Question

Find the concentration of the conduction electrons.
(Given, $E=5{\mu V}/{cm}\;,j=200{A}/{c{{m}^{2}}}\;,\,\,B=1T$)
$A.\,2.5\times {{10}^{20}}{{m}^{-3}}$
$B.\,2.5\times {{10}^{26}}{{m}^{-3}}$
$C.\,2.5\times {{10}^{28}}{{m}^{-3}}$
$D.\,2.5\times {{10}^{30}}{{m}^{-3}}$

Answer 1

In this question, we will equate the magnetic and the electric force equations. Using this condition, we will obtain the expression for the number of electrons. Then, we will substitute the given values in this equation to find the concentration of the conduction electrons.
Formula used:

& {{F}_{m}}=e{{v}_{d}}B \\\ & {{F}_{e}}=eE \\\ & I=neA{{v}_{d}} \\\ \end{aligned}$$ **Complete step-by-step solution:** The formulae being used are as follows. The magnetic force is given as follows. $${{F}_{m}}=e{{v}_{d}}B$$ Where e is the charge on the electron, B is the magnetic field and $${{v}_{d}}$$is the drift velocity of the electrons. The electric force is given as follows. $${{F}_{e}}=eE$$ Where e is the charge on the electron and E is the electric field. The current is given as follows. $$I=neA{{v}_{d}}$$ Where n is the number of electrons, e is the charge on the electron, A is the area of the conductor, and $${{v}_{d}}$$is the drift velocity of the electrons. Represent the above equation in terms of the drift velocity. $$\begin{aligned} & {{v}_{d}}=\dfrac{I}{neA} \\\ & \Rightarrow {{v}_{d}}=\dfrac{j}{ne} \\\ \end{aligned}$$ The above term ‘j’ represents the current density. At the equilibrium, the magnetic force equals the electric force. So, we have, $$\begin{aligned} & {{F}_{m}}={{F}_{e}} \\\ & \Rightarrow e{{v}_{d}}B=eE \\\ \end{aligned}$$ Now cancel out the common terms and substitute the expression of the drift velocity. $$\begin{aligned} & \dfrac{j}{ne}B=E \\\ & \Rightarrow n=\dfrac{jB}{eE} \\\ \end{aligned}$$ Substitute the values in the above equation. $$\begin{aligned} & n=\dfrac{200\times {{10}^{-4}}\times 1}{1.6\times {{10}^{-19}}\times 5\times {{10}^{-4}}}\\\ & \Rightarrow n=2.5\times {{10}^{28}}{{m}^{-3}} \\\ \end{aligned}$$ **As the value of the concentration of the conduction electrons is obtained to be equal to $$2.5\times {{10}^{28}}{{m}^{-3}}$$, thus, the option (C) is correct.** **Note:** The given question is a direct one, but the way to obtain the expression to find the same is a bit confusing. As the electron is given to be conducting, so, we have made use of the electric force and the magnetic force.