Energy of the Electromagnetic Field

Abstract

This chapter deals with the energy associated with the electric and magnetic fields. In all branches of Physics, the concept of energy is present. The introduction of the idea of energy in electromagnetism provides a way of understanding many different phenomena, such as the properties of magnetic matter at different temperatures, circuits and networks, thermal electromagnetic radiation of bodies, and electromechanical machines. Due to the wide range of applications we can find, we have chosen for this chapter a viewpoint as simple as possible but, at the same time, accurate in order to explain the most important ideas involved.

Appendices

Solved Problems

Problems A

9.1. :

A hydrogen atom is formed by a proton of charge 1.602$\times$10$^{-19}$C and an electron of charge –1.602$\times$10$^{-19}$C. Supposing that the separation between these two charges is 0.5 $\times$10$^{-10}$ m, calculate the electrostatic energy of the atom.

Solution

As the charges and distances are known, Eq. (9.2) is applicable:

$$ U_e =\frac{q_{1} q_{2}}{4\pi \varepsilon _0 r_{12} }=-\frac{{1}.{6}0{2}\times {1}0^{-{19}}\times {1}.{6}0{2}\times {1}0^{-{19}}}{4\pi 8.8542\times 10^{-12}\times 0.{5}\times {1}0^{-{1}0}}\mathrm{J}=-4.61\times {1}0^{-{18}}\,{\mathrm{J}}. $$

9.2. :

Calculate the energy of a system of three particles, each with charge q and located on the vertices of an equilateral triangle of side L.

Solution

Suppose that the three charges are initially located on the vertices of the triangle. The work that the charge $q_{3}$ can give to exterior charges in its displacement d r, with $q_{1 }$ and $q_{2}$ remaining fixed, is, by application of the principle of superposition of forces and of (9.1),

$$ dW=\mathbf{F}.d\mathbf{r}=(\mathbf{F}_1 +\mathbf{F}_2 ).d\mathbf{r}=\mathbf{F}_1 .d\mathbf{r}+\mathbf{F}_2 .d\mathbf{r}=\frac{q_1 q_3 }{4\pi \varepsilon _0 r_{13} ^{2}}dr_{13} +\frac{q_2 q_3 }{4\pi \varepsilon _0 r_{23} ^{2}}dr_{23}, $$

where dr $_{12}$ and dr $_{13}$ are the components of the displacement d r in the directions of the respective forces. Applying (9.2) to the displacement of $q_{3}$ to infinity gives

$$ U_{e3} =\frac{q_{1} q_{3}}{4\pi \varepsilon _{0} r_{13}}+\frac{q_{2}q_{3}}{4\pi \varepsilon _{0}r_{23}}. $$

Charge $q_{2}$ then moves to infinity, with $q_{1}$ remaining fixed. The energy provided by this charge is, in agreement with (9.2),

$$ U_{e2} =\frac{q_1 q_2 }{4\pi \varepsilon _0 r_{12} }, $$

then the energy of the system of the three charges is

$$ U_{e3} =\frac{q_{1}q_{2}}{4\pi \varepsilon _{0}r_{12}}+\frac{q_{1}q_{3}}{4\pi \varepsilon _{0}r_{13} }+\frac{q_{2}q_{3}}{4\pi \varepsilon _{0}r_{23}}=\frac{q_{1}q_{2}+q_{1}q_{3}+q_{2}q_{3}}{4\pi \varepsilon _{0} L}, $$

where it has been taken into account that $r_{12}=r_{13}=r_{23}=L$.

Equation (9.4) could have been applied directly and would give

$$\begin{aligned} U_e&=\sum _{i=1}^{3-1} {\sum _{j=i+1}^3 {\frac{q_i q_j }{4\pi \varepsilon _0 r_{ij} }} } =\sum _{j=1+1}^3 {\frac{q_1 q_j }{4\pi \varepsilon _0 r_{1j} }+} \sum _{j=2+1}^3 {\frac{q_2 q_j }{4\pi \varepsilon _0 r_{2j} }}\\&=\frac{q_1 q_2 }{4\pi \varepsilon _0 r_{12} }+\frac{q_1 q_3 }{4\pi \varepsilon _0 r_{13} }+\frac{q_2 q_3 }{4\pi \varepsilon _0 r_{23} }, \end{aligned}$$

which can be also expressed as

$\begin{array}{l} \displaystyle U_e =\frac{1}{2}\sum \limits _{i=1}^N {q_i V_i } =\frac{1}{2}\left[ {q_1 \left( {\frac{q_2 }{4\pi \varepsilon _0 r_{12} }+\frac{q_3 }{4\pi \varepsilon _0 r_{13} }} \right) +q_2 \left( {\frac{q_1 }{4\pi \varepsilon _0 r_{12} }+\frac{q_3 }{4\pi \varepsilon _0 r_{23} }} \right) +q_3 \left( {\frac{q_1 }{4\pi \varepsilon _0 r_{13} }+\frac{q_2 }{4\pi \varepsilon _0 r_{23} }} \right) } \right] \\[16pt] \displaystyle =\frac{q_1 q_2 }{4\pi \varepsilon _0 r_{12} }+\frac{q_1 q_3 }{4\pi \varepsilon _0 r_{13} }+\frac{q_2 q_3 }{4\pi \varepsilon _0 r_{23}}. \end{array}$

Note how this result agrees with the previous results.

9.3. :

A parallel-plate capacitor has a separation between the plates of $z=$ 5 mm, the area of each plate is 200 cm$^{2}$, the gap is filled with a dielectric of relative permittivity 5 and the potential between the plates is 300 V. Calculate the energy that it accumulates.

Solution

Applying (9.5) and taking into account the formula of the capacitance of a parallel-plate capacitor gives

$$\begin{aligned} U_e&=\frac{1}{2}CV^{2}=\frac{1}{2}\frac{\varepsilon _r \varepsilon _0 S}{z}V^{2}\\&=\frac{1}{2}\frac{5\times 8.854\times 10^{-12}\times 200\times 10^{-4}}{5\times 10^{-3}}300^{2}\,{\mathrm{J}}=7.969\times 10^{-12}\,{\mathrm{J}}. \end{aligned}$$

9.4. :

A system is formed by two flat, opposed plates each of area S, and separated by a small distance d. This system is introduced into a dielectric liquid of relative permittivity $\varepsilon _{r}$. The plates are connected to a battery of electromotive force V, and they slowly move apart due to the application of an exterior force F $_{\mathrm{e}}$ to one of plates until their separation is $d'.$ Calculate: (a) the increase of energy of the system; (b) the work of the exterior force; (c) relate the force F from one plate on the other and the electrostatic energy of the capacitor.

