00:05numerical number two of unit 19 electromagnetic induction is a coil with an inductance of 0.5
00:14henry experiences a rate of change of current of 2 ampere per second calculate the induced
00:22dmf in the coil. यहाँ पर single coil का जखिर कर रहा है तो means self inductance. सबसे पहले हम
00:32data बनाएंगे
00:36self inductance का value means L का value हमें obtained है which is 0.5 henry
00:44और rate of change of current delta i upon delta t का value है वो हमें obtained है which is
00:522 ampere upon
00:53second. यहाँ पर delta ip upon delta t नहीं लिखेंगे कि यहाँ पर single coil है. तो उसके लिए हम
01:00just delta i upon delta t लिखेंगे. और हमें क्या find करनी है और हमें उस coil में
01:11जो emf induced हो रही है same coil में वो calculate करनी है means
01:16zi is equal to what. So for solution or calculation we will use the formula to calculate
01:29induced emf in same coil as zi is equal to minus L delta i upon delta t. यहाँ पर जो
01:40negative sign है वो due to lenses lie. So now by putting the values zi will be
01:50equal to minus the value of L is 0.5 multiplied with the value of delta i upon delta t
01:59is 2 ampere
02:01per second. So 0.5 into 2 it will be minus 1 volt. So this is our required answer.
