Solution Found!
In Fig. 26-30, current is set up through a truncated right
Chapter , Problem 35(choose chapter or problem)
In Fig. 26-30, current is set up through a truncated right circular cone of resistivity \(731 \Omega \cdot \mathrm{m}\), left radius a = 2.00 mm, right radius b = 2.30 mm, and length L = 1.94 cm. Assume that the current density is uniform across any cross section taken perpendicular to the length.What is the resistance of the cone?
Questions & Answers
QUESTION:
In Fig. 26-30, current is set up through a truncated right circular cone of resistivity \(731 \Omega \cdot \mathrm{m}\), left radius a = 2.00 mm, right radius b = 2.30 mm, and length L = 1.94 cm. Assume that the current density is uniform across any cross section taken perpendicular to the length.What is the resistance of the cone?
Step 1 of 4
The electric field can be determined as,
\(E = \frac{{i\rho }}{{\pi {r^2}}}\)
Here, \(\rho\) is the resistivity, and r is the radius.
Step 2 of 4
From Figures 26-30, the radius of the cone increases linearly with x. Hence, it can be written as follows:
\(r = {c_1} + {c_2}x\)
Here, \({c_1}\;{\rm{and}}\;{c_2}\) are the constants.
When x is 0, the value of r is equal to a. Hence,
\(a = {c_1} + {c_2}\left( 0 \right)\)
\({c_1} = a\)
When x is L, then the value of r is equal to b. Hence,
\(b = a + {c_2}L\)
\({c_2} = \frac{{b - a}}{L}\)
Substitute the values to solve the equation for r as
\(r = a + \frac{{b - a}}{L}x\)
Substitute the values to so