A block in the shape of a rectangular solid has a cross-sectional area of 3.50cm²across its width, a front-to-rear length of 15.8cm , and a resistance of $\mathbf{935}\mathbf{\Omega }$. The block’s material contains $\mathbf{5}\mathbf{.}\mathbf{33}\mathbf{\times }{\mathbf{10}}^{\mathbf{22}}$conduction electrons/m³. A potential difference of 35.8 Vis maintained between its front and rear faces. (a) What is the current in the block? (b) If the current density is uniform, what is its magnitude? What are (c) the drift velocity of the conduction electrons and (d) the magnitude of the electric field in the block?

1. Cross sectioned area of block$A=3.50{\mathrm{cm}}^2$
2. Front-to-rear length,$L=15.8\mathrm{cm}\mathrm{or}0.158\mathrm{m}$
3. Resistance$R=935\mathrm{\Omega }$
4. Density of block$n=5.33\times {10}^{22}\mathrm{electrons}/{\mathrm{m}}^3$
5. Potential difference $V=35.8\mathrm{V}$

Ohm’s law states that the current is directly proportional to the voltage across the two terminals. The opposition to the flow of current is resistance. The current density is the time rate of flow of charges per unit area. Drift velocity is the velocity achieved by the charges in the electric field. Using the formulae for current density and the drift velocity as well as Ohm’s law, we can find the required quantities.

Formulae:

The voltage equation from Ohm’s law, $V=iR$ …(i)

Here, $V$is potential difference, $i$is current, $R$is resistance.

The current density of the material, $J=\frac{i}{A}$ …(ii)

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

The drift velocity of the electrons, $V_d=\frac{J}{ne}$ …(iii)

Here, J is the current density,V_d is the drift velocity,n is the number of electrons, e is the charge on the electron.

The magnitude of the electric field in terms of potential difference $E=\frac{V}{L}$ …(iv)

Here, E is the electric field, V is voltage, and L is separation.

Substituting the given data in equation (i), we can get the value of the current in the block as follows:

$$\begin{gathered} i=\frac{V}{R} \\ =\frac{35.8\mathrm{V}}{935\mathrm{\Omega }} \\ =3.83\times {10}^{-2}\mathrm{A} \end{gathered}$$

Hence, the value of the current is $3.83\times {10}^{-2}\mathrm{A}$.

Now, using the current value and the given area in equation (ii), we can get the magnitude of the uniform current density as follows:

$$\begin{gathered} J=\frac{3.83\times {10}^{-2}\mathrm{A}}{3.50\times {10}^{-4}{\mathrm{m}}^2} \\ =109\mathrm{A}/{\mathrm{m}}^2 \end{gathered}$$

Hence, the value of the current density is $109\mathrm{A}/{\mathrm{m}}^2$.

Using the given data in equation (iii), we can get the drift velocity of the conducting electron as follows:

$$\begin{gathered} V_d=\frac{109\mathrm{A}/{\mathrm{m}}^2}{\left[\left(5.33\times {10}^{22}\right)\left(1.6\times {10}^{-19}\mathrm{C}\right)\right]} \\ =1.28\times {10}^{-2}\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of the drift velocity is $1.28\times {10}^{-2}\mathrm{m}/\mathrm{s}$.

Using the given data in equation (iv), we can get the strength value of the electric field as follows:

$$\begin{gathered} E=\frac{35.8\mathrm{V}}{0.158\mathrm{m}} \\ =227\mathrm{V}/\mathrm{m} \end{gathered}$$

Hence, the value of the field is $227\mathrm{V}/\mathrm{m}$.