The figure below represents a section of a circular conductor of nonuniform diam
ID: 1391236 • Letter: T
Question
The figure below represents a section of a circular conductor of nonuniform diameter carrying a current of I = 6.00 A. The radius of cross-section A1 is r1 = 0.390 cm.
(a) What is the magnitude of the current density across A1?
??? A/m2
The radius r2 at A2 is larger than the radius r1 at A1.
(b) Is the current at A2 larger, smaller, or the same?
The current is larger.
The current is smaller.
The current is the same.
(c) Is the current density at A2 larger, smaller, or the same?
The current density is larger.
The current density is smaller.
The current density is the same.
Assume A2 = 3A1.
(d) Specify the radius at A2.
??? mm
(e) Specify the current at A2.
??? A
(f) Specify the current density at A2.
??? A/m2
Explanation / Answer
Given that,
I = 6 A ; r1 = 0.390 cm = 0.0039 meters
(a) We know that the current density is given by:
J = I/A , where J is the current density, I the the flowing current and A is the Area of cross section.
J1 = I /A1
A1 = pi (ri)2 = 3.14 x (0.0039)2 = 4.78 x 10-5 m2
J1 = 6 / 4.78 x 10-5 = 1.26 x 105 A/m2
Hence, J1 = 1.26 x 105 A/m2
(b) The current is the same.
Because, the charge entering the conductor must be the same, as the charge leaving the conductor.
(c)The current density is smaller.
Because, As we know that, J is inversely related to Area of cross section, and we see in the figure that A2>>A1, implies that the value of J2= I/A2 will be smaller.
(d)Condition given, A2 = 3 A1 = 3 (4.78 x 10-5) = 14.34 x 10-5 m2
pi (r2)2 = 14.34 x 10-5 m2 => r2 = sqrt ( 14.34 x 10-5 / 3.14)
r2 = 0.00676 m
Hence, r2 = 6.76 mm
(e) Current at A2 as dicussed in part (b) will be same as A1
hence,I2= 6 Amperes (I1 = I2 = I= 6 A)
(f) Again using the realtion,
J2 = I / A2 = 6 / 14.34 x 10-5 = 0.418 x 105 = 4.2 x 104 A/m2
Hence, J2 = 4.2 x 104 A/m2
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