Solution

(a) According to (9.5), the initial energy of the capacitor is given by

$$ U_e =\frac{1}{2}CV^{2}=\frac{1}{2}\frac{\varepsilon _r \varepsilon _0 S}{d}V^{2}. $$

The energy that the capacitor has when the distance between the plates is $d'$ can be expressed as

$$ U_{e}{}^{\prime }=\frac{1}{2}\frac{\varepsilon _r \varepsilon _0 S}{d^{\prime }}V^{2}, $$

therefore the increase of energy is

$$ \Delta U_e =U_{e}{}^{\prime }-U_e =\frac{1}{2}\frac{\varepsilon _r \varepsilon _0 S}{d^{\prime }}V^{2}-\frac{1}{2}\frac{\varepsilon _r \varepsilon _0 S}{d}V^{2}=-\frac{1}{2}\varepsilon _r \varepsilon _0 SV^{2}\frac{d^{\prime }-d}{dd^{\prime }}. $$

Note that the energy of the capacitor has diminished, because $d'$ > d.

(b) Capacitance decreases with the separation of the plates, but the electric voltage remains constant, therefore the charge, $q=CV$, decreases. The charge variation is

$$ \Delta q=q^{\prime }-q=C^{\prime }V-CV=\frac{\varepsilon _r \varepsilon _0 S}{d^{\prime }}V-\frac{\varepsilon _r \varepsilon _0 S}{d}V=-\varepsilon _r \varepsilon _0 S\frac{d^{\prime }-d}{dd^{\prime }}V. $$

Therefore, the energy that the battery provides is

$$ W_{bat} =\Delta q.V=-\varepsilon _r \varepsilon _0 S\frac{d^{\prime }-d}{dd^{\prime }}V^{2}. $$

If $W_{\mathrm{ex}}$ is the work provided by the external force, then the principle of conservation of energy can be written thus:

$$ \begin{array}{l} \displaystyle W_{bat} +W_{ex} =\Delta U_e \quad \Rightarrow \quad -\varepsilon _r \varepsilon _0 S\frac{d^{\prime }-d}{dd^{\prime }}V^{2}+W_{ex} =-\frac{1}{2}\varepsilon _r \varepsilon _0 SV^{2}\frac{d^{\prime }-d}{dd^{\prime }} \\ \Rightarrow \quad W_{ex} =\frac{1}{2}\varepsilon _r \varepsilon _0 SV^{2}\frac{d^{\prime }-d}{dd^{\prime }}. \\ \end{array} $$

Note that both the work of the external force and the reduction of the energy of the capacitor have gone into increasing the energy of the battery, that is, in recharging the battery. Furthermore

$$ W_{bat} =2\Delta U_e. $$

(c) For a infinitesimal displacement dx under forces exterior $F_{\mathrm{e}}$ and interior F the principle of conservation of energy gives

$$\begin{aligned} dW_{bat} +dW_{ex} =dU_e \quad \Rightarrow \quad 2dU_{e}&+F_{e} dx=dU_e \quad \Rightarrow \quad F_e dx=-dU_e =-\frac{\partial U_e }{\partial x}dx \\&F_e =-\frac{\partial U_e }{\partial x}. \end{aligned}$$

In equilibrium, the electrostatic force is opposite to the external force, $F=-F_{e}$, therefore, the electric force between the two plates is

$$ F=+\frac{\partial U_e }{\partial x}. $$

Note the plus sign in this case of constant potential. In (9.11), at constant charge, the sign was minus.

9.5. :

In a region of space that contains no matter, there is an electrostatic field E $=$ ay u $_{\mathrm{x}}$. Calculate the energy inside a cube of side L, which supports three edges on the positive part of the coordinate axes. Solve for: $a=$ 20000 V/m$^{2}$ and $L=$ 0.4 m.

Solution

Application of (9.8) gives

$$\begin{aligned} U_e&=\frac{\varepsilon _0 }{2}\int _v {\mathbf{E}.\mathbf{E}dv} =\frac{\varepsilon _0 }{2}\int _{cube} {ay\mathbf{i}.ay\mathbf{i}dv} =\frac{a^{2}\varepsilon _0 }{2}\int _{x=0}^{x=L} {\int _{y=0}^{x=L} {\int _{z=0}^{x=L} {y^{2}dxdydz} } }\\&=\frac{a^{2}\varepsilon _0 }{2}L^{2}\frac{1}{3}L^{3}=\frac{a^{2}\varepsilon _0 }{6}L^{5}. \end{aligned}$$

With the numeric data provided, this becomes

$$ U_e =\frac{a^{2}\varepsilon _0 }{6}L^{5}=\frac{20000^{2}\times 8.8542\times 10^{-12}}{6}0.4^{5}{}\,{\mathrm{J}=60.44}\times {10}^{{-7}}{}\,{\mathrm{J}}. $$

Fig. 9.4

Capacitor and the Gaussian surface

9.6. :

Calculate the energy of a flat capacitor with capacitance C and charge q by means of the application of (9.7).

Solution

The electric field outside the capacitor is dismissed by considering that the separation between the plates is small, and it is null within each plate since it is a conductor, therefore the only region of space to consider is that between the plates. Consider a rectangular parallelepiped, such as that shown in Fig. 9.4, that includes the inner face of the positive plate, of charge q; the flux of the electric field through its surface is

$$ \oint _{paral.} \mathbf{E}.d\mathbf{S} =\int _{S} {EdS} =ES, $$

where S is the area of the plate. Applying Gauss’s theorem to the parallelepiped gives

$$ ES=\frac{q}{\varepsilon _0 }\quad \Rightarrow \quad E=\frac{q}{\varepsilon _0 S}. $$

Applying (9.8) to the space between the plates, with a separation of z, gives

$$ U_e =\frac{\varepsilon _0 }{2}\int _v {\mathbf{E}.\mathbf{E}dv} =\frac{\varepsilon _0 }{2}\frac{q}{\varepsilon _0 S}\frac{q}{\varepsilon _0 S}Sz=\frac{1}{2}\frac{q^{2}}{\varepsilon _0 S}z. $$

Remembering the formula of the capacity of the flat capacitor , $C=\varepsilon _{0}S/z,$ finally gives

$$ U_e =\frac{1}{2}\frac{q^{2}}{C}, $$

in agreement with (9.5).

Fig. 9.5

Capacitor and a closed surface (dotted lines)

9.7. :

Obtain the expression of the energy of a capacitor by applying (9.6).

Solution

Consider the box of volume v, drawn with dotted lines, in the diagram of the capacitor, shown in Fig. 9.5; all the charges of the capacitor are in its interior. Let q be the charge of the upper plate and $V_{1}$ its potential, and – q and $V_{2}$ be those corresponding to the lower plate. The integral that appears in (9.6) can be broken down into three integrals, one extended to the space surrounding the interior face of the upper plate, the other extended to the lower plate, and the third to the space between the plates which has no charge:

$$\begin{aligned}&U_e =\frac{1}{2}\int _v {\rho Vdv} =\frac{1}{2}\int _{v1} {\rho _1 V_1 dv} +\frac{1}{2}\int _{v2} {\rho _2 V_2 dv} +\frac{1}{2}\int _v {0Vdv} =\frac{1}{2}V_1 \int _{v1} {\rho _1 dv} +\frac{1}{2}V_2 \int _{v2} {\rho _2 dv} \\&=\frac{1}{2}V_1 q+\frac{1}{2}V_2 (-q)=\frac{1}{2}qV, \end{aligned}$$

where $V=V_{1}-V_{2}$, i.e. the potential difference between the potential of the upper plate and that of the lower plate.

