Question

When a current I is set up in a wire of radius r, the drift velocity is ${v_d}$. If the same current us set up through a wire of radius 2r, the drift velocity will be
A) 4${v_d}$
B) 2${v_d}$
C) ${v_d}$/2
D) ${v_d}$/4

Answer 1

Hint:The question is from the section drift velocity and mobility. Using the relation between current (I) and drift velocity (${v_d}$) we can solve this problem.

Formula Used:
\[{v_d} = \dfrac{I}{{neA}}\]
Where \[{v_d}\]= drift velocity, I = current flow, n = free electron density, e = charge of an electron and A = cross sectional area.

Complete answer:
The relation between drift velocity and current is given below.
\[{v_d} = \dfrac{I}{{neA}}\]
From the equation, we can find that drift velocity (\[{v_d}\]) is inversely proportional to cross sectional area (A).
\[{v_d} \propto \dfrac{1}{A}\]
In the first case with radius r,
\[{({v_d})_1} \propto \dfrac{1}{{\pi {r^2}}}\]
In the second case with radius 2r,
\[{({v_d})_2} \propto \dfrac{1}{{\pi {{(2r)}^2}}} = \dfrac{1}{{4\pi {r^2}}}\]
By comparing the two cases,
\[{({v_d})_2} = \dfrac{1}{4}{({v_d})_1}\]
Hence, the correct option is Option (D) ${v_d}$/4.

Additional Information:

Drift velocity is the average velocity attained by charged particles, (electrons) in a material due to an electric field (\[\overrightarrow E \]).
Current density is defined as the total amount of current passing through a unit cross-sectional conductor in unit time. The relation between drift velocity (\[{v_d}\]) and Current Density is given below.
\[J = \dfrac{I}{A} = \dfrac{{nA{v_d}e}}{A} = n{v_d}e\]
\[J = n{v_d}e\]
From this equation we can find that drift velocity and current density are directly proportional to each other.



Note: The drift velocity and current flowing through the conductor both increase as the intensity of the electric field increases. Drift velocity is directly proportional to electric field intensity.