Express amperes law in

I) differential form

Ii) integral form

# Electromagnetic

**PUERTOGALERA**#3

- Intuitively, the divergence was based on the idea of the div-meter, a/1. The corresponding device for measuring the curliness of a field is the curl-meter, a/2. If the field is curly, then the torques on the charges will not cancel out, and the wheel will twist against the resistance of the spring. If your intuition tells you that the curlmeter will never do anything at all, then your intuition is doing a beautiful job on static fields; for nonstatic fields, however, it is perfectly possible to get a curly electric field.

Gauss’ law in differential form relates divEdivE , a scalar, to the charge density, another scalar. Ampère’s law, however, deals with directions in space: if we reverse the directions of the currents, it makes a difference. We therefore expect that the differential form of Ampère’s law will have vectors on both sides of the equal sign, and we should be thinking of the curl-meter’s result as a vector. First we find the orientation of the curl-meter that gives the strongest torque, and then we define the direction of the curl vector using the right-hand rule shown in figure a/3.

To convert the div-meter concept to a mathematical definition, we found the infinitesimal flux, dΦdΦ through a tiny cubical Gaussian surface containing a volume dvdv . By analogy, we imagine a tiny square Ampèrian surface with infinitesimal area dAdA . We assume this surface has been oriented in order to get the maximum circulation. The area vector dAdA will then be in the same direction as the one defined in figure a/3. Ampère’s law is

dΓ=4πkc2dIthrough.

dΓ=4πkc2dIthrough.

We define a current density per unit area, jj , which is a vector pointing in the direction of the current and having magnitude j=dI/|dA|j=dI/|dA| . In terms of this quantity, we have

dΓ=4πkc2j|j||dA|dΓ|dA|=4πkc2|j|

dΓ=4πkc2j|j||dA|dΓ|dA|=4πkc2|j|

With this motivation, we define the magnitude of the curl as

|curl B|=dΓ|dA|.

|curl B|=dΓ|dA|.

Note that the curl, just like a derivative, has a differential divided by another differential. In terms of this definition, we find Amp To convert the div-meter concept to a mathematical definition, we found the infinitesimal flux, dΦdΦ through a tiny cubical Gaussian surface containing a volume dvdv . By analogy, we imagine a tiny square Ampèrian surface with infinitesimal area dAdA . We assume this surface has been oriented in order to get the maximum circulation. The area vector dAdA will then be in the same direction as the one defined in figure a/3. Ampère’s law is {e}re’s law in differential form:

curl B=4πkc2j

curl B=4πkc2j

The complete set of Maxwell’s equations in differential form is collected on page 914.

11.4.2 Properties of the curl operator

The curl is a derivative.

b / The coordinate system used in the following examples.

As an example, let’s calculate the curl of the field x^x^ shown in figure c.

c / The field x^x^ .

For our present purposes, it doesn’t really matter whether this is an electric or a magnetic field; we’re just getting out feet wet with the curl as a mathematical definition. Applying the definition of the curl directly, we construct an Ampèrian surface in the shape of an infinitesimally small square. Actually, since the field is uniform, it doesn’t even matter very much whether we make the square finite or infinitesimal. The right and left edges don’t contribute to the circulation, since the field is perpendicular to these edges. The top and bottom do contribute, but the top’s contribution is clockwise, i.e., into the page according to the right-hand rule, while the bottom contributes an equal amount in the counterclockwise direction, which corresponds to an out-of-the-page contribution to the curl. They cancel, and the circulation is zero. We could also have determined this by imagining a curl-meter inserted in this field: the torques on it would have canceled out.

It makes sense that the curl of a constant field is zero, because the curl is a kind of derivative. The derivative of a constant is zero.

The curl is rotationally invariant.

d / The field y^y^ .

Figure d looks just like figure c, but rotated by 90 degrees. Physically, we could be viewing the same field from a point of view that was rotated. Since the laws of physics are the same regardless of rotation, the curl must be zero here as well. In other words, the curl is rotationally invariant. If a certain field has a certain curl vector, then viewed from some other angle, we simply see the same field and the same curl vector, viewed from a different angle. A zero vector viewed from a different angle is still a zero vector.