Note: the exterior field E has not been considered because the distance between the plates is assumed to be small.

9.8. :

In the cube described in Problem 9.5, in addition to the indicated electric field, there is a uniform magnetic field of B $=0.001$ u $_{\mathrm{x}}$ T. Calculate the magnetic energy stored inside the cube.

Solution

The energy contained in the cube due to the existence of the electric field is, as was seen in Problem 9.5,

$$ U_e =\frac{a^{2}\varepsilon _0 }{6}L^{5}={60.44}\times {10}^{{-7}}{}\,{\mathrm{J}}. $$

The magnetic energy is, according to (9.22),

$$\begin{aligned} U_m&=\frac{1}{2}\int \limits _v {\mathbf{B}.\mathbf{H}dv} =\frac{1}{2}\int \limits _v {BHdv} =\frac{1}{2}\int \limits _v {\frac{B^{2}}{\mu _0 }dv}\\&=\frac{1}{2}\frac{B^{2}}{\mu _0 }L^{3}=\frac{1}{2}\frac{0.001^{2}}{4\pi \times 10^{-7}}0.4^{3}{}\,{\mathrm{J}}=25.46\times 10^{-3}{}\,{\mathrm{J}}. \end{aligned}$$

Therefore, the total energy contained in the cube is

$$ U={60.44}\times {10}^{{-7}}{}\,{\mathrm{J}}+25.46\times 10^{-3}{}\,{\mathrm{J=}}25.47\times 10^{-3}{}\,{\mathrm{J}}. $$

Problems B

9.9. :

A spherical region of space of radius R has a uniform charge density of $\rho$ (Fig. 9.6). The charge is made in stages such that in an intermediate stage, when a sphere of radius r has been charged, then a layer of charges of thickness dr is added to it, and so on. Calculate: (a) The electric field $E_{e}$ outside the sphere of radius r in the intermediate stage. (b) The work necessary to place the charged layer on the spherical surface of radius r. (c) The energy of the charged sphere of radius R. (d) The energy of this sphere if $R=$ 20 cm and $\rho$ $=$ 0.3 C/m$^{3}$.

Fig. 9.6

Spherical region of charge density $\rho$

Solution

(a) By applying Gauss’s theorem to the spherical surface of radius $r_{e}$ ($r_{e} > r$), the outer electric field E $_{e}$ is obtained:

$$\begin{aligned}&\oint \mathbf{E}_{\mathrm{e}}.d\mathbf{S} =\oint {E_{\mathrm{e}} dS} =E_{\mathrm{e}} 4\pi r_{e}^{2}=\frac{q_{\mathrm{int}}}{\varepsilon _{0}}=\frac{1}{\varepsilon _{0}}\rho \frac{4}{3}\pi r^{3} \\&\Rightarrow \quad E_{\mathrm{e}} =\frac{\rho r^{3}}{3\varepsilon _0 }\frac{1}{r_{e}^{2}}, \end{aligned}$$

where it has been taken into account that the charge is located inside the sphere of radius r.

(b) As the charge of the layer is $dq =$ $\rho { dv} = {\rho }4{\pi {r}}^{2}$ dr, the work necessary to approach it from infinity is 

$$\begin{aligned}&dW=\int _{r_e =\infty }^{r_e =r} {d\mathbf{F}_{ex} .d\mathbf{r}_e } =-\int _{r_e =\infty }^{r_e =r} {dqE_e dr_e } =-\int _{r_e =\infty }^{r_e =r} {\rho {4}\pi r^{{2}}dr\frac{\rho r^{3}}{3\varepsilon _0 }\frac{1}{r_e ^{2}}dr_e } \\&=-\frac{{4}\pi \rho ^{2}}{3\varepsilon _0 }r^{{5}}dr\int _{r_e =\infty }^{r_e =r} {\frac{dr_e }{r_e ^{2}}} =\frac{{4}\pi \rho ^{2}}{3\varepsilon _0 }r^{{5}}dr\frac{1}{r}=\frac{{4}\pi \rho ^{2}}{3\varepsilon _0 }r^{{4}}dr. \end{aligned}$$

(c) The energy of the charged sphere is the sum of the work needed for the transport of all the layers from the first, $r=$ 0, to the last, $r=R$:

$$ U_e =\int _{r=0}^{r=R} {\frac{{4}\pi \rho ^{2}}{3\varepsilon _0 }r^{{4}}dr} =\frac{{4}\pi \rho ^{2}}{3\varepsilon _0 }\frac{1}{5}R^{5}=\frac{{4}\pi \rho ^{2}R^{5}}{15\varepsilon _0 }. $$

(d) The application of this result to the data provided gives

$$ U_e =\frac{{4}\pi \rho ^{2}R^{5}}{15\varepsilon _0 }=\frac{{4}\pi 0.3^{2}\times 0.20^{5}}{15\times 8.854\times 10^{-12}}\,{\mathrm{J}=2.725}\times {10}^{6}\,{\mathrm{J}}. $$

9.10. :

Calculate the energy of the charged sphere of the previous problem by applying (9.7).

Solution

The electric field E inside the sphere of radius r is radial, due to the symmetry.

Applying Gauss’s theorem it is deduced:

$$ \oint {\mathbf{E}.d\mathbf{S}} =\oint {EdS} =E4\pi r^{2}=\frac{q_{\mathrm{int}} }{\varepsilon _0 }=\frac{1}{\varepsilon _0 }\rho \frac{4}{3}\pi r^{3}\quad \Rightarrow \quad E=\frac{\rho r}{3\varepsilon _0 }. $$

