A long Iron slab of width w and height h emerges from a furnace, as shown in Figure 19.79. Because the end of the slab near the furnace is hot and the other end Is cold, the electron mobility increases significantly with the distance x. The electron mobility is $\mathit{u}\mathbf{=}{\mathit{u}}_{\mathbf{0}}\mathbf{+}\mathit{k}\mathit{x}$where u₀is the mobility of the iron at the hot end of the slab. There are n iron atoms per cubic meter, and each atom contributes one electron to the sea of the mobile electron (we can neglect the small thermal expansion of the iron). A steady state conventional current runs through the slab from the hot end towards cold end, and an ammeter (not shown) measures the current to have a magnitude I in amperes. A voltmeter is connected to two locations a distance d apart, as shown. (a) Show the electric field inside the slab at two locations marked with ×. Pay attention to the relative magnitudes of the two vectors that you draw. (b) Explain why the magnitude of the electric field is different at these two locations. (c) At a distance x from the left voltmeter connection, what is the magnitude of the electric field in terms x and the given quantities w,h,d,u₀,k,l, and n ( and fundamental constants)? (d) What is the sign of potential difference displayed on the voltmeter? Explain briefly. (e) In terms of the given quantitiesw,h,d,u₀,k,l, and n and ( and fundamental constants), what is the magnitude of the voltmeter reading? Check your work. (f) What is the resistance of this length of the iron slab?

The given data can be listed below as,

• A long Iron slab has a w width and h height, h, which has a hot end and cold end. n is the number of iron atoms per cubic meter.
• The electron mobility is given as$u=u_0+kx$ where u₀ is the mobility of the iron at the hot end of the slab, and the mobility increases as we move away from the left end with respect to distance x.
• The steady-state current is given as I. A voltmeter is connected to the circuit to measure the potential drop between two locations that are at a finite distance d.

The relation between electric field and value of potential in the circuit can be written as,

$\overrightarrow{\mathit{E}}\mathit{=}\mathit{-}\frac{\mathit{d}\mathit{V}}{\mathit{d}\mathit{r}}$ (1)

The negative of the change in the potential with respect to the measured distance can be written as the produced electric field.

The magnitude of the electric field at a distance x from the left end is$\frac{I}{newh\left(u_0+kx\right)}$. (refer to SID: 875865-19-61 P-c)

Substitute the value in equation 1, and we get,

$\frac{I}{newh\left(u_0+kx\right)}=-\frac{dV}{dx}$

We are measuring the voltage between 0 to a distance d so that we can write the above expression as,

$\int dV=-\underset{0}{\overset{d}{\int }}\frac{Idx}{newh\left(u_0+kx\right)}$

Now we can solve it as,

$\begin{array}{rcl}V& =& -\frac{I}{newh}\underset{0}{\overset{d}{\int }}\frac{dx}{\left(u_0+kx\right)}\\ & =& -\frac{I}{newhk}{\left(\log \left(u_0+kx\right)\right)}_0^d\\ & =& -\frac{I}{newhk}\left(\log \left(u_0+kd\right)-\log \left(u_0\right)\right)\\ V& =& -\frac{I}{newhk}\left(\log \left(\frac{u_0+kd}{u_0}\right)\right)\\ & & \end{array}$

Let's say we are connecting both terminals to the same point, which means distance d is zero; according to its working principle, the reading should be zero; let's solve the above expression as,

$\begin{array}{rcl}V& =& -\frac{I}{newhk}\left(\log \left(\frac{u_0+k\times 0}{u_0}\right)\right)\\ V& =& -\frac{I}{newhk}\left(\log \left(\frac{u_0}{u_0}\right)\right)\\ V& =& -\frac{I}{newhk}\left(\log \left(1\right)\right)\\ & =& 0\\ & & \end{array}$

That means this expression is correct.

Thus, the magnitude of the voltage reading is $\frac{I}{newhk}\left(\log \left(\frac{u_0+kd}{u_0}\right)\right)$.