As a less trivial example, let’s compute the curl of the field F=xy^F=xy^ shown in figure e, at the point (x=0,y=0)(x=0,y=0) .

e / The field xy^xy^ .

The circulation around a square of side ss centered on the origin can be approximated by evaluating the field at the midpoints of its sides,

xxxx=s/2=0=−s/2=0yyyy=0=s/2=0=−s/2FFFF=(s/2)y^=0=−(s/2)y^=0s1⋅Fs2⋅Fs3⋅Fs4⋅F=s2/2=0=s2/2=0,

x=s/2y=0F=(s/2)y^s1⋅F=s2/2x=0y=s/2F=0s2⋅F=0x=−s/2y=0F=−(s/2)y^s3⋅F=s2/2x=0y=−s/2F=0s4⋅F=0,

which gives a circulation of s2s2 , and a curl with a magnitude of s2/area=s2/s2=1s2/area=s2/s2=1 . By the right-hand rule, the curl points out of the page, i.e., along the positive zz axis, so we have

curl xy^=z^.

curl xy^=z^.

Now consider the field −yx^−yx^ , shown in figure f.

f / The field −yx^−yx^ .

This is the same as the previous field, but with your book rotated by 90 degrees about the zz axis. Rotating the result of the first calculation, z^z^ , about the zz axis doesn’t change it, so the curl of this field is also z^z^ .

Scaling

When you’re taking an ordinary derivative, you have the rule

ddx[cf(x)]=cddxf(x).

ddx[cf(x)]=cddxf(x).

In other words, multiplying a function by a constant results in a derivative that is multiplied by that constant. The curl is a kind of derivative operator, and the same is true for a curl.

Example 14: Multiplying the field by −1−1 .

▹▹ What is the curl of the field −xy^−xy^ at the origin?

g / Example 14.

▹▹ Using the scaling property just discussed, we can make this into a curl that we’ve already calculated:

curl(−xy^)=−curl(xy^)=−z^

curl(−xy^)=−curl(xy^)=−z^

This is in agreement with the right-hand rule.

The curl is additive.

We have only calculated each field’s curl at the origin, but each of these fields actually has the same curl everywhere. In example 14, for instance, it is obvious that the curl is constant along any vertical line. But even if we move along the xx axis, there is still an imbalance between the torques on the left and right sides of the curl-meter. More formally, suppose we start from the origin and move to the left by one unit. We find ourselves in a region where the field is very much as it was before, except that all the field vectors have had one unit worth of y^y^ added to them. But what do we get if we take the curl of −xy^+y^−xy^+y^ ? The curl, like any god-fearing derivative operation, has the additive property

curl(F+G)=curlF+curlG,

curl(F+G)=curlF+curlG,

so

curl (−xy^+y^)=curl (−xy^)+curl (y^).

curl (−xy^+y^)=curl (−xy^)+curl (y^).

But the second term is zero, so we get the same result as at the origin.

Example 15: A field that goes in a circle

▹▹ What is the curl of the field xy^−yx^xy^−yx^ ?

h / Example 15.

▹▹ Using the linearity of the curl, and recognizing each of the terms as one whose curl we have already computed, we find that this field’s curl is a constant 2z^2z^ . This agrees with the right-hand rule.

Example 16: The field inside a long, straight wire

▹▹ What is the magnetic field inside a long, straight wire in which the current density is jj ?

▹▹ Let the wire be along the zz axis, so j=jz^j=jz^ . Ampère’s law gives

curl B=4πkc2jz^.

curl B=4πkc2jz^.

In other words, we need a magnetic field whose curl is a constant. We’ve encountered several fields with constant curls, but the only one that has the same symmetry as the cylindrical wire is xy^−yx^xy^−yx^ , so the answer must be this field or some constant multiplied by it,

B=b(xy^−yx^).

B=b(xy^−yx^).

The curl of this field is 2bz^2bz^ , so

2band thusB=4πkc2j,=2πkc2j(xy^−yx^).

