Question

A small but measurable current of 5.8 × 10-10 A exists in a copper wire whose diameter is 3.0 mm. The number of charge carriers per unit volume is 8.49 × 1028 m-3. Assuming the current is uniform, calculate the (a) current density and (b) electron drift speed.

Answer 1

Answer:

a) The current density ,J = 2.05×10^-5

b) The drift velocity Vd= 1.51×10^-15

Explanation:

The equation for the current density and drift velocity is given by:

J = i/A = (ne)×Vd

Where i= current

A = Are

Vd = drift velocity

e = charge ,q= 1.602 ×10^-19C

n = volume

Given: i = 5.8×10^-10A

Raduis,r = 3mm= 3.0×10^-3m

n = 8.49×10^28m^3

a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]

J = (5.8×10^-10) /(2.83×10^-5)

J = 2.05 ×10^-5

b) Drift velocity, Vd = J/ (ne)

Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)

Vd = (2.05×10^-5)/(1.36 ×10^10)

Vd = 1.51× 10^-5

Answer 2

Answer: a) J = 8.2×10^-5A/m^2

b) V = 6.07×10^-15m/s

Explanation:

The current density is the ratio of current to area of the wire.

J = i/A

Drift speed is inversly proportional to the square of the radius of the wire.v α 1/r^2

Please find the attached file for the solution