Question

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 speed
• B. drift speed only
• C. current and drift speed
• D. current only

Answer 1

The Correct Option is D

When a steady current flows in a metallic conductor of non-uniform crosssection then the drift speed is $V _{ d }= I /$ neA and the electric field $E = I / pA$ $\Rightarrow V _{ d } \propto \frac{1}{ A }$ and $E \propto \frac{1}{ A }$ $\Rightarrow$ Only current remains constant. In a metallic conductor of non-uniform cross-section, only the current remains constant along the entire length of the conductor.

Answer 2

Consider a continuous current flowing through a non-uniform cross-sectional metallic conductor. The definition of steady current is that the current is uniform across the conductor. As a result, it will remain constant throughout the conductor's length.As a result, current (i) flowing through a conductor is independent of its cross-sectional area.
When a current flows through a conductor, electrons travel at a velocity known as drift velocity.
Drift velocity is the average uniform (constant) velocity of the conductor's free electrons.
The relation between drift velocity and current is given as \(i=neAv_d\).
Therefore,\(v_d=\frac{i}{neA}\).
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 that i is constant. Therefore, drift velocity is directly proportional to the cross-sectional area i.e.
 vd∝\(\frac{1}{A}\).
As a result, if the cross-sectional area is non-uniform, the electron drift velocity is not constant over the whole length of the wire.
The electric field produced when a potential difference is created across the conductor is what causes the current.
Consider the length dl of the conductor.
Let the potential difference across this dl length be dV and the magnitude of the electric field due to dV be E.
Then it is given that, dV=Ed l⇒ E=\(\frac{dV}{dl}\)……(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=\(\frac{\rho l}{A}\).
ρ 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 is A. Let the resistance of the element be dR.
Therefore, dR=ρ\(\frac{dl}{A}\)
⇒ \(dl\)=\(\frac{AdR}{\rho}\).
Substitute the value of dl in equation (i).
⇒ \(E=\frac{dV}{\frac{AdR}{\rho}}\)
⇒ \(E=\frac{\rho dV}{AdR}\)
And dV = idR.
⇒ \(E=\frac{\rho idR}{AdR}\)
⇒ \(E=\frac{\rho i}{A}\).
ρi is constant in this example. As a result, E is inversely proportional to the conductor's cross-sectional area. As a result, it will not remain constant with conductor length.
As a result, in the given scenario, just the current is constant.