Applying this result in (9.8), gives, for the interior of the sphere of radius R, 

$$ U_{ein} =\frac{\varepsilon _0 }{2}\int _v {\mathbf{E}.\mathbf{E}dv} =\frac{\varepsilon _0 }{2}\int _v {E^{2}dv} =\frac{\varepsilon _0 }{2}\int _0^R {\frac{\rho ^{2}r^{2}}{3^{2}\varepsilon _0 ^{2}}4\pi r^{2}dr} =\frac{2\pi \rho ^{2}R^{5}}{45\varepsilon _0 }, $$

which does not agree with the result of the previous problem. Why? Equation (9.7) refers to the whole of the space where there is an electric field , not only to the zone where there are charges. Therefore the energy outside the sphere of radius R must be added. The electric field outside this sphere is

$$ \oint {\mathbf{E}_{ex} .d\mathbf{S}} =\oint {E_{ex} dS} =E_{ex} 4\pi r^{2}=\frac{q_{\mathrm{int}} }{\varepsilon _0 }=\frac{1}{\varepsilon _0 }\rho \frac{4}{3}\pi R^{3}\quad \Rightarrow \quad E_{ex} =\frac{\rho R^{3}}{3\varepsilon _0 r_{ex} ^{2}}, $$

and the energy outside is

$$\begin{aligned} U_{eex}&=\frac{\varepsilon _0 }{2}\int _{vex} {\mathbf{E}_{ex} .\mathbf{E}_{ex} dv} =\frac{\varepsilon _0 }{2}\int _{vex} {E_{ex}{}^{2}dv} \\&=\frac{\varepsilon _0 }{2}\frac{\rho ^{2}R^{6}}{3^{2}\varepsilon _{0}^{2}}\int _{r_{ex} =R}^{r_{ex} =\infty } {\frac{1}{r_{ex}{}^{4}}4\pi r_{ex}{}^{2}dr_{ex} } =\frac{2\pi \rho ^{2}R^{5}}{9\varepsilon _0 }. \end{aligned}$$

The total energy is

$$ U_e =U_{ein} +U_{eex} =\frac{2\pi \rho ^{2}R^{5}}{45\varepsilon _0 }+\frac{2\pi \rho ^{2}R^{5}}{9\varepsilon _0 }=\frac{4\pi \rho ^{2}R^{5}}{15\varepsilon _0 }, $$

which coincides, as would be expected, with that of Problem 9.9.

9.11. :

The capacitor described in Problem 9.3 becomes disconnected from the generator that has charged it. It is wished to calculate the electrostatic force exerted by the plates. Take into account that if, once charged, an external force F is applied to move a plate a distance dz from the other, then the capacitance and the energy vary.

Solution

As the plates of the capacitor have charges of opposite signs, they are attracted to each other. In order to quantify the electrostatic force, we make use of energy methods.

The capacity of a flat capacitor is ${C}=\varepsilon$ S/z, therefore its energy is, according to (9.5),

$$ E_{p} =\frac{1}{2}\frac{q^{2}}{C}=\frac{1}{2}\frac{q^{2}}{\varepsilon S}z. $$

When the distance is increased by dz by the application of external force F, which is just the minimum force necessary to keep the plates of the capacitor in equilibrium, then the charge does not change upon being disconnected from the generator, although the energy increases, according to the equation above, by

$$ dE_{p} =\frac{1}{2}\frac{q^{2}}{\varepsilon S}dz. $$

The energy is supplied to the capacitor by means of the external force F and is Fdz, which, by application of the principle of conservation of energy, must be equal to the increase of energy (given by the previous expression), that is:

$$ Fdz=dE_{p} =\frac{1}{2}\frac{q^{2}}{\varepsilon S}dz, $$

and hence

$$\begin{aligned}&F=\frac{1}{2}\frac{q^{2}}{\varepsilon S}=\frac{1}{2}\frac{q^{2}}{\varepsilon _r \varepsilon _0 S}=\frac{1}{2}\frac{C^{2}V^{2}}{\varepsilon _r \varepsilon _0 S}=\frac{1}{2}\frac{\varepsilon _r \varepsilon _0 SV^{2}}{z^{2}} \\&=\frac{1}{2}\frac{5\times 8.854\times 10^{-12}\times 200\times 10^{-4}\times 300^{2}}{(5\times 10^{-3})^{2}}\text {N}=1.59\times 10^{-3}\text {N}. \\ \end{aligned}$$

It should be noted that the mechanical and electrostatic forces are of equal magnitude and in opposite direction. The result obtained above is coincident with that given by (9.11).

Fig. 9.7

Capacitor connected to a generator

9.12. :

Suppose that the capacitor described in Problem 9.3 remains connected to the generator that has charged it. It is wished to calculate the force exerted by the plates.

Solution

In this case, when F changes the distance between the plates by dz (Fig. 9.7), then capacity of the capacitor changes by dC, but since the tension remains constant, the charge will change by dq. Therefore:

(a) The exterior supplies the mechanical energy Fdz,

(b) The generator supplies the energy

$$ dU_g =Vdq=V^{2}dC=-\frac{V^{2}\varepsilon S}{z^{2}}dz, $$

where $C=\varepsilon$ S /z, z being the distance between the two plates, and ${dC}=(-\varepsilon$ S /$z^{2})dz$.

(c) The capacitor changes its energy by

$$ dU_e =d\left( {\frac{1}{2}CV^{2}} \right) =\frac{1}{2}V^{2}dC=-\frac{1}{2}\frac{V^{2}\varepsilon S}{z^{2}}dz. $$

The principle of conservation of energy establishes that

$$ Fdz+dU_g =dU_e, $$

which can be written as:

$$ Fdz-\frac{V^{2}\varepsilon S}{z^{2}}dz=-\frac{1}{2}\frac{V^{2}\varepsilon S}{z^{2}}dz\quad \Rightarrow \quad F=\frac{1}{2}\frac{V^{2}\varepsilon S}{z^{2}}=\frac{1}{2}\frac{q^{2}}{\varepsilon S}. $$

Note that the result is equal to that of Problem 9.11. This equality should be evident, since the distribution of charges is the same in both problems and, therefore, the forces must be equal.

Fig. 9.8

Parallel-plate capacitor and a dielectric plate

Observe that, if the distance between the plates increases, ${dz}>0$, then: (a) the energy supplied to the system from the external force is positive; (b) the energy provided by the generator is negative, that is, the generator is recharged; (c) the capacitor reduces its energy. That is, both the exterior and the capacitor collaborate in recharging the generator.

9.13. :

The attached figure represents a parallel-plane capacitor with charge q. Each plate has a size a $\times$ b and the separation between them is t, which is small. A dielectric plate, of length a, depth b and thickness t, with dielectric constant $\varepsilon _{r}$, is introduced a known distance x between the plates, as shown in Fig. 9.8. Calculate the force that the plates exert on the dielectric, discounting friction and neglecting edge effects.

Solution

On introducing the dielectric, it is polarised as indicated in the figure, and hence the polarisation charges are attracted by those of the plates and, due to the symmetry, the resultant force is expected to be towards the right.

Fig. 9.9

Parallel-plate capacitor. Observe the sense of the force

The electrostatic force acting on the dielectric slab is going to be calculated through the energy of the system and application of (9.11), where charges are kept constant.