2b=4πkc2j,and thusB=2πkc2j(xy^−yx^).

The curl in component form

Now consider the field

FxFyF=ax+by+c=dx+ey+f, i.e.,=axx^+byx^+cx^+dxy^+eyy^+fy^.

Fx=ax+by+cFy=dx+ey+f, i.e.,F=axx^+byx^+cx^+dxy^+eyy^+fy^.

The only terms whose curls we haven’t yet explicitly computed are the aa , ee , and ff terms, and their curls turn out to be zero (homework problem 50). Only the bb and dd terms have nonvanishing curls. The curl of this field is

curl F=curl (byx^)+curl (dxy^)=bcurl (yx^)+dcurl (xy^) [scaling] =b(−z^)+d(z^) [found previously]=(d−b)z^

curl F=curl (byx^)+curl (dxy^)=bcurl (yx^)+dcurl (xy^) [scaling] =b(−z^)+d(z^) [found previously]=(d−b)z^

But any field in the x−yx−y plane can be approximated with this type of field, as long as we only need to get a good approximation within a small region. The infinitesimal Ampèrian surface occurring in the definition of the curl is tiny enough to fit in a pretty small region, so we can get away with this here. The dd and bb coefficients can then be associated with the partial derivatives ∂Fy/∂x∂Fy/∂x and ∂Fx/∂y∂Fx/∂y . We therefore have

curlF=(∂Fy∂x−∂Fx∂y)z^

curlF=(∂Fy∂x−∂Fx∂y)z^

for any field in the x−yx−y plane. In three dimensions, we just need to generate two more equations like this by doing a cyclic permutation of the variables xx , yy , and zz :

(curlF)x=∂Fz∂y−∂Fy∂z (curlF)y=∂Fx∂z−∂Fz∂x (curlF)z=∂Fy∂x−∂Fx∂y

(curlF)x=∂Fz∂y−∂Fy∂z (curlF)y=∂Fx∂z−∂Fz∂x (curlF)z=∂Fy∂x−∂Fx∂y

i / A cyclic permutation of xx , yy , and zz .

Example 17: A sine wave

▹▹ Find the curl of the following electric field

E=(sinx)y^,

E=(sinx)y^,

and interpret the result.

j / Example 17.

▹▹ The only nonvanishing partial derivative occurring in this curl is

(curlE)z=∂Ey∂x=cosx, socurl E=cosz^

(curlE)z=∂Ey∂x=cosx, socurl E=cosz^

This is visually reasonable: the curl-meter would spin if we put its wheel in the plane of the page, with its axle poking out along the zz axis. In some areas it would spin clockwise, in others counterclockwise, and this makes sense, because the cosine is positive in some placed and negative in others.

This is a perfectly reasonable field pattern: it the electric field pattern of a light wave! But Ampère’s law for electric fields says the curl of E is supposed to be zero. What’s going on? What’s wrong is that we can’t assume the static version of Ampère’s law. All we’ve really proved is that this pattern is impossible as a static field: we can’t have a light wave that stands still.

Figure k is a summary of the vector calculus presented in the optional sections of this book. The first column shows that one function is a related to another by a kind of differentiation. The second column states the fundamental theorem of calculus, which says that if you integrate the derivative over the interior of a region, you get some information about the original function at the boundary of that region.

k / A summary of the derivative, gradient, curl, and divergence.

**PUERTOGALERA**#4

2)onsider, again, an infinite straight wire aligned along the $z$-axis and carrying a current $I$. The field generated by such a wire is written

\begin{displaymath}

B_\theta = \frac{\mu_0 I}{2\pi r}

\end{displaymath} (257)

in cylindrical polar coordinates. Consider a circular loop $C$ in the $x$-$y$ plane which is centred on the wire. Suppose that the radius of this loop is $r$. Let us evaluate the line integral $\oint_C {\bf B} \cdot d{\bf l}$. This integral is easy to perform because the magnetic field is always parallel to the line element. We have

\begin{displaymath}

\oint_C {\bf B} \cdot d{\bf l} = \oint B_\theta r d\theta= \mu_0 I.

