Question

Electric field \( E \) is applied across a metallic wire of length \( l \) and area of cross-section \( A \). Obtain the formula of the relationship between the drift velocity (\( v_d \)) of free electrons of the conductor and electric field \( E \) in vector form.

Hint:

The drift velocity is proportional to the electric field. The stronger the field, the higher the drift velocity of the electrons. The drift velocity is also influenced by the relaxation time, which depends on the material of the conductor.

Answer 1

When an electric field \( E \) is applied across a metallic wire, the free electrons experience a force that causes them to move in the direction opposite to the electric field, resulting in a drift velocity \( v_d \). The relationship between the drift velocity and the electric field can be derived using the following steps:
Step 1: Force on an electron due to electric field.
The force \( F \) on an electron of charge \( e \) due to the electric field is given by:
\[ F = eE \] Where:
- \( F \) is the force on the electron,
- \( e \) is the charge of the electron (\( e = 1.6 \times 10^{-19} \, \text{C} \)),
- \( E \) is the electric field.
Step 2: Acceleration of an electron.
From Newton's second law, the acceleration \( a \) of the electron is given by:
\[ a = \frac{F}{m_e} = \frac{eE}{m_e} \] Where:
- \( m_e \) is the mass of the electron.
Step 3: Drift velocity.
The drift velocity \( v_d \) is the average velocity that the electron acquires due to the electric field. This velocity is reached after a characteristic time called the relaxation time \( \tau \), which is the average time between collisions of the electron with the atoms in the wire. The drift velocity is related to the acceleration by:
\[ v_d = a \times \tau = \frac{eE}{m_e} \times \tau \] Thus, the drift velocity \( v_d \) of the electrons is given by:
\[ v_d = \frac{eE \tau}{m_e} \] This equation gives the relationship between the drift velocity \( v_d \), the electric field \( E \), and the properties of the electron (charge and mass).