Question

Consider a steady current flow in a metallic conductor of non-uniform cross section. The quantity/quantities remaining constant along the whole length of the conductor is/are:
A. current, electric field and drift velocity
B. drift speed only
C. current and drift speed
D. current only

Answer 1

Hint: The current will remain the constant. Use the formula $i=neA{{v}_{d}}$ and find the relation between cross sectional area A and drift velocity. Then consider an element of the conductor of length dl. For these elements, use the formulas V = iR, dV=Edl and $R=\rho \dfrac{l}{A}$ and find the relation between electric field and cross sectional area.

Formula used:
$i=neA{{v}_{d}}$
V = iR
$dV=Edl$
$R=\rho \dfrac{l}{A}$

Complete step by step answer:
Consider a steady current flowing in a metallic conductor of non-uniform cross section. The meaning of the steady current is that the current is uniform all over the conductor. Therefore, it will be constant throughout the length of the conductor.
Hence, current (i) flowing in a conductor is independent of the cross sectional area of the conductor.
When there is a current in a conductor, the electrons move with a velocity called drift velocity (${{v}_{d}}$). Drift velocity is the average uniform (constant) velocity acquired by the electrons by the free electrons of the conductor.
The relation between drift velocity and current is given as $i=neA{{v}_{d}}$.
Therefore, ${{v}_{d}}=\dfrac{i}{neA}$.
n is the number of electrons in one unit of volume, e is the charge of an electron and A is the cross sectional area of the conductor.
For a given conductor, n is constant. And e is also a constant.
Here, it is given i is constant. Therefore, drift velocity is directly proportional to the cross sectional area i.e.
${{v}_{d}}\propto \dfrac{1}{A}$.
Hence, if the area of the cross section is non-uniform, the drift velocity of the electrons is not constant for the whole length of the conductor.
The reason for the current is the electric field that is produced when a potential difference is created across the conductor.
Consider an element of length dl of the conductor.
Let the potential difference across this dl length be dV and the magnitude of electric field due to dV be E.
Then it is given that, $dV=Edl\Rightarrow E=\dfrac{dV}{dl}$ ……(i).
According to Ohm’s law, V = iR. Here R is the resistance of the conductor.
The value of R is given to be $R=\rho \dfrac{l}{A}$.
$\rho$ is the resistivity of the conductor and it is constant for a given material, l and A are the length and cross sectional area of the conductor respectively.
So, for the element of dl length, the cross sectional area be A. Let the resistance of the element be dR.
Therefore, $dR=\rho \dfrac{dl}{A}$
$\Rightarrow dl=\dfrac{AdR}{\rho }$.
Substitute the value of dl in equation (i).
$\Rightarrow E=\dfrac{dV}{\dfrac{AdR}{\rho }}$
$\Rightarrow E=\dfrac{\rho dV}{AdR}$.
And dV = idR.
$\Rightarrow E=\dfrac{\rho idR}{AdR}$
$\Rightarrow E=\dfrac{\rho i}{A}$.
In this case, $\rho i$ is constant. Therefore, E is inversely proportional to the cross sectional area of the conductor. Hence, it will also not remain constant with length of the conductor.
Therefore, in the given case only the current remains constant.
Hence, the correct option is D.

Note: The relation between drift velocity and cross sectional area can be explained in this way. Drift velocity is produced due to the electric force and collisions between electrons.
Where the cross sectional area is more, more electrons will be present in these sections. Hence, the collisions will increase and thus the drift velocity will decrease. That is why drift velocity is inversely proportional to the cross sectional area.