Current Density Problems and Solutions

 Problem #1 

The (United States) National Electric Code, which sets maximum safe currents for insulated copper wires of various diameters, is given (in part) in the table. Plot the safe current density as a function of diameter. Which wire gauge has the maximum safe current density? (“Gauge” is a way of identifying wire diameters, and 1 mil = 10^─3 in.)
 
Answer;
Plot the safe current density as a function of diameter

we use, area A = πr² and current density is J = I/A, then

Calculate the current density as follows:

Hence, Gauge 21 having maximum current density

Problem #2 
A beam contains 2.0 x 10⁸ doubly charged positive ions per cubic centimeter, all of which are moving north with a speed of 1.0 x 10⁵ m/s. What are the (a) magnitude and (b) direction of the current density J (c) What additional quantity do you need to calculate the total current i in this ion beam?

Answer;
Known:
amount positive ions, n =  2.0 x 10⁸ x 2/cm³ = 4.0 x 10¹⁴/m³
speed, v = 1.0 x 10⁵ m/s
charged positive, e = 1.6 x 10^─19 C

(a) magnitude of the current density J is

J = nev

  = (4.0 x 10¹⁴/m³)(1.6 x 10^─19 C)(1.0 x 10⁵ m/s)

J = 6.4 A/m²

(b) direction of J current density to the north.

(c) the additional amount you need to calculate the total current i in this ion file is to add the conductor cross-sectional area.

Problem #3 
Certain cylindrical wire carries current. We draw a circle of radius r around its central axis in Fig. 02a to determine the current I within the circle. Figure 02b shows current i as a function of r². The vertical scale is set by is i_s = 4.0 mA, and the horizontal scale is set by r_s² = 4.0 mm². (a) Is the current density uniform? (b) If so, what is its magnitude?

Fig.02

Answer;
(a) from the figure (b) the current i varies uniformly as a function of r² because the plot of current versus r² is a straight line.
Yes, current density in the cylindrical wire is uniform
(b) the value of the each line on vertical scale is,
i = i_s/4 = 4.0 mA/4 = 1.0 mA
therefore , the value of current at r_s² is
i = 5 x 1.0 mA = 5.0 mA
the slope of straight line in figure (b) is,

i/r² = (i ─ 0)/(r_s² ─ 0)

      = 5.0 mA/(4.0 mm²)

i/r² = 1250 A/m²

Problem #4 
A fuse in an electric circuit is a wire that is designed to melt, and thereby open the circuit, if the current exceeds a predetermined value. Suppose that the material to be used in a fuse melts when the current density rises to 440 A/cm². What diameter of cylindrical wire should be used to make a fuse that will limit the current to 0.50 A?

Answer;
Known:
current density, J = 440 A/cm² = 440 x 10⁴ A/m²
current, i = 0.50 A

By the definition of the current density we have

J = I/A = i/(πd²/4) = 4i/(πd²)

here, i is the current flowing through the wire, A is the cross-sectional area of the wire, d is the diameter of the wire

From this formula we can find the diameter of the wire that should be used to make a fuse that will limit the current
0.50 A:

440 x 10⁴ A/m² = 4 x 0.50 A/(πd²)

d² = 1.4476 x 10^─7 m²

d = 3.8 x 10^─4 m = 0.38 mm

Share :