Question

The wires leading to and from a 0.12-mm-diameter lightbulb filament are 1.5 mm in diameter. The wire to the filament carries a current with a current density of 4.9×105 A/m2 Part A What is the current in the filament? Express your answer to two significant figures and include the appropriate units. Part B What is the current density in the filament? Express your answer to two significant figures and include the appropriate units.

Answer 1

Current density is given by:

J = I/A

Given that current density from wire to the filament = 4.9*10^5 A/m^2

diameter of filament = 1.5 mm

radius of filament = 1.5/2 = 0.75 mm

So,

I = J*A

I = J*pi*r^2

I = 4.9*10^5*pi*(0.75*10^-3)^2

I = 0.87 Amp = current in the filament

Part B.

Now diameter of the wire = 0.12 mm

radius of wire, r1 = 0.06 mm

So, Current density in filament will be:

J1 = I1/A1

I1 = Current in filament = 0.87 Amp

A1 = pi*r1^2

So,

J1 = 0.87/(pi*(0.06*10^-3)^2)

J1 = 7.7*10^7 A/m^2

Please Upvote.