The system shown in Fig. 9.9 can be considered as two capacitors in parallel. Each of the capacitors has the respective areas, charges, capacities and energies:

$$\begin{aligned}&S_1 =bx,\quad \quad \quad \; q_1 ,\quad \quad C_1 =\frac{\varepsilon _r \varepsilon _0 S_1 }{t}=\frac{\varepsilon _r \varepsilon _0 b}{t}x,\quad \quad \;U_{e1} =\frac{1}{2}\frac{q_1 ^{2}}{C_1 }, \\&S_2 =b(a-x),\quad q_2 ,\quad \quad C_2 =\frac{\varepsilon _0 S_2 }{t}=\frac{\varepsilon _0 b}{t}(a-x),\;\quad U_{e2} =\frac{1}{2}\frac{q_2 ^{2}}{C_2 }. \\ \end{aligned}$$

The energy is additive, therefore the energy of the system is the sum of both energies,

$$ U_{\mathrm{e}} =U_{\mathrm{e1}} +U_{\mathrm{e2}}. $$

To calculate the resulting force, it is sufficient to find the component of the force in the x-direction i.e. in the direction towards the right, by applying (9.11). A difficulty remains: to find the values of $q_{1}$ and $q_{2}$. To this end, use is made of the principle of conservation of electric charge

$$ q=q_{1} +q_{2}, $$

and of the fact that, being conducting plates, the potential is the same at all the points of the upper plate, as it is also in the lower plate. For this reason, the difference of potential V is identical in both capacitors, and can be written

$$ \frac{q_{1}}{C_{1}}=V=\frac{q_{2}}{C_{2}}\quad \Rightarrow \quad \frac{q_{1}}{q_{2}}=\frac{C_{1}}{C_{2}}. $$

These last two equations allow the calculation of $q_{1}$ and $q_{2}$:

$$ q_{1} =\frac{qC_{1}}{C_{1}+C_{2}},\quad q_{2} =\frac{qC_{2}}{C_{1} +C_{2}}. $$

Replacing both values in the previous equations, the energy is given by

$$ U_{e} =\frac{q^{2}t}{2\varepsilon _0 b[(\varepsilon _{r}-1)x+a]}. $$

The force is calculated by

$$ F=-\frac{\partial U_{e}}{\partial x}=\frac{q^{2}t}{2\varepsilon _{0} b[(\varepsilon _{r} -1)x+a]^{2}}(\varepsilon _{r} -1). $$

Applying this formula to the particular case $x=a$, which corresponds to the circumstance where the dielectric is totally inserted, gives

$$ F=\frac{q^{2}t}{2\varepsilon _r ^{2}\varepsilon _0 ba^{2}}(\varepsilon _r -1). $$

That is, when $x=a$, F is not null. However, if the figure is considered when the dielectric is inserted, then the plane perpendicular to that of the drawing which passes through the centre of the plates is symmetric, therefore it is impossible that there is a force towards the right, because the same reasoning would lead to the result that the force is towards the left. As the argument of symmetry is more basic than that used in the solution of the problem, the conclusion is reached that the force is not properly calculated. In the demonstration of the calculation of F it is supposed that the electric field is limited to the parallelepiped delimited by the plates, i.e. edge effects have been neglected. However, this assumption is not in complete agreement with the laws of Electromagnetism, therefore, it can be taken as only an approximation of the reality, and therefore the formula obtained for the force is only approximate.

Problems C

9.14. :

Take a capacitor which is equal to that in the previous problem except it is connected to a battery of electromotive force ${\mathcal {E}}$ (Fig. 9.10). Calculate the force on the dielectric.

Fig. 9.10

Capacitor connected to an external source

Fig. 9.11

The forces on the dielectric material

Solution

To establish equilibrium and obtain the solution, the force F $_{\mathrm{ex}}=-$ F is added to the electric force F of the system (Fig. 9.11). In a state of equilibrium, the difference of potential between the upper and the lower plates is $V= {\mathcal {E}}$ and it remains constant for any value of x because the battery gives, or removes, charges to/from the plates. Let us consider that the dielectric moves the small distance dx towards the right, the energy conservation principle (First Law of Thermodynamics) states that the energy provided by the external force F $_{\mathrm{ex}}$, plus the energy provided by the battery is equal to the increase of energy of the system:

$$ dW_{ex} +dW_{bat} =dU_e. $$

Furthermore:

$$ dW_{ex} =\mathbf{F}_{ex} .d\mathbf{r}=-\mathbf{F}.d\mathbf{r}=-Fdx, $$

where F is the component on the axis OX of the force F. By definition, the electromotive force is

$$ dW_{bat} ={\mathcal {E}} dq\quad \Rightarrow \quad dW_{bat} ={\mathcal {E}}^{2}dC. $$

The energy is additive, therefore the energy of the system formed by both capacitors is

$$\begin{aligned} U_{\mathrm{e}}&=U_{\mathrm{e1}} +U_{\mathrm{e2}} =\frac{1}{2}C_{1} {\mathcal {E}}^{2}+\frac{1}{2}C_{2} {\mathcal {E}}^{2}=\frac{1}{2}{\mathcal {E}}^{2}(C_{1} +C_{2} )=\frac{1}{2}{\mathcal {E}}^{2}C\\&\Rightarrow \quad dU_{\mathrm{e}} =\frac{1}{2}{\mathcal {E}}^{2}dC, \end{aligned}$$

where $C=C_{1}+C_{2}$ is the capacitance of the system.

Therefore,

$$ -Fdx+{\mathcal {E}}^{2}dC=\frac{1}{2}{\mathcal {E}}^{2}dC\quad \Rightarrow \quad Fdx=\frac{1}{2}{\mathcal {E}}^{2}dC. $$

The capacity C of the system is

$$\begin{aligned} C=C_{1} +C_{2}&=\frac{\varepsilon _{r} \varepsilon _{0} bx}{t}+\frac{\varepsilon _{0} b(a-x)}{t}\\&\Rightarrow \quad dC=\frac{\varepsilon _{r} \varepsilon _{0} b}{t}dx-\frac{\varepsilon _{0} b}{t}dx=\frac{\varepsilon _{0} b}{t}(\varepsilon _{r} -1)dx. \end{aligned}$$

From the two last equations it is deduced that

$$\begin{aligned} Fdx&=\frac{1}{2}{\mathcal {E}}^{2}\frac{\varepsilon _0 b}{t}(\varepsilon _r -1)dx\\&\Rightarrow \quad F=\frac{1}{2}{\mathcal {E}}^{2}\frac{\varepsilon _0 b}{t}(\varepsilon _r -1)=\frac{q^{2}t}{2\varepsilon _0 b[(\varepsilon _r -1)x+a]^{2}}(\varepsilon _r -1), \end{aligned}$$

where it has been taken into account that $C=q/V=q$/${\mathcal {E}}$. This result agrees with that of the previous problem.

It should be noted that in this case of constant voltage dW $_{bat}=$ 2dU $_{e}$ and, therefore, Fdx $=$ dU $_{e}$. Hence, the electrostatic force can be expresses as

$$ F=+\frac{\partial U_e }{\partial x}, $$

where it is assumed that the voltage between the two plates is kept constant. This equation also leads to the result obtained above for F.

