Question

if the rate of current of 2 A/s induces an emf of 10mV in a solenoid what is the self inductance of the solenoid?

Answer 1

Dear Student,

$$\begin{gathered} \frac{\mathrm{dI}}{\mathrm{dt}}=2\mathrm{A}/\mathrm{s} \\ \mathrm{emf},\mathrm{e}=10\mathrm{mV}=10\times {10}^{-3}\mathrm{V} \\ \mathrm{e}=\mathrm{L}\frac{\mathrm{dI}}{\mathrm{dt}} \\ 10\times {10}^{-3}=\mathrm{L}\times 2 \\ \mathrm{L}=5\times {10}^{-3}=5\mathrm{mH} \end{gathered}$$