Question
In Fig. $28-8,$ show that the ratio of the Hall electric field magnitude $E$ to the magnitude $E_{c}$ of the electric field responsible for moving charge (the current) along the length of the strip is$$\frac{E}{E_{C}}=\frac{B}{n e \rho}$$where $\rho$ is the resistivity of the material and $n$ is the number density of the charge carriers. (b) Compute this ratio numerically for Problem 13.
Step 1
We can write $J$ as $n e V_D$, where $n$ is the number density of charge carriers, $e$ is the charge of the electron, and $V_D$ is the drift velocity of the charge carriers. So, we have $E_C = \rho n e V_D$. Show more…
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In Fig. $28-8$, show that the ratio of the Hall electric field magnitude $E$ to the magnitude $E_{C}$ of the electric field responsible for moving charge (the current) along the length of the strip is $$ \frac{E}{E_{C}}=\frac{B}{n e \rho} $$ where $\rho$ is the resistivity of the material and $n$ is the number density of the charge carriers. (b) Compute this ratio numerically for Problem 13. (See Table 26-1.)
(a) In Fig. $28-8$ show that the ratio of the Hall clectric ficld magnitude $E$ to the magnitude $E_{C}$ of the electric field responsible for moving charge (the current) along the length of the strip is $$\frac{E}{E_{C}}=\frac{B}{\operatorname{ne} \rho}$$ $\begin{array}{l}{\text { where } \rho \text { is the resistivity of the material and } n \text { is the number den- }} \\ {\text { sity of the charge carriers. (b) Compute this ratio numerically for }} \\ {\text { Problem } 13 . \text { (Sce Table } 26-1 . )}\end{array}$
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