\end{displaymath} (258)

However, we know from Stokes’ theorem that

\begin{displaymath}

\oint_S {\bf B} \cdot d{\bf l} = \int_S \nabla\times {\bf B} \cdot d{\bf S},

\end{displaymath} (259)

where $S$ is any surface attached to the loop $C$.

Let us evaluate $\nabla\times {\bf B} $ directly. According to Eq. (254),

$\displaystyle (\nabla\times{\bf B})_x$ $\textstyle =$ $\displaystyle \frac{\partial B_z}{\partial y} - \frac{\partial B_y}

{\partial z} = 0,$ (260)

$\displaystyle (\nabla\times{\bf B})_y$ $\textstyle =$ $\displaystyle \frac{\partial B_x}{\partial z} - \frac{\partial B_z}

{\partial x} = 0,$ (261)

$\displaystyle (\nabla\times {\bf B})_z$ $\textstyle =$ $\displaystyle \frac{\partial B_y}{\partial x} - \frac{\partial B_x}

{\partial y…

…rac{1}{r^2} - \frac{2 x^2}{r^4} +\frac{1}{r^2} -\frac{2 y^2}{r^4}\right) =0,$ (262)

where use has been made of $\partial r/\partial x = x/r$, etc. We now have a problem. Equations (258) and (259) imply that

\begin{displaymath}

\oint_S \nabla\times {\bf B} \cdot d{\bf S} = \mu_0 I.

\end{displaymath} (263)

But, we have just demonstrated that $\nabla\times{\bf B} = {\bf0}$. This problem is very reminiscent of the difficulty we had earlier with ${\nabla}\cdot {\bf E}$. Recall that $\int_V {\nabla}\cdot {\bf E} dV= q/\epsilon_0$ for a volume $V$ containing a discrete charge $q$, but that $\nabla\cdot {\bf E}=0$ at a general point. We got around this problem by saying that $\nabla\cdot {\bf E}$ is a three-dimensional delta-function whose spike is coincident with the location of the charge. Likewise, we can get around our present difficulty by saying that $\nabla\times {\bf B} $ is a two-dimensional delta-function. A three-dimensional delta-function is a singular (but integrable) point in space, whereas a two-dimensional delta-function is a singular line in space. It is clear from an examination of Eqs. (260)-(262) that the only component of $\nabla\times {\bf B} $ which can be singular is the $z$-component, and that this can only be singular on the $z$-axis (i.e., $r=0$). Thus, the singularity coincides with the location of the current, and we can write

\begin{displaymath}

\nabla\times{\bf B} = \mu_0 I \delta(x) \delta(y) \hat{\bf z}.

\end{displaymath} (264)

The above equation certainly gives $(\nabla\times{\bf B})_x

=(\nabla\times{\bf B})_y = 0$, and $(\nabla\times{\bf B})_z=0$ everywhere apart from the $z$-axis, in accordance with Eqs. (260)-(262). Suppose that we integrate over a plane surface $S$ connected to the loop $C$. The surface element is $d{\bf S} = dx dy \hat{\bf z}$, so

\begin{displaymath}

\int_S \nabla\times{\bf B} \cdot d{\bf S} = \mu_0 I \int\int

\delta(x) \delta(y) dx dy

\end{displaymath} (265)

where the integration is performed over the region $\sqrt{x^2+ y^2} \leq r$. However, since the only part of $S$ which actually contributes to the surface integral is the bit which lies infinitesimally close to the $z$-axis, we can integrate over all $x$ and $y$ without changing the result. Thus, we obtain

\begin{displaymath}

\int_S \nabla\times{\bf B} \cdot d{\bf S} = \mu_0 I \int_{…

…elta(x) dx\int_{-\infty}^{\infty}

\delta(y) dy = \mu_0 I,

\end{displaymath} (266)

which is in agreement with Eq. (263).

