What would have to be the self-inductance of a solenoid for it to store 10.0 J of energy when a 2.00-A current runs through it? (b) If this solenoid’s cross-sectional diameter is 4.00 cm, and if you could wrap its coils to a density of 10 coils/mm, how long would the solenoid be? (See Exercise 30.15.) Is this a realistic length for ordinary laboratory use?

The formula for the Energy value that is stored in the solenoid as follows $\mathbf{U}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{Li}$.The solenoid self-inductance based on the length I number of turns N and cross sectional are A is $\mathbf{L}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{AN}}^{\mathbf{2}}}{\mathbf{I}}$

The Energy stored value is given as U -10.0 J, the Current that passes through the solenoid, I = 2.00A The solenoid self-inductance is given by the formula as,$\mathrm{L}=\frac{2{\mathrm{U}}_{\mathrm{c}}}{{\mathrm{I}}^2}$Substitute the values in the given equation

$$\begin{gathered} \mathrm{L}=\frac{2\left(10.0\mathrm{J}\right)}{{\left(2.00\mathrm{A}\right)}^2} \\ =5.0\mathrm{H} \end{gathered}$$

The Density of the coins value is given as n = 10 coils/mm .On substituting the values we get,

$$\begin{gathered} =\left(10\mathrm{coils}/\mathrm{mm}\right)\left(\frac{{10}^3\mathrm{mm}}{1\mathrm{m}}\right) \\ =10\times {10}^3\mathrm{coils}/\mathrm{m} \end{gathered}$$

The Self-inductance value of the solenoid is given as $\mathrm{L}=\frac{{\mathrm{\mu }}_0{\mathrm{AN}}^2}{\mathrm{I}}$which can be written as role="math" localid="1664266340830" ${\mathrm{\mu }}_0{\mathrm{AN}}^2\mathrm{I}$.Diameter of cross-section, D = 4.00 cm

$$\begin{gathered} =\left(4.00\mathrm{cm}\right)\left(\frac{{10}^{-2}\mathrm{m}}{1\mathrm{cm}}\right) \\ =4.00\times {10}^{-2}\mathrm{m} \end{gathered}$$

Then area of cross-section is

$$\begin{gathered} A=\frac{{\mathrm{\pi D}}^2}{4} \\ =\mathrm{\pi }\frac{{\left(4.00\times {10}^{-2}\mathrm{m}\right)}^2}{4} \\ =1.256\times {10}^{-3}{\mathrm{m}}^2 \end{gathered}$$

Then length of the solenoid is$\mathrm{I}=\frac{\mathrm{L}}{{\mathrm{\mu }}_0{\mathrm{AN}}^2}$.Substitute the values we get,

$$\begin{gathered} \mathrm{I}=\frac{5.00\mathrm{H}}{\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{Wb}/\mathrm{A}.\mathrm{m}\right){\left(10\times {10}^3\mathrm{coils}/\mathrm{m}\right)}^2\left(1.256\times {10}^{-3}{\mathrm{m}}^2\right)} \\ =31.69\mathrm{m} \end{gathered}$$

The Number of turns of the solenoid is N is$3.169\times {10}^5$

The Circumference value of one turn is$C=\mathrm{\pi D}$.Therefore,

$$\begin{gathered} \mathrm{C}=3.14\times 4.00\mathrm{cm} \\ =0.1256\mathrm{m} \end{gathered}$$

The wire length is determined by the formula asrole="math" localid="1664267083503" ${\mathrm{I}}_{\mathrm{wire}}=\mathrm{NC}$. Substitute the values we get,

$$\begin{gathered} {\mathrm{I}}_{\mathrm{wire}}=3.169\times {10}^5\times 0.1256\mathrm{m} \\ =39.8\mathrm{m} \end{gathered}$$

Thus, it is long wire and it has very high resistance thus there will electrical energy loss.