Chapter 18, Problem 19Q

To determine

To explain:

The effects on the electron drift speed by doubling the length, the radius and the potential drop one by one across a resistor.

Answer to Problem 19Q

Solution:

Correct answers are:

(a)The electron drift speed becomes half

(b)The electron drift speed is unchanged

(c)The electron drift speed becomes double

Formula used:

  •The electric field $\left(E\right)$ in a conductor is correlated to its length $\left(l\right)$ and the electric potential $\left(V\right)$ applied across it, which is given as:

         $E=\frac{V}{l}$

  •Average drift velocity depends on the structural properties of the material and the external electric field $\left(\overrightarrow{E}\right)$, as

         $\overrightarrow{v_d}=-\frac{e\tau }{m}\overrightarrow{E}$

    Where, e is the charge over one electron and $v_d$ is drift velocity of the free electrons in the wire at that particular temperature, τ is the average relaxation time between two collisions of free electrons and m is the mass of one electron.

Explanation of Solution

a)     When the length is doubled keeping V and r unchanged, from the formula $E=\frac{V}{l}$, we can say that the value of electric field (E)will become half.

    Also as the material is not changed so the average relaxation time between two collisions of free electrons will remain unchanged. Thus in this case, we can say from the formula $\overrightarrow{v_d}=-\frac{e\tau }{m}\overrightarrow{E}$, that drift velocity will become half, i.e.,

         $v_{d_2}=\frac{v_{d_1}}{2}$

b)     By changing r while keeping V and l unchanged, neither E nor τ will change. Therefore the drift velocity will also remain unchanged, i.e.,

       $v_{d_2}=v_{d_1}$

c)    When the potential is doubled keeping l and r unchanged, from the formula $E=\frac{V}{l}$, we can say that the value of electric field will become twice.

      Also, as the material is not changed, the average relaxation time between the two collisions of free electrons will remain unchanged. Thus in this case, we can say from the formula $\overrightarrow{v_d}=-\frac{e\tau }{m}\overrightarrow{E}$, that drift velocity will become two times, i.e.,

       $v_{d_2}=2v_{d_1}$