Question

74. II The inductor in FIGURE P30.74 is a $9.0-\text{cm}$-long, $2.0-\text{cm}-$diameter solenoid wrapped with 300 turns. What is the current in the circuit $10\mu s$after the switch is moved from a to b ?

Short answer

Hence, the required current is$0.093\text{A}$

Step-by-step solution

Step 1. Given information

The number of turns if 300, and length of inductor is 9 cm

The radius of the coil is,

$\begin{array}{l}r=\frac{d}{2}\\ r=\frac{2\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)}{2}=0.01\text{m}\end{array}$

Step 2. Explanation

The inductance of the coil is,

$\begin{array}{l}L=\frac{{\mu }_0{N}^{2}A}{l}\\ L=\frac{{\mu }_0{N}^2\pi r^2}{l}\\ L=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right){\left(300\right)}^2\pi {(0.01\text{m})}^2}{9\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)}\\ =3.94\times {10}^{-4}\text{H}\end{array}$

According to ohm's law, the initial current is,

$\begin{array}{l}{I}_0=\frac{V_0}{R}\\ I_0=\frac{6\text{V}}{30\Omega }=0.2\text{A}\end{array}$

The expression for the current in the coil is,

$\begin{array}{l}I=I_0{e}^{\frac{-t}{\tau }}\\ I=I_0{e}^{\frac{-t}{\left(\frac{L}{R}\right)}}=I_0{e}^{\frac{-Rt}{L}}\\ I=I_0{e}^{\frac{-Rt}{L}}\\ I=(0.2\text{A})e^{-\left(30\Omega \right)\left(10\mu \text{s}\right)\left(\frac{{10}^{-4}\text{s}}{1/\text{s}}\right)}\\ =0.093\text{A}\end{array}$

Hence, the required current is$0.093\text{A}$