Potentiometer at null pointer! Consider the situation when a cell of an unknown emf is being measured using a potentiometer. We slide the jockey so as to obtain the null point. Now, is there any current in the potentiometer wire at the null point? Since we know that there is no current in the arm containing the unknown cell, its terminals have acquired equal potentials,how is it possible that there is any current in the potentiometer wire when that is in parallel to that cell.
Potential difference across AB= Potential difference across CD?

 A: If you connect a battery with a voltage $V$ between two points that already have a potential difference of $V$, then no current will flow out of the battery. Of course if you remove the other source of the potential difference, then the battery will start to push out the current needed to keep that voltage drop across the load.
When you say that the the cell is in parallel with the potentiometer you make a mistake: the cell is in parallel with the part of potentiometer and with the DC source (which is in series with the remaining part of the potentiometer).
A: 
Now, is there any current in the potentiometer wire at the null point?

If by potentiometer wire, you mean the resistance element, the answer is yes.
Let the current, from left to right, through the section of the potentiometer resistance from $A$ to $B$, $R_{AB}$ be labelled $I_{AB}$.
Let the current, from left to right, through the remainder of the potentiometer resistance $R_B$ be labelled $I_B$.
Clearly, if $I_{AB} = I_B$, there is no current through the arm of the potentiometer but this does not imply that these currents are zero.
The voltage across $R_{AB}$ is the unknown voltage $V_{CD}$ thus, by Ohm's law
$$I_{AB} = \frac{V_{CD}}{R_{AB}}$$
By KVL, the voltage $V_B$ across $R_B$ is
$$V_B = 3V - V_{CD}$$
and thus, by Ohm's law
$$I_B = \frac{V_B}{R_B}  = \frac{3V - V_{CD}}{R_B}$$
Setting both currents equal yields
$$\frac{V_{CD}}{R_{AB}} =  \frac{3V - V_{CD}}{R_B}$$
which can be solved for $V_{CD}$
$$V_{CD} = 3V \frac{R_{AB}}{R_B + R_{AB}}$$
But this is just, by voltage division, the voltage across $R_{AB}$ if the unknown voltage source were removed which, if you think about it clearly, must be the case for there to be zero current in the wiper arm.

Finally, if the total resistance of the potentiometer resistance element is $R$, we have
$$R_{AB} = R \chi$$
and
$$R_B = R (1 - \chi)$$
where
$$0 \le \chi \le 1$$
Thus, we can write
$$V_{CD} = \chi 3V $$
