In Fig. 26-20, a wire that carries a current consists of three sections with different radii. Rank the sections according to the following quantities, greatest first: (a) current, (b) magnitude of current density, and (c) magnitude of electric field.

Figure 26-20 is the wire with different radii.

The rate of flow of electrons per unit time is called electric current. The current density is electric current per unit cross-section area at a given point.To explain the electrostatic force between the two charges, we assume that the charges create an electric field around them. The magnitude of electric field E set up by the electric charge q at a distance r is given as,

$\mathit{E}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}}$

We can use the formulae for current, current density, and electric fields. Inserting the given values of radii in it, we can rank the sections of wire accordingly.

Formulae:

The current flowing through the circuit in the given time, $I=\frac{Q}{t}$ …(i)

Here,I is current,Q is the charge, and t is time.

The current density due to the uniform current, $J=\frac{I}{A}$ …(ii)

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

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

Here,J is the current density,E is the electric field, and$\rho$ is charge density.

From equation (i),the current is charge flow per unit time means; it doesn’t depend on area or radius; thus, the current for all regions will be same.

Hence, the rank of the sections is $I_A=I_B=I_C$.

Substituting the area value of the wire,in equation (ii), we can get the current density as follows:

$J=\frac{I}{\pi R^2}\left(\because A=\pi R^2\right)$ …(iv)

For region A, we get the current density using equation (iv) as follows:

$$\begin{gathered} J_A=\frac{I}{\pi {\left(2R_0\right)}^2} \\ =0.25\left(\frac{I}{\pi 2{R^2}_0}\right) \end{gathered}$$ …(v)

For region B, we get the current density using equation (iv) as follows:

$$\begin{gathered} J_B=\frac{I}{\pi {\left(R_0\right)}^2} \\ =\left(\frac{I}{\pi {R_0}^2}\right) \end{gathered}$$ …(vi)

For region C, we get the current density using equation (iv) as follows:

$$\begin{gathered} J_C=\frac{I}{\pi {\left(1.5R_0\right)}^2} \\ =\frac{I}{2.25\pi R_0^2} \\ =0.44\left(\frac{I}{\pi R_0^2}\right) \end{gathered}$$ …(vii)

From equation (v), (vi), and (vii), we can write the rank as,$J_B>J_C>J_A$.

Using the equation (iii), we can get the electric field as follows:

$E=J_{\mathrm{\rho }}$

Here, resistivity of the material is the same, so the electric field depends on only the current density.

From part b), we can write the rank as,$E_B>E_C>E_A$.