But, why have we gone to so much trouble to prove something using vector field theory which can be demonstrated in one line via conventional analysis [see Eq. (258)]? The answer, of course, is that the vector field result is easily generalized, whereas the conventional result is just a special case. For instance, it is clear that Eq. (266) is true for any surface attached to the loop C, not just a plane surface. Moreover, suppose that we distort our simple circular loop $C$ so that it is no longer circular or even lies in one plane. What now is the line integral of ${\bf B}$ around the loop? This is no longer a simple problem for conventional analysis, because the magnetic field is not parallel to a line element of the loop. However, according to Stokes’ theorem,

\begin{displaymath}

\oint_C {\bf B} \cdot d{\bf l} = \int_S \nabla\times{\bf B} \cdot d{\bf S},

\end{displaymath} (267)

with $\nabla\times {\bf B} $ given by Eq. (264). Note that the only part of $S$ which contributes to the surface integral is an infinitesimal region centered on the $z$-axis. So, as long as $S$ actually intersects the $z$-axis, it does not matter what shape the rest the surface is, and we always get the same answer for the surface integral: namely,

\begin{displaymath}

\oint_C {\bf B}\cdot d{\bf l} = \int_S \nabla\times{\bf B} \cdot d{\bf S}

=\mu_0 I.

\end{displaymath} (268)

Thus, provided the curve $C$ circulates the $z$-axis, and, therefore, any surface $S$ attached to $C$ intersects the $z$-axis, the line integral $\oint_C {\bf B} \cdot d{\bf l}$ is equal to $\mu_0 I$. Of course, if $C$ does not circulate the $z$-axis then an attached surface $S$ does not intersect the $z$-axis and $\oint_C {\bf B} \cdot d{\bf l}$ is zero. There is one more proviso. The line integral $\oint_C {\bf B} \cdot d{\bf l}$ is $\mu_0 I$ for a loop which circulates the $z$-axis in a clockwise direction (looking up the $z$-axis). However, if the loop circulates in an anti-clockwise direction then the integral is $-\mu_0 I$. This follows because in the latter case the $z$-component of the surface element $d{\bf S}$ is oppositely directed to the current flow at the point where the surface intersects the wire.

Let us now consider $N$ wires directed along the $z$-axis, with coordinates ($x_i$, $y_i$) in the $x$-$y$ plane, each carrying a current $I_i$ in the positive $z$-direction. It is fairly obvious that Eq. (264) generalizes to

\begin{displaymath}

\nabla\times{\bf B} = \mu_0\sum_{i=1}^N I_i \delta(x-x_i) \delta(y-y_i)

\hat{\bf z}.

\end{displaymath} (269)

If we integrate the magnetic field around some closed curve $C$, which can have any shape and does not necessarily lie in one plane, then Stokes’ theorem and the above equation imply that

\begin{displaymath}

\oint_C {\bf B}\cdot d{\bf l} = \int_S \nabla\times{\bf B} \cdot d{\bf S}

= \mu_0 {\cal I},

\end{displaymath} (270)

where ${\cal I}$ is the total current enclosed by the curve. Again, if the curve circulates the $i$th wire in a clockwise direction (looking down the direction of current flow) then the wire contributes $I_i$ to the aggregate current ${\cal I}$. On the other hand, if the curve circulates in an anti-clockwise direction then the wire contributes $-I_i$. Finally, if the curve does not circulate the wire at all then the wire contributes nothing to ${\cal I}$.

Equation (269) is a field equation describing how a set of $z$-directed current carrying wires generate a magnetic field. These wires have zero-thickness, which implies that we are trying to squeeze a finite amount of current into an infinitesimal region. This accounts for the delta-functions on the right-hand side of the equation. Likewise, we obtained delta-functions in Sect. 3.4 because we were dealing with point charges. Let us now generalize to the more realistic case of diffuse currents. Suppose that the $z$-current flowing through a small rectangle in the $x$-$y$ plane, centred on coordinates ($x$, $y$) and of dimensions $dx$ and $dy$, is $j_z(x,y) dx dy$. Here, $j_z$ is termed the current density in the $z$-direction. Let us integrate $(\nabla\times{\bf B})_z$ over this rectangle. The rectangle is assumed to be sufficiently small that $(\nabla\times{\bf B})_z$ does not vary appreciably across it. According to Eq. (270), this integral is equal to $\mu_0$ times the total $z$-current flowing through the rectangle. Thus,

\begin{displaymath}

(\nabla\times{\bf B})_z dx dy = \mu_0 j_z dx dy,

\end{displaymath} (271)

which implies that

\begin{displaymath}

(\nabla\times{\bf B})_z = \mu_0 j_z.

