Question

$5.0\mu \mathrm{s}$after the switch of FIGURE is moved from $\mathrm{a}$to $b$, the magnetic energy stored in the inductor has decreased by half. What is the value of the inductance $L$?

Short answer

Inductance,$L=0.72\mathrm{mH}$

Step-by-step solution

Inductor

An inductor appears to be a non-active electrical component with a strong magnetic field. An inductance is a wire loop or coil in the most basic form. The number of turns on a coil influences its inductance.

Find Final stored energy

Given:

The final energy is half the energy of the initial one at time $t=5\mu \mathrm{s}$

Although the voltage differential across the coils produces an induced emf, the inductor's magnetic field stores energy. The below is the equation for the stored energy:

$U_{\mathrm{L}}=\frac{1}{2}L{I}^2$

Where $L$is the inductance of the coil. Initially, when the switch is at$\left(a\right)$the current is maximum $I_o$and the stored energy is

$U_{\mathrm{i}}=\frac{1}{2}L{I}_o^2$

After time $t=5\mu$when the switch is moved to $\left(b\right)$the current decays according to the equation

$I=I_o{e}^{-{}_L{}^{R}t}$

The final stored energy is,

$U_{\mathrm{f}}=\frac{1}{2}L{I}^2=\frac{1}{2}L{\left(I_o{e}^{-\frac{R}{L}t}\right)}^2$$=\frac{1}{2}L{I}_o^2{e}^{-\frac{2R}{L}t}$

Step3: 

The final energy is half the initial energy, therefore the expression for $L$by

$U_{\mathrm{f}}=\frac{1}{2}{U}_{\mathrm{i}}$

$\frac{1}{2}L{I}_o^2{e}^{-\frac{2R}{L}t}=\frac{1}{2}\left(\frac{1}{2}L{I}_o^2\right)$

$e^{-\frac{2R}{L}t}=\left(\frac{1}{2}\right)$

$-\frac{2R}{L}t=\ln 0.5$

$L=-\frac{2Rt}{\ln 0.5}$

put the values for $R$and $t$into the above equation to get $L$

$L=-\frac{2Rt}{\ln 0.5}=-\frac{2\left(50\Omega \right)\left(5\times {10}^{-6}\mathrm{s}\right)}{\ln 0.5}$

$=0.72\times {10}^{-3}\mathrm{H}$

$=$$0.72\mathrm{mH}$