In Figure, current is set up through a truncated right circular cone of resistivity $\mathbf{731}\mathbf{\Omega }\mathbf{.}\mathbf{m}$, left radius a = 2.00mm, right radius b = 2.30mm, and lengthL = 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?

a) The resistivity of the circular cone is $\mathrm{\rho }=731\mathrm{\Omega }.\mathrm{m}$

b) The left radius of the circular cone is $\mathrm{a}=2.00\mathrm{mm}\mathrm{or}2.00\times {10}^{-3}\mathrm{m}$

c) The right radius of the circular cone is $b=2.30\mathrm{mm}\mathrm{o}2.30\times {10}^{-3}\mathrm{m}$

d) The length of the cone is$L=1.94cm\mathrm{or}1.94\times {10}^{-2}\mathrm{m}$

The current is related to the flow of charge. The rate of flow of charge with respect to time is called current. The current density is equal to the current per unit cross-section area of the conductor. We can use the formula of current density in terms of electric field and current. By using Ohm’s law, we can find the resistance.

Formulae:

The current density due to the current flow through the wire, $J=\frac{i}{A}$ …(i)

Here, I is current, J is current density, and A is area of the cross-section.

The current density due to the electric field, $J=\frac{E}{\mathrm{\rho }}$ …(ii)

Here, J is the current density,E is the electric field,$\rho$ is resistivity of the material.

The potential difference due to the differential equation, $V=-{\int }_0^{L}Edx$ …(iii)

Here, E is the electric field, V is voltage, dx is small distance.

The voltage equation using Ohm’s law,V = IR …(iv)

Here, V is the potential difference, I is current, R is resistance.

The cross-sectional area of the wire, $A=\pi r^2$ …(v)

Here, A is area of cross-section, r is radius.

The current i is flowing from left to right in the positive x direction. The current density at each value of x can be given using equation (v) in equation (i) as follows:

$J=\frac{i}{\pi r^2}$ …(vi)

Where, A is the cross-sectional area at a particular value of x.

Now, comparing equations (ii) and (vi), we can get that

$\frac{i}{\pi r^2}=\frac{E}{\mathrm{\rho }}$ …(vii)

The radius of the cone increases linearly with x so

$r=c_1+c_{2}x$ …(viii)

Consider that the origin is at the left end of cone, and we chose the coefficient $c_1$such that r = a,

So, when x = 0, we get

$c_1=a$ …(ix)

Also, the coefficient of $c_2$must be chosen at the right-side end of the cone so that r = b .

Thus, the value of the second coefficient is given as:

$c_2=\frac{\left(b-a\right)}{L}$

Now, substituting the above value and value of coefficient from equation (ix) in equation (viii), we can get the value of the radius as follows:

$r=a+\left(\frac{b-a}{L}\right)x$

Now, substituting the value of radius in equation (vii), we can get the value of the electric field as follows:

$E=\frac{i\mathrm{\rho }}{\pi }{\left[a+\left(\frac{b-a}{L}\right)x\right]}^{-2}$

Thus, the potential difference between the two ends of the cone using the above value in equation (iii) can be given as follows:

$$\begin{gathered} V=-{\int }_0^L\frac{i\mathrm{\rho }}{\pi }{\left[a+\left(\frac{b-a}{L}\right)x\right]}^{-2}dx \\ =\frac{i\mathrm{\rho }}{\pi }\frac{L}{b-a}{\left[{\left(a+\left(\frac{b-a}{L}\right)x\right)}^{-1}\right]}_0^L \\ =\frac{i\mathrm{\rho }}{\pi }\frac{L}{b-a}\left(\frac{1}{a}-\frac{1}{b}\right) \\ =\frac{i\mathrm{\rho }}{\pi }\frac{L}{b-a}\left(\frac{b-a}{ab}\right) \\ =\frac{i\mathrm{\rho L}}{\pi ab} \end{gathered}$$

Now, using the above value of potential difference in equation (iv), we can get the value of the resistance of the cone from the substituted values as follows:

$$\begin{gathered} R=\frac{\left(\frac{i\mathrm{\rho L}}{\pi ab}\right)}{i} \\ =\frac{\mathrm{\rho }L}{\pi ab} \\ =\frac{731\Omega .m\times \left(1.94\times {10}^{-2}\right)m}{3.14\times \left(2.00\times {10}^{-3}\right)m\times \left(2.30\times {10}^{-3}\right)m} \\ =9.81\times {10}^5\Omega \end{gathered}$$

Hence, the value of the resistance is $9.81\times {10}^5\Omega$.