The self-inductance of the coil can be determined using the formula for the induced emf in a coil due to a changing current. The formula is given by: E=−L‌
dI
dt
Where: E is the induced emf L is the self-inductance of the coil ‌
dI
dt
is the rate of change of current We are given the following values:
E=2µV=2×10−6V Initial current, I1=3A Final current, I2=5A Time interval, ∆t=0.2s The rate of change of current can be calculated as:
‌
dI
dt
=‌
I2−I1
∆t
=‌
5A−3A
0.2s
=‌
2A
0.2s
=10A/s
Now, using the formula for induced emf: 2×10−6V=L×10A∕s Solving for L : ‌L=‌
2×10−6V
10A∕s
‌L=2×10−7H ‌L=0.2µH Therefore, the self-inductance of the coil is 0.2µH. So, the correct option is: Option D: 0.2µH