Question

A potential difference of 5.00 V will be applied to a 47.00 m length of 18-gauge tungsten wire (diameter = 0.0400 in). Calculate the current.

Calculate the magnitude of the current density.

Calculate the magnitude of the electric field within the wire.

Calculate the rate at which thermal energy will appear in the wire.

Tungsten: 5.28e-8

Answer 1

Part A.

Resistance of a wire is given by:

R = rho*L/A

L = 47.00 m

A = pi*d^2/4

d = diameter = 0.0400 in = 0.001016 m (Since 1 inch = 0.0254 m)

Using ohm's law:

V = I*R

I = V/R = V*A/(rho*L) = V*pi*d^2/(4*rho*L)

Using given values:

I = 5.00*pi*0.001016^2/(4*5.28*10^-8*47.00)

I = Current in wire = 1.633 Amp

Part B.

Current density is given by:

J = Current/Cross-section Area = I/A

I = Current in wire = 1.633 A

A = Cross-sectional area of wire = pi*d^2/4

So,

J = 1.633*4/(pi*0.001016^2)

J = 2014229.2 A/m^2 = 2.01*10^6 A/m^2

Part C.

Relation between current density and Electric field in wire is given by:

J = E

= conductivity of material = 1/resistivity of material = 1/(5.28*10^-8)

So,

E = J/

E = 2.01*10^6*5.28*10^-8

E = 0.106 N/C

Part D.

rate at which thermal energy appears in the wire will be equal to power dissipated, So

P = V^2/R = V*I

P = 5.00*1.633

P = 8.165 W

Let me know if you've any query.