Question

A straight conductor of uniform cross-section carries a current $I$. Let $S$ be the specific charge of an electron. The momentum of all the free electron per unit length of the conductor due to their drift velocity only, is
(A) $I.S$
(B) $\dfrac{I}{S}$
(C) $\sqrt {\dfrac{I}{S}}$
(D) ${\left( {\dfrac{I}{S}} \right)^2}$

Answer 1

Hint We need to find the momentum of all the free electrons per unit length of the conductor due to their drift velocity. We know that momentum is a product of mass and velocity. The momentum of all the free electrons per unit length of the conductor is the product of free electrons per unit length, mass and velocity (drift velocity).

Complete step by step answer
Momentum is given by the product of mass and acceleration of the object in which it is in movement.
$\Rightarrow P = mv$
Where,
P is the momentum
m is the mass
v is the velocity
Given,
The current through the conductor is $I$
Specific charge of an electron is S
We know that the specific charge of an electron is $S = \dfrac{e}{m}{\text{ }} \to {\text{1}}$
Let $L$ be the length of the conductor
$A$ be the area of the conductor
$e$ be the charge of the electron
${v_d}$ be the drift velocity
m be the mass of the electron
When an electric field is applied free electrons are drifted towards the positive terminal of the conductor. The velocity with which free electrons are drifted towards the positive terminal is known as the drifted velocity
We can derive the value of drift velocity using many formulas
Here in the question they have provided the current, so here we can use the formula which relates current and drift velocity which is:
$\Rightarrow {V_d} = \dfrac{I}{{neA}}{\text{ }} \to 2$
${v_d}$ is the drift velocity
The free electron per unit length of the conductor is given by
$\Rightarrow N = \dfrac{{nAL}}{L}$
$\Rightarrow N = nA{\text{ }} \to 3$
Where,
N is the free electron per unit length of the conductor
n is the total number of free electrons in the conductor
A is the area of the conductor
L is the length of the conductor
Now, the momentum of the free electron per unit length of the conductor is
$\Rightarrow P = Nm{v_d}{\text{ }} \to 4$ (The velocity experienced by the electrons is drift velocity)
P is the momentum
N is the free electron per unit length of the conductor
m is the mass
${v_d}$ is the drift velocity
We have to find the momentum of all the free electrons per unit length of the conductor due to their drift velocity only.
We know that
The momentum of the free electron per unit length of the conductor is
$\Rightarrow P = Nm{v_d}$
Substitute the value of the N and ${v_d}$ from equation 2 and 3 in above equation
$\Rightarrow P = (nA)m\left( {\dfrac{I}{{neA}}{\text{ }}} \right)$
$\Rightarrow P = m\left( {\dfrac{I}{e}} \right)$
We can write the above equation as
$\Rightarrow P = \dfrac{I}{{\dfrac{e}{m}}}$
From equation 1 we know that $\dfrac{e}{m}$ is the specific charge S of the electron
$\Rightarrow P = \dfrac{I}{S}$

Hence the correct answer is option B) $\dfrac{I}{S}$

Note The specific charge is the charge per unit mass of the particle. I.e. every particle has an individual specific charge. Electron protons and other different particles have different specific charges according to their charge and mass.