Answer:
0.712 mH
Step-by-step explanation:
The emf produced by the coil = 2.6 V
Current is changes from -29 mA to 33 mA so change in current di =33-(-29)=62 mA
The time duration that is dt=17 ms
The emf is given by
![e=L(di)/(dt)](https://img.qammunity.org/2020/formulas/physics/college/yt0mglg3yrbdh633lbmliofu2r3bagnznj.png)
![2.6=L(62)/(17* 10^(-3))](https://img.qammunity.org/2020/formulas/physics/college/k7zs239bfbb7dedi3e45f2n1q2dtc4ae7n.png)
![L=0.712* 10^(-3)H=0.712mH](https://img.qammunity.org/2020/formulas/physics/college/4q8cd8gcitmamtwfprdct0z2ct5gnxa2dl.png)
So the value of inductance will be 0.712 mH