Answer:
Self inductance, L = 0.259 Henry
Explanation:
It is given that,
Increase in current, [tex]\dfrac{dI}{dt}=6.35\times 10^{-2}\ A/s[/tex]
Self induced emf, [tex]\epsilon=1.65\times 10^{-2}\ V[/tex]
Let L is the self-inductance of the inductor. It is given by :
[tex]\epsilon=L\dfrac{dI}{dt}[/tex]
[tex]L=\dfrac{\epsilon}{dI/dt}[/tex]
[tex]L=\dfrac{1.65\times 10^{-2}}{6.35\times 10^{-2}}[/tex]
L = 0.259 Henry
So, the self inductance of the inductor is 0.259 Henry. Hence, this is the required solution.
