Question

A solenoid that is 85 cm long has a cross-sectional area of 17.0${\mathbf{cm}}^{\mathbf{2}}$. There are 950of wire carrying a current of 6.60 A. (a) Calculate the energy density of the magnetic field inside the solenoid. (b) Find the total energy stored in the magnetic field there (neglect end effects).

Short answer

$$\begin{gathered} a)u_b=34.2\frac{\mathrm{J}}{{\mathrm{m}}^3} \\ \mathrm{b}){\mathrm{u}}_{\mathrm{b}}=4.94\times {10}^{-2}\mathrm{J} \end{gathered}$$

Step-by-step solution

Given

i) Length of solenoid$L=85.0\mathrm{cm}=85.0\times {10}^{-2}\mathrm{m}$

ii) Number of turns N = 950

iii) Current in the wire i = 6.60 A

iv) Cross-sectional area of the solenoid$A=17.0c{m}^2=17.0\times {10}^{-4}{m}^2$

Understanding the concept

From the given total number of turns of wire, we can find the number of turns per meter. Using this number of turns into the formula of magnetic energy density, we can find the energy density of the magnetic field inside the solenoid. As the magnetic field is uniform inside the solenoid, using this calculated magnetic energy density and the given cross-sectional area and length of the solenoid, we can find the energy stored in the solenoid.

Formula:

$$\begin{gathered} u_B=\frac{1}{2}{\mathrm{\mu }}_0{\mathrm{n}}^2{\mathrm{i}}^2 \\ {\mathrm{U}}_{\mathrm{B}}={\mathrm{u}}_{\mathrm{B}}\mathrm{V} \end{gathered}$$

(a) Calculate the energy density of the magnetic field inside the solenoid

Theenergy density of the magnetic field inside the solenoidis given by

$u_B=\frac{1}{2}{\mathrm{\mu }}_0{n}^2{i}^2$

Here, we have

$$\begin{gathered} n=\frac{N}{L}=\frac{950\mathrm{turns}}{85.0\times {10}^{-2}\mathrm{m}}=1.118\times {10}^3{\mathrm{m}}^{-1} \\ \mathrm{Therefore} \\ {\mathrm{u}}_{\mathrm{B}}=\frac{1}{2}\left(4\mathrm{\pi }\times {10}^{-7}\frac{\mathrm{H}}{\mathrm{m}}\right){\left(1.118\times {10}^3{\mathrm{m}}^{-1}\right)}^2{\left(6.60\mathrm{A}\right)}^2 \\ {\mathrm{u}}_{\mathrm{B}}=34.2\frac{\mathrm{J}}{{\mathrm{m}}^3} \end{gathered}$$

Therefore, the energy density of the magnetic field inside the solenoid is$34.2\mathrm{J}/{\mathrm{m}}^3$

(b) Find the total energy stored in the magnetic field there 

The energy stored in the solenoid is given by

$$\begin{gathered} U_B=u_{B}V \\ V=AL=\left(17.0\times {10}^{-4}{\mathrm{m}}^2\right)\left(85.0\times {10}^{-2}\mathrm{m}\right) \\ V=1.45\times {10}^{-3}{\mathrm{m}}^3 \\ {\mathrm{u}}_{\mathrm{B}}=\left(34.2\frac{\mathrm{J}}{{\mathrm{m}}^3}\right)\left(1.45\times {10}^{-3}{\mathrm{m}}^3\right) \\ {\mathrm{u}}_{\mathrm{B}}=4.94\times {10}^{-2}\mathrm{J} \end{gathered}$$

Therefore, the total energy stored in the magnetic field is $4.94\times {10}^{-2}\mathrm{J}$.