9.15. :

The coil represented in Fig. 9.12 is of N tightened turns, of small cross-sectional area S, and great mean length l. The circulating current intensity is I. The core material is magnetically linear and of relative permeability $\mu _{r}$. Calculate the magnetic energy that it contains. Apply: $N=$ 2000, $S=$ 1 cm$^{2}$, $l=$ 30 cm, $I=$ 4 A, and $\mu _{r}=$ 10.

Fig. 9.12

Coil of N turns

Fig. 9.13

Electric circuit

Solution

Ampère’s law applied to the inner circumference of length l gives

$$ \oint {\mathbf{H}.d\mathbf{l}} =Hl=NI\quad \Rightarrow \quad H=\frac{NI}{l}. $$

Although H depends on l, if the solenoid is very thin, then the length of all the inner circumferences are almost equal, and hence H can be assumed independent of the inner point considered. Of course the circulation of H in outer circumferences is null and, therefore, outside the solenoid the magnetic field is null. 

Therefore the magnetic field B in the interior is

$$ B=\mu _r \mu _0 H=\mu _r \mu _0 \frac{NI}{l}. $$

Hence, the energy per unit volume (9.22) is

$$ u_m =\frac{1}{2}{} \mathbf{B}.\mathbf{H}=\frac{1}{2}BH=\frac{1}{2}\mu _r \mu _0 \frac{NI}{l}\frac{NI}{l}=\frac{1}{2}\mu _r \mu _0 \frac{N^{2}I^{2}}{l^{2}}. $$

As the volume is $v=Sl$, then the magnetic energy is

$$ U_m =u_m Sl=\frac{1}{2}\mu _r \mu _0 \frac{N^{2}I^{2}}{l^{2}}Sl=\frac{1}{2}\mu _r \mu _0 \frac{N^{2}I^{2}S}{l}. $$

Using the numerical data provided yields:

$$\begin{aligned}&u_m = \frac{1}{2}\mu _r \mu _0 \frac{N^{2}I^{2}}{l^{2}}=\frac{1}{2}10\times 4\pi \times 10^{-7}\frac{2000^{2}\times 4^{2}}{0.30^{2}}\frac{\mathrm{J}}{\mathrm{m}^{{3}}}=4468\frac{\mathrm{J}}{\mathrm{m}^{{3}}}, \\ U_m =&\frac{1}{2}\mu _r \mu _0 \frac{N^{2}I^{2}}{l^{2}}Sl=\frac{1}{2}10\times 4\pi \times 10^{-7}\frac{2000^{2}\times 4^{2}}{0.30^{2}}10^{-4}\times 0.30\;{\mathrm{J}}=0.134\;{\mathrm{J}}. \end{aligned}$$

9.16. :

The capacitor in Fig. 9.13 has a charge $q_{0}$. The switch is closed and it is wished to know the energy of the capacitor and the coil, of null resistance, at instant t.

Solution

The charge that the capacitor contains at any instant must be known in order to calculate the energy stored in it, and the current intensity that circulates around the coil must be determined to obtain its stored energy.

When closing the circuit, a certain current of intensity I will be established at t, and the electromotive force, evaluated around the circuit following the direction attributed to d l, will be $-LdI/dt$, then the difference of potential between the upper and lower terminal of the coil is LdI / dt. This value agrees with the difference of potential between the upper plate of charge q and lower of charge – q, that is,

$$ L\frac{dI}{dt}=V=\frac{q}{C}. $$

As the current is equal to the charge that crosses a section of the conductor per unit of time, it must agree, by the principle of conservation of charge, with the reduction of charge q per unit of time: $I=-dq/dt$. Therefore:

$$ L\frac{d}{dt}\left( {-\frac{dq}{dt}} \right) =\frac{1}{C}q\quad \Rightarrow \quad \frac{d^{2}q}{dt^{2}}+\frac{1}{LC}q=0. $$

As this equation is mathematically equal to that which regulates the harmonic movement of a block attached to a spring, the solution must be equal, that is:

$$ q=A\cos (\omega t+\varphi )\quad \Rightarrow \quad \frac{dq}{dt}=-A\omega \sin (\omega t+\varphi )\quad \Rightarrow \quad \frac{d^{2}q}{dt^{2}}=-A\omega ^{2}\cos (\omega t+\varphi ). $$

Since, in $t=$ 0, the charge is $q=q_{0}$, the first of these equalities gives

$$ q_0 =A\cos \varphi . $$

From the second equality it is deduced that

$$ I=-\frac{dq}{dt}=A\omega \sin (\omega t+\varphi ). $$

As at the initial moment, $t=$ 0, $I=$ 0, therefore

$$ 0=A\omega \sin (0+\varphi )=A\omega \sin \varphi \quad \Rightarrow \quad \varphi =0. $$

Therefore $q_0 =A\cos 0=A$, which implies:

$$ q=q_0 \cos (\omega t)\quad \Rightarrow \quad I=q_0 \omega \sin (\omega t). $$

The angular frequency $\omega$ can be obtained by substituting q and I in the expression $LdI/dt=q/C$, giving

$$ \omega =\frac{1}{\sqrt{LC}}\quad \Rightarrow \quad f=\frac{1}{2\pi \sqrt{LC}}, $$

where f is the frequency in Hertz.

With these results it is possible to deduce the energies stored by the capacitor and the coil:

$$ U_e =\frac{1}{2}\frac{q^{2}}{C}=\frac{q_0 ^{2}}{2C}\cos ^{2}(\omega t), $$

$$ U_m =\frac{1}{2}LI^{2}=\frac{1}{2}Lq_0 ^{2}\omega ^{2}\sin ^{2}(\omega t)=\frac{q_0 ^{2}}{2C}\sin ^{2}(\omega t). $$

Figure 9.14 represents the charge of the capacitor, the current intensity and the energies of the capacitor and the coil as functions over time. Note how the intensity varies with time (oscillating circuit) and how the capacitor and the coil transfer the energy.

Fig. 9.14

Different functions depending on time

9.17. :

A ring of linear magnetic material , of relative permeability $\mu _{r}$, has mean length l and its cross section is a circle of area S (Fig. 9.15). There are N windings of a conductor wire on the ring. It is wished to know the magnetic energy in the device when the current intensity through the conductor is I.

Solution

Ampère’s law around the circumference of mean length l gives

$$ \oint {\mathbf{H}.d\mathbf{l}} =Hl=NI\quad \Rightarrow \quad H=\frac{NI}{l}. $$

Therefore, magnetic field B is

$$ B=\mu _r \mu _0 H=\mu _r \mu _0 \frac{NI}{l}. $$

The flux through the whole circuit is

$$ \Phi =BNS=\mu _r \mu _0 \frac{NI}{l}NS\quad \Rightarrow \quad d\Phi =\mu _r \mu _0 \frac{N^{2}S}{l}dI. $$

Applying (9.14) gives

$$ dU_m =Id\Phi =\mu _r \mu _0 \frac{N^{2}S}{l}IdI\quad \Rightarrow \quad U_m =\frac{\mu _r \mu _0 N^{2}S}{2l}I^{2}. $$

9.18. :

A toroidal solenoid has an outer radius $R_{e}=$ 40 cm, an inner radius $R_{i}=$ 30 cm, and $N=$ 1200 windings, through which circulates a current of intensity $I=$ 20 A. (a) Calculate the energy that it contains. (b) Calculate the energy that it would contain if it were filled successively with a diamagnetic, paramagnetic and ferromagnetic material , of relative permeability 0.9, 1.1, and 50,000 respectively.

