When an electric field \( E \) is applied to the conductor, the free electrons experience a force and accelerate, attaining an average drift velocity \( v_d \). The equation of motion for the drift velocity is given by:
\[ v_d = \frac{e E \tau}{m} \]
The current density \( J \) is given by the product of the number density of free electrons, the electron charge, and the drift velocity:
\[ J = n e v_d \]
Substituting the expression for \( v_d \):
\[ J = n e \left( \frac{e E \tau}{m} \right) = \frac{n e^2 \tau}{m} E \]
The relationship between current density \( J \) and the applied electric field \( E \) is given by Ohm's Law:
\[ J = \sigma E \]
where \( \sigma \) is the conductivity of the material. Comparing the two expressions for \( J \), we can equate the coefficients of \( E \) to find the conductivity \( \sigma \):
\[ \sigma = \frac{n e^2 \tau}{m} \]
Resistivity \( \rho \) is the reciprocal of conductivity. Thus:
\[ \rho = \frac{1}{\sigma} = \frac{m}{n e^2 \tau} \]
The resistivity \( \rho \) is given by:
\[ \rho = \frac{m}{n e^2 \tau} \]
A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is :
If \(\begin{vmatrix} 2x & 3 \\ x & -8 \\ \end{vmatrix} = 0\), then the value of \(x\) is: