The correct option is C i>0.5 A
As the sliding contact is being pulled, the current in the circuit changes.
An induced emf, E=−Ldidt is produced across the inductor.
The net emf in the circuits is,
Enet=6 V−Ldidt
And hence the current is,
i=EnetR=6−Ldidt12 ...(i) [At the instant shown]
Now the resistance in the circuit is increasing, so the current is decreases and so didt is negative.
Thus, the numerator of (i) is more than 6 and hence i is greater than 612=0.50 A.
Thus, option (B) is the correct answer.