Fig. 9.15

Ring of a magnetic material

Solution

The stored energy is calculated from the density of magnetic energy and the volume of the solenoid.

To calculate the field H inside the solenoid, Ampère’s law is applied to the mean circumference, giving:

$$ \oint {\mathbf{H}.d\mathbf{l}=I_{total} } \quad \Rightarrow \quad HL=NI\quad \Rightarrow \quad H=\frac{NI}{L}. $$

The length of the solenoid can be estimated with the average radius

$$ R=\frac{R_e +R_i }{2}\quad \Rightarrow \quad L=2\pi R=\pi (R_e +R_i ). $$

Therefore

$$ H=\frac{NI}{\pi (R_e +R_i )}. $$

The magnetic field $B =\upmu _{r}\upmu _{0}H$, therefore the density of energy is

$$ u_m =\frac{1}{2}{} \mathbf{B}.\mathbf{H}=\frac{1}{2}BH=\frac{\mu _{\mathrm{r}} \mu _{0}}{2}H^{2}==\frac{\mu _{\mathrm{r}} \mu _0 }{2}\left( {\frac{NI}{\pi (R_e +R_i )}} \right) ^{2}=\frac{\mu _{\mathrm{r}} \mu _0 }{2}\frac{N^{2}I^{2}}{\pi ^{2}(R_e +R_i )^{2}}. $$

The diameter of the circle that produces a cross section is $R_{e}-R_{i}$, then the area of the circle is

$$ S=\pi \frac{(R_e -R_i )^{2}}{4}, $$

and hence the volume of the solenoid can be estimated thus

$$ v=SL=\pi \frac{(R_e -R_i )^{2}}{4}\pi (R_e +R_i ), $$

and therefore the energy stored with any of the materials is

$$ U_m =u_m v=\frac{\mu _{\mathrm{r}} \mu _0 }{2}\frac{N^{2}I^{2}}{\pi ^{2}(R_e +R_i )^{2}}\pi \frac{(R_e -R_i )^{2}}{4}\pi (R_e +R_i )=\mu _{\mathrm{r}} \mu _0 \frac{N^{2}I^{2}}{8}\frac{(R_e -R_i )^{2}}{R_e +R_i }. $$

For the diamagnetic material this is

$$ U_m =\mu _{\mathrm{r}} \mu _0 \frac{N^{2}I^{2}}{8}\frac{(R_e -R_i )^{2}}{R_e +R_i }=0.9\times 4\pi \times 10^{-7}\frac{1200^{2}20^{2}}{8}\frac{(0.4-0.3)^{2}}{0.4+0.3_i }\,{\mathrm{J}}=1.16 {}\,{\mathrm{J}}. $$

For the paramagnetic material with $\mu _{r}=$ 1.1 it is $U_{m}=$ 1.42 J and for the ferromagnetic material with $\mu _{r}=$ 50,000 it is $U_{m}=64.4\times 10^{3}$ J.

Note the large amount of energy accumulated when the introduced material is ferromagnetic.

9.19. :

At instant $t=$ 0, terminal 1 is connected to terminal 2 by means of the switch drawn in Fig. 9.16. After a long time, 1 is disconnected from 2 and immediately connected with 3. (a) During the first operation, calculate the current for the coil, the energy it stores and the power released as heat. (b) Calculate these magnitudes in the second operation.

Fig. 9.16

Electric circuit

Fig. 9.17

Electric circuit when connecting 1 and 2

Solution

(a) In the first stage, Kirchhoff ’s second law (Fig. 9.17) gives

$$ {\mathcal {E}}-L\frac{dI}{dt}=2RI\quad \Rightarrow \quad \frac{dI}{{\mathcal {E}}-2RI}=\frac{dt}{L}\quad \Rightarrow \quad -\frac{1}{2R}\ln \left( {{\mathcal {E}}-2RI} \right) =\frac{t}{L}+k. $$

As at the initial instant, $t=$ 0, $I=$ 0 therefore:

$$ I=\frac{{\mathcal {E}}}{2R}\left( {1-e^{-\frac{2R}{L}t}} \right) . $$

Note that the current is null when beginning the first stage, and when finishing the first at t $\approx \infty$, it becomes

$$ I=\frac{{\mathcal {E}}}{2R}. $$

The energy stored in the coil, (9.15), is

$$ U_m =\frac{1}{2}LI^{2}=\frac{1}{2}L\frac{{\mathcal {E}}^{2}}{2^{2}R^{2}}\left( {1-e^{-\frac{2R}{L}t}} \right) ^{2}=\frac{L{\mathcal {E}}^{2}}{8R^{2}}\left( {1-e^{-\frac{2R}{L}t}} \right) ^{2}. $$

The energy when finishing the first stage ( $t \approx \infty )$ is

$$ U_m =\frac{L{\mathcal {E}}^{2}}{8R^{2}}. $$

Note how the stored energy grows until reaching this final value.

The calorific power released is

$$ \frac{dQ}{dt}=2RI^{2}=2R\frac{{\mathcal {E}}^{2}}{2^{2}R^{2}}\left( {1-e^{-\frac{2R}{L}t}} \right) ^{2}=\frac{{\mathcal {E}}^{2}}{2R}\left( {1-e^{-\frac{2R}{L}t}} \right) ^{2}. $$

Note that this power begins as null and finishes as ${\mathcal {E}}^{2}$/(2R).

Fig. 9.18

Circuit composed by a resistance R and an autoinductance L

(b) In the second stage, the clock restarting from 0 s at the start of this stage, terminal 1 is connected to terminal 3, see Fig. 9.18. From Ohm’s Law , we have for the voltages

$$ -L\frac{dI}{dt^{\prime }}=RI\quad \Rightarrow \quad \frac{dI}{I}=-\frac{R}{L}dt^{\prime }\quad \Rightarrow \quad \ln I=-\frac{R}{L}t^{\prime }+k^{\prime }, $$

where time is denoted by $t'$.

At $t'= 0$, then $I = {\mathcal {E}}$/(2R), as demonstrated in section (a), and hence ln(${\mathcal {E}}$/(2R)) $= k'$ and therefore

$$ \ln I=-\frac{R}{L}t^{\prime }+\ln \frac{{\mathcal {E}}}{2R}\quad \Rightarrow \quad \ln \frac{2RI}{{\mathcal {E}}}=-\frac{R}{L}t^{\prime }\quad \Rightarrow \quad I=\frac{{\mathcal {E}}}{2R}e^{-\frac{R}{L}t^{\prime }}. $$

It can be seen how the intensity diminishes over time $t'$ from the final value of the first operation to zero.

