Q6Q - Fundamentals Of Physics

Question

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.

Step-by-Step Solution

Answer

a)   The ranking the sections of wire according to the current is $I_A=I_B=I_C$.

b)   The ranking the sections of wire according to the magnitude of the current density is $J_B>J_C>J_A$.

c)   The ranking the sections of wire according to the magnitude of the electric field is $E_B>E_C>E_A$.

Step 1: The given data

Figure 26-20 is the wire with different radii.

Step 2: Understanding the concept of the current, current density, and electric field

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.

Step 3: (a) Calculation of the rank of the sections according to the current

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 $I_A=I_B=I_C$.

Step 4: (b) Calculation of the rank of the sections according to the current density

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,$J_B>J_C>J_A$.

Step 5: (c) Calculation of the ranking according to the magnitude of electric field

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,$E_B>E_C>E_A$.