Question

: A solenoid$\mathbf{25}\mathit{c}\mathit{m}$long and with a cross-sectional area of$\mathbf{0}\mathbf{.}\mathbf{5}\mathit{c}\mathit{m}$contains$\mathbf{400}$turns of wire and carries a current of$\mathbf{80}\mathit{A}$. Calculate: (a) the magnetic field in the solenoid; (b) the energy density in the magnetic field if the solenoid is filled with air; (c) the total energy contained in the coil’s magnetic field (assume the field is uniform); (d) the inductance of the solenoid.

Short answer

a)$B=0.161T$

b)$u=1.03\times {10}^{4}J/m^3$

c)$U=129mJ$

d)$L=4.02\times {10}^{-5}H$

Step-by-step solution

The formulas used in the given problem

The magnetic field in a solenoid is given by $\mathit{B}\mathbf{=}{\mathit{\mu }}_{\mathbf{0}}\mathit{n}\mathit{I}$.

The energy stored in a solenoid is given by $\mathit{U}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{L}{\mathbf{I}}^{\mathbf{2}}}$.

The energy density is given by $\frac{{\mathbf{B}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{\mu }}_{\mathbf{0}}}$or it is simple energy per volume.

Calculate the magnetic field

The magnetic field is given by

$$\begin{gathered} B={\mu }_{0}nI \\ B=\frac{{\mu }_{0}NI}{l} \\ B=\frac{4\pi \times {10}^{-7}\times 400\times 80}{0.25} \\ B=0.161T \end{gathered}$$

Calculate the energy density

The energy density is given by

$$\begin{gathered} u=\frac{B^2}{2{\mu }_0} \\ u=\frac{0.{161}^2}{2\times 4\pi \times {10}^{-7}} \\ u=1.03\times {10}^{4}J/m^3 \end{gathered}$$

Calculate the total energy stored in the solenoid

The total energy can be given by the product of energy density and volume.

$$\begin{gathered} V=lA \\ V=0.25\times 0.5\times {10}^{-4} \\ V=1.25\times {10}^{-5}{m}^3 \end{gathered}$$

The total energy will be

$$\begin{gathered} U=uV \\ U=1.03\times {10}^4\times 1.25\times {10}^{-5} \\ U=129mJ \end{gathered}$$

Calculate the self-inductance

The energy is given by $U=\frac{1}{2}L{I}^2$. Use this formula to calculate the self-inductance.

$$\begin{gathered} L=\frac{2U}{I^2} \\ L=\frac{2\times 0.129}{{80}^2} \\ L=4.02\times {10}^{-5}H \end{gathered}$$