The figure below represents a section of a circular conductor of nonuniform diameter carrying a current of 4.50 A. The radius of cross-section A1 is 0.300 cm.
(a)What is the magnitude of the current density across A1?
A: 1.59155e2 using J=I/((r^2)pi)
(b) If the current density across A2 is one-fourth the value across A1, what is the radius of the conductor at A2?
I cannot figure out part b.
b. I/(pi*r^2) = 1.59e2/4. r = ?.
To solve part (b), we need to find the radius of the conductor at A2 when the current density across A2 is one-fourth the value across A1. We can use the formula for current density:
J = I / (r^2 * π)
We know that the current density across A1 is J1 = 1.59155e2 A/m^2.
Let's assume that the radius of cross-section A2 is r2, and the current density across A2 is J2. We can set up the following equation:
J2 = (1/4) * J1
Substituting in the values, we have:
(1/4) * J1 = I / (r2^2 * π)
Now, we can rearrange the equation to isolate r2:
r2^2 = I / ((1/4) * J1 * π)
To find the radius, we take the square root of both sides:
r2 = √(I / ((1/4) * J1 * π))
Substituting in the known values:
r2 = √(4.50 A / ((1/4) * 1.59155e2 A/m^2 * π))
Calculating the expression inside the square root:
r2 = √(4.50 A / (3.978875e1 A/m^2 * π))
r2 = √(4.50 A / (3.978875e1 A/m^2 * 3.14159))
Simplifying further:
r2 = √(0.03564825 m^2)
r2 = 0.188962 m
Therefore, the radius of the conductor at A2 is approximately 0.189 m.
To find the radius of the conductor at A2, we can use the formula:
J = I / (r^2 * π)
where J is the current density, I is the current, and r is the radius.
We are given the current density across A1 (J1) and we know that the current density across A2 (J2) is one-fourth the value across A1. Let's assume the radius of A2 is r2.
We can set up the following equation:
J2 = J1 / 4
Substituting the formula for current density (J) into the equation, we have:
I / (r2^2 * π) = I / (0.3^2 * π) / 4
Cancelling out the common terms, we get:
1 / (r2^2 * π) = 1 / (0.3^2 * π) / 4
To simplify the equation further, we can cross-multiply:
(r2^2 * π) = (0.3^2 * π) / 4
Now we can solve for r2:
r2^2 = (0.3^2 * π) / 4π
r2^2 = 0.3^2 / 4
Taking the square root of both sides:
r2 = √(0.3^2 / 4)
r2 = 0.3 / 2
r2 = 0.15 cm
Therefore, the radius of the conductor at A2 is 0.15 cm.