An 8-gauge copper wire (diameter = 3.665 mm)carries a current of 40 A.The current density of the wire is A)1.52 $\times$ 10⁷ A/m² B)3.79 $\times$ 10⁶ A/m² C)1.90 $\times$ 10⁶ A/m² D)1.21 $\times$ 10⁶ A/m² E)None of these is correct.

Question

Quizplus · Accepted Answer

The formula for current density (J) is:

J = I / A

where I is the current and A is the cross-sectional area of the wire.

The cross-sectional area of an 8-gauge wire is given by:

A = πr^2 = π(d/2)^2

where r is the radius (diameter/2) and d is the diameter.

Substituting the given values, we get:

A = π(3.665/2)^2 = 10.561 mm^2 = 1.0561×10^-5 m^2

Now, we can calculate the current density:

J = I / A = 40 / 1.0561×10^-5 = 3.787×10^6 A/m^2

Therefore, the current density of the wire is approximately 3.79×10^6 A/m^2, which corresponds to choice B.

An 8-Gauge Copper Wire (Diameter = 3 ×\times× 107 A/m2 B)3 ×\times×

An 8-Gauge Copper Wire (Diameter = 3 $\times$ 10⁷ A/m²
B)3 $\times$