Question

The current density varies with radial distance \[r\] as \[J = a{r^2}\], in a cylindrical wire of radius \[R\]. The current passing through the wire between the radial distance \[\dfrac{R}{3}\] and \[\dfrac{R}{2}\] is?
(A) \[\dfrac{{65\pi \alpha {R^4}}}{{2592}}\]
(B) \[\dfrac{{25\pi \alpha {R^4}}}{{72}}\]
(C) \[\dfrac{{65\pi {\alpha ^2}{R^3}}}{{2938}}\]
(D) \[\dfrac{{81\pi {\alpha ^2}{R^4}}}{{144}}\]

Answer 1

Hint: First of all, differentiate the area of the cross section, to find a relation between the small change in both area and radius. Find the total current by integrating the product of current density and the infinitesimal area. Take the upper and the lower limits and manipulate accordingly.

Complete step by step answer:
In the given problem, we are given the current density which varies with the radial distance by the following relation:
\[J = a{r^2}\] …… (1)
Where,
\[J\] indicates current density.
\[r\] indicates radial distance.
Again, we know that, area of the cross section of the wire (area of cross section is circular in shape) is:
\[A = \pi {r^2}\] …… (2)
Now, we differentiate the above equation, with respect to radius, we get:
$
A = \pi {r^2} \\
\Rightarrow\dfrac{{dA}}{{dr}} = \dfrac{d}{{dr}}\left( {\pi {r^2}} \right) \\
\Rightarrow\dfrac{{dA}}{{dr}} = 2\pi r \\
\Rightarrow dA = 2\pi rdr \\
$
Since, we were given the current density, so to find the total current, we have to integrate the current over the whole area.
Mathematically, we can write:
\[I = \int {JdA} \] …… (3)
We take the lower limit for radial distance as \[\dfrac{R}{3}\] and the upper limit as \[\dfrac{R}{2}\] .
Now, we substitute \[J = a{r^2}\] and \[dA = 2\pi rdr\] in equation (3), we get:
$
I = \int_{R/3}^{R/2} {JdA} \\
\Rightarrow I = \int_{R/3}^{R/2} {a{r^2} \times 2\pi rdr} \\
\Rightarrow I = 2\pi a\int_{R/3}^{R/2} {{r^3}dr} \\
\Rightarrow I = 2\pi a\left[ {\dfrac{{{r^4}}}{{3 + 1}}} \right]_{R/3}^{R/2} \\
$
Now, we manipulate the expression further:
$
I = 2\pi a\left[ {\dfrac{{{r^4}}}{4}} \right]_{R/3}^{R/2} \\
\Rightarrow I = 2\pi a \times \dfrac{1}{4}\left[ {{r^4}} \right]_{R/3}^{R/2} \\
\Rightarrow I = \dfrac{{\pi a}}{2}\left[ {{{\left( {\dfrac{R}{2}} \right)}^4} - {{\left( {\dfrac{R}{3}} \right)}^4}} \right] \\
\Rightarrow I = \dfrac{{\pi a}}{2}\left[ {\dfrac{{{R^4}}}{{{2^4}}} - \dfrac{{{R^4}}}{{{3^4}}}} \right] \\
$
Now, simplifying the expression:
$
I = \dfrac{{\pi a}}{2}\left[ {\dfrac{{{R^4}}}{{16}} - \dfrac{{{R^4}}}{{81}}} \right] \\
\Rightarrow I = \dfrac{{\pi a}}{2}\left[ {\dfrac{{81{R^4} - 16{R^4}}}{{1296}}} \right] \\
\Rightarrow I = \dfrac{{\pi a}}{2}\left[ {\dfrac{{65{R^4}}}{{1296}}} \right] \\
\therefore I = \dfrac{{65\pi a{R^4}}}{{2592}} \\
$
Hence, the current passing through the wire between the radial distance \[\dfrac{R}{3}\] and \[\dfrac{R}{2}\] is found to be \[\dfrac{{65\pi a{R^4}}}{{2592}}\] .
So, the correct option is (A).

Additional information:
In electromagnetism, current density is the volume of charge that passes through a unit area of a selected cross section per unit time. The electrical current density in SI base units is calculated in amperes per square metre. The rate of flow of electric charges past a point is an electrical current. When a net flow of electric charge through an area occurs, an electric current is assumed to exist. Electric charges are borne by charged particles, so a stream of charged particles is an electric current. Depending on the conductor, the travelling particles are called charge carriers, and may be one of many kinds of particles.

Note: To solve this problem, you should be well aware of integrating an expression. Just keep in mind that to find the whole quantity integration comes handy. In this problem, while taking the integration limits, always keep in mind that lower limit is the smaller value while the upper limit is the larger value. Changing the orders, will affect the result.