91Ó°ÊÓ

$$\begin{gathered} R=0.75\mathrm{Î\copyright } \\ {\mathrm{Ï•}}_B=26mWb=26×{10}^{-3}Wb \\ i=5.5A \end{gathered}$$

An inductor is a device that can be used to produce a magnetic field in aspecified region. If a current iis established through each of the N windings of an inductor, a magneticflux${\mathit{Ï•}}_{\mathbf{B}}$links those windings. The inductance Lof the inductor is given by equations 30-28. After that using the equation 30-41 for finding the total time required for current rises from 0 to 2.5.

Formula

$$\begin{gathered} L=\frac{{\mathrm{Φ}}_B}{i} \\ i=\frac{\mathrm{Î\mu }}{R}\left(1-e^{-\frac{Rt}{L}}\right) \end{gathered}$$

By using the equation 30-28 to find the inductance in the coil is

$$\begin{gathered} L=\frac{{\mathrm{Φ}}_B}{i} \\ \mathrm{So}, \\ L=\frac{(26×{10}^{-3})}{5.5} \\ L=4.7×{10}^{-3}\mathrm{H} \end{gathered}$$

Ifa constantis introduced into a single loop circuitcontaining a resistance Rand an inductance L, the current rises to an equilibrium value of$\mathrm{Î\mu }/R$

So the rise of current is

$i=\frac{\mathrm{Î\mu }}{R}\left(1-e^{-\frac{Rt}{L}}\right)$

Rearrange this equation to find the time

$$\begin{gathered} i\frac{\mathrm{Î\mu }}{R}=\left(1-e^{-\frac{Rt}{L}}\right) \\ {e}^{-\frac{Rt}{L}}=1-i\frac{R}{\mathrm{Î\mu }} \end{gathered}$$

Taking natural log of both side

$$\begin{gathered} -\frac{Rt}{L}=\ln \left(1-i\frac{R}{\mathrm{Î\mu }}\right) \\ t=-\frac{L}{R}\ln \left(1-i\frac{R}{\mathrm{Î\mu }}\right) \end{gathered}$$

By substituting the value

$$\begin{gathered} t=-\frac{4.7×{10}^{-3}}{0.75}\ln \left(1-2.5\frac{0.75}{6.0}\right) \\ t=-6.27×{10}^{-3}\ln (0.6875) \\ t=2.3×{10}^{-3}s \end{gathered}$$