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A current flows in a wire of circular cross section with the free electrons travelling with drift velocity $\vec{v}$. If an equal current flows in a wire of the same material and twice the radius, the new drift velocity is:
A. $\vec{v}$
B. $\dfrac{\vec{v}}{2}$
C. $\dfrac{\vec{v}}{4}$
D. $2\vec{v}$

Answer
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Hint: Try to deduce the relationship between the drift velocity of the electrons and the cross-sectional area of the conductor. First start by deducing the total number of electrons in the entire volume of the conductor, then arrive at a relation for the time taken by an electron to travel the entire length of the conductor. Do not forget that the velocity with which they travel here will be the drift velocity since they move under the influence of the applied field. Then using the definition of current as the charge per unit time flowing through the conductor determines the expression for the current in terms of drift velocity and area, and this should help you determine the interdependency.

Formula used:
The current flowing through a conductor $ I = nAqv_d$, where n is the number of electrons per unit volume of the conductor, A is the cross sectional area of the conductor, q is the electron charge and $v_d$ is the drift velocity of the electrons in the conductor upon the application of an electric field across the conductor.

Complete answer:
Let us begin by understanding what the drift velocity of an electron in a current carrying conductor means.
In terms of electrons, the drift velocity is defined as the average velocity attained by the electrons in a conductor due to an electric field applied across the conductor. Under no directional fields, an electron in a conductor will propagate randomly within the conductor with a velocity resulting in an average velocity of zero. Applying an electric field sets this random motion into a net flow in one direction. This is called the drift, since the electrons are forced to move in a direction under the influence of an external field, and gets superimposed on the random motion of free electrons.
Now, consider a conductor of length l and area of cross section A. Let q be the charge of an electron.
When a n electric field is applied, let the drift velocity of the electrons be $\vec{v}$.
Let the number of free electrons per unit volume of the conductor be n.
Therefore, the total number of free electrons in the entire volume of the conductor will be $N = n(Al)$, where the volume of the conductor is given as $area \times length$.
The total charge in the conductor due to N number of electrons will be $Nq$
If the time taken by this charge to travel the conductor is t, it is given by $t=\dfrac{l}{v}$
We know that the current produced in a conductor is given by
$I =\dfrac{q}{t} = \dfrac{Nq}{\dfrac{l}{v}} = nAlq \times \dfrac{v}{l} \Rightarrow I = nAqv$
Now let us bring the question into context.
Let$ \vec{v}$ be the drift velocity. And if r is the radius of this wire then let A be the area of this conductor= $\pi r^2$
Now, we have another wire where the same amount of current passes through but it has twice the radius, which means that
$A = \pi (2r)^2 = \pi 4r^2$
Let the drift velocity of electrons through this wire be $\vec{V}$
From the relation that we derived above, we have
$\vec{v} \propto \dfrac{1}{A} \Rightarrow \dfrac{\vec{v}}{\vec{V}} = \dfrac{ \pi 4r^2}{ \pi r^2} = \dfrac{4}{1}$
$\Rightarrow \vec{V} = \dfrac{\vec{v}}{4}$

So, the correct answer is “Option C”.

Note:
Always remember that when an electric field is applied over the ends of a conductor, the electrons drift between successive collisions in a direction that is opposite to the applied field.
You may also come across another relation for the drift velocity other than what we have derived and it looks something like
Drift velocity $v = \left(\dfrac{-\mu E}{m}\right)T$, where $\mu$ is the electron mobility, E is the electric field applied, m is the mass of the electron and T is the relaxation time, which is the time required by an electron to return to its initial equilibrium state. The negative sign indicates that the electrons flow opposite to the applied electric field direction.
Mobility of an electron is defined as the drift velocity of an electron for a unit electric field applied:
$\mu = \dfrac{v}{E}$.