Now, from Ohm's Law:

V = IR

R = V/I

where,

R = Resistance of wire = ?

I = Current = 3 A

Therefore,

R = 3.8 V/3 A

R = 1.27 Ω

Now, the resistance of a wire can be given as:

R = ρL/A

where,

ρ = resistivity of material = ?

L = Length = 44 m

A = Cross-sectional area = πr² = π(0.002 m)² =  1.25 x 10⁻⁵ m²

Therefore,

1.27 Ω = ρ*44 m/1.25 x 10⁻⁵ m²

(1.27 Ω)(1.25 x 10⁻⁵ m²)/44 m = ρ

ρ = 3.6 x 10⁻⁷ Ωm