The energy stored in the coil at this stage is

$$ U_m =\frac{1}{2}LI^{2}=\frac{1}{2}L\frac{{\mathcal {E}}^{2}}{2^{2}R^{2}}e^{-\frac{2R}{L}t^{\prime }}=\frac{L{\mathcal {E}}^{2}}{8R^{2}}e^{-\frac{2R}{L}t^{\prime }}. $$

Which indicates that the initial energy in the second stage agrees with the energy at the end of the first, L ${\mathcal {E}}^{2}$/(8$R^{2})$, and the energy at the end of the whole process is null.

The power released as heat is

$$ \frac{dQ}{dt}=RI^{2}=R\frac{{\mathcal {E}}^{2}}{2^{2}R^{2}}e^{-\frac{2R}{L}t^{\prime }}=\frac{{\mathcal {E}}^{2}}{4R}e^{-\frac{2R}{L}t^{\prime }}, $$

beginning with ${\mathcal {E}}^{2}$/(4R) and becoming null at the end of the experiment.

Fig. 9.19

Hysteresis loop

To sum up, the energy provided by the generator in the first stage is partly stored in the coil and the rest is transmitted to the outside in heat form by the resistance. In the second stage, the electrical energy stored in the coil is transformed into thermal energy of the resistance of the circuit of Fig. 9.18 and released as heat.

9.20. :

A ferromagnetic material has such a simple behaviour that its hysteresis loop is in the form of a rhomboid, two sides of which are: segment 1–2 on the line $B=c$ and segment 2–3 on the line $B=sH+c$, where c and s are known positive constants (Fig. 9.19). Calculate the magnetic energy dissipated by each cycle described and per unit volume.

Solution

With the data provided, the complete hysteresis loop 1–2–3–4–1 can be drawn in diagram $B-H$. For an intermediate state, for example on side 4–1 of the loop, a certain H and a certain B exist in the material. If an infinitesimal change of B occurs, dB, the magnetic energy provided to the system per unit volume is du $_{m}{=HdB}$.

Applying this change to the four sections of the loop gives:

Section 1–2. This is described by the line $B=c$, therefore

$$ u_{21} =\int _{B_1 }^{B_2 } {Hd} B=0, $$

since dB $=$ 0.

Section 2–3. Its points belong to the line $B=sH+c$. The point 3 is at $B=-c$, therefore $H_{3}=-$2c / s, and $H_{2}$=0. Hence,

$$ u_{32} =\int _{B_2 }^{B_3 } {Hd} B=\int _0^{-2c/s} {H(sdH+0)} =2c^{2}/s. $$

Section 3–4. This section is contained in the line $B=-c$, therefore

$$ u_{43} =\int _{H_3 }^{H_4 } {Hd} B=0. $$

Section 4–1. This is defined by $B=sH-c$. Therefore

$$ u_{14} =\int _{H_4 }^{H_1 } {HdB} =\int _0^{2c/s} {H(sdH+0)} =2c^{2}/s. $$

The total energy supplied in one cycle and per unit volume is the sum of the variations in each section. Therefore

$$ u_{\mathrm{mcicle}} =0+2c^{2}/s+0+2c^{2}/s=4c^{2}/s. $$

However, as the system has recovered its initial magnetic state and therefore its initial magnetic energy, this energy appears as thermal internal energy , which is transmitted to the exterior in the form of heat. That is, the system has “lost” the supplied energy 4$c^{2}$/s per unit volume by the fact of following a hysteresis loop.

It should be noted that the area of the loop is 4$c^{2}$/s.

Fig. 9.20

Experimental set-up

Fig. 9.21

Hysteresis loop

9.21. :

The magnetic circuit shown in Fig. 9.20 has N windings. It is connected to an alternating current generator such that an alternating current of intensity $I = I_{0}$sin($\omega$ t) circulates. The material is ferromagnetic with a hysteresis loop for the said current as drawn (Fig. 9.21). (a) Estimate the energy dissipated per cycle. (b) Estimate the dissipated power. (c) Calculate the dissipated power, using: $a =$ 40 cm, $b =$ 30 cm, $c =$ 20 cm, $d =$ 10 cm, $e =$ 0.005 T, $N =$ 1000, $I_{0 }=$ 6 A, and current frequency of 50 Hz.

Solution

(a) Ampère’s law establishes that

$$ \oint {\mathbf{H}.d\mathbf{l}=I_{total} } =NI\quad \Rightarrow \quad HL=NI\quad \Rightarrow \quad H=\frac{NI}{L}. $$

The mean length can be estimated, from the figure of the ferromagnetic core, as

$$ L=2(a-d)+2(b-d)=2a+2b-4d, $$

therefore

$$ H=\frac{NI}{L}=\frac{NI}{2a+2b-4d}=\frac{NI_0 }{2a+2b-4d}\sin (\omega t). $$

The maximum value of H is given when sin($\omega$ t) is one and must coincide with the value $H=h$, that is,

$$ h=\frac{NI_{0}}{2a+2b-4d} \,.$$

Given the data of the hysteresis loop , a change of B gives rise to an injection of energy per unit volume of H.dB, and therefore for one cycle of the loop, the energy per unit of volume introduced is

$$\begin{aligned} u_{mcicle}&=\oint {H dB=\int _{1}^{2} {H dB}} +\int _{2}^{3} {H dB} +\int _{3}^{4} {H dB} +\int _{4}^{1} {H dB} \\&=0-h(-e-e)+0+h(e+e)=4eh. \end{aligned}$$

Therefore the power dissipated per unit volume of material is 4eh, which coincides with the “area” of the loop, as expected. Substituting the value calculated for h gives

$$ u_{mcicle} =4e\frac{NI_0 }{2a+2b-4d}=\frac{2eNI_0 }{a+b-2d}. $$

As the volume of the material is

$$ v=2(acd)+2((b-2d)cd)=2acd+2bcd-4cd^{2}, $$

the energy given off per cycle in heat form is

$$ Q=4eNI_0 \frac{acd+bcd-2cd^{2}}{a+b-2d}. $$

(b) As frequency f represents the number of cycles per unit time, the dissipated power is

$$ P={4}eNI_0 f\frac{acd+bcd-2cd^{2}}{a+b-2d}. $$

(c) Application

$$ \begin{array}{l} P=4\times 0.005\times 1000\times 6\times 50\frac{0.40\times 0.20\times 0.10+0.30\times 0.20\times 0.10-2\times 0.20\times 0.10^{2}}{0.40+0.30-2\times 0.10}\;{\mathrm{W}} \\ =120\;{\mathrm{W}}.\\ \end{array} $$

This is the dissipated power.