\end{displaymath} (272)

Of course, there is nothing special about the $z$-axis. Suppose we have a set of diffuse currents flowing in the $x$-direction. The current flowing through a small rectangle in the $y$-$z$ plane, centred on coordinates ($y$, $z$) and of dimensions $dy$ and $dz$, is given by $j_x(y,z) dy dz$, where $j_x$ is the current density in the $x$-direction. It is fairly obvious that we can write

\begin{displaymath}

(\nabla\times{\bf B})_x = \mu_0 j_x,

\end{displaymath} (273)

with a similar equation for diffuse currents flowing along the $y$-axis. We can combine these equations with Eq. (272) to form a single vector field equation which describes how electric currents generate magnetic fields,

\begin{displaymath}

\nabla\times{\bf B} = \mu_0 {\bf j},

\end{displaymath} (274)

where ${\bf j} = (j_x, j_y, j_z)$ is the vector current density. This is the third Maxwell equation. The electric current flowing through a small area $d{\bf S}$ located at position ${\bf r}$ is ${\bf j}({\bf r})

\cdot d{\bf S}$. Suppose that space is filled with particles of charge $q$, number density $n({\bf r})$, and velocity ${\bf v}({\bf r})$. The charge density is given by $\rho({\bf r})= q n$. The current density is given by ${\bf j}({\bf r})

= q n {\bf v}$, and is obviously a proper vector field (velocities are proper vectors since they are ultimately derived from displacements).

If we form the line integral of ${\bf B}$ around some general closed curve $C$, making use of Stokes’ theorem and the field equation (274), then we obtain

\begin{displaymath}

\oint_C {\bf B} \cdot d{\bf l} = \mu_0 \int_S {\bf j} \cdot d{\bf S}.

\end{displaymath} (275)

In other words, the line integral of the magnetic field around any closed loop $C$ is equal to $\mu_0$ times the flux of the current density through $C$. This result is called Ampère’s circuital law. If the currents flow in zero-thickness wires then Ampère’s circuital law reduces to Eq. (270).

The flux of the current density through $C$ is evaluated by integrating ${\bf j}\cdot d{\bf S}$ over any surface $S$ attached to $C$. Suppose that we take two different surfaces $S_1$ and $S_2$. It is clear that if Ampère’s circuital law is to make any sense then the surface integral $\int_{S_1} {\bf j}

\cdot d{\bf S} $ had better equal the integral $\int_{S_2} {\bf j}

\cdot d{\bf S} $. That is, when we work out the flux of the current though $C$ using two different attached surfaces then we had better get the same answer, otherwise Eq. (275) is wrong (since the left-hand side is clearly independent of the surface spanning C). We saw in Sect. 2 that if the integral of a vector field ${\bf A}$ over some surface attached to a loop depends only on the loop, and is independent of the surface which spans it, then this implies that $\nabla\cdot {\bf A} = 0$. The flux of the current density through any loop $C$ is calculated by evaluating the integral $\int_S {\bf j} \cdot d{\bf S} $ for any surface $S$ which spans the loop. According to Ampère’s circuital law, this integral depends only on $C$ and is completely independent of $S$ (i.e., it is equal to the line integral of ${\bf B}$ around $C$, which depends on $C$ but not on $S$). This implies that $\nabla\cdot{\bf j} = 0$. In fact, we can obtain this relation directly from the field equation (274). We know that the divergence of a curl is automatically zero, so taking the divergence of Eq. (274), we obtain

\begin{displaymath}

\nabla\cdot{\bf j} = 0.

\end{displaymath} (276)

We have shown that if Ampère’s circuital law is to make any sense then we need $\nabla\cdot{\bf j} = 0$. Physically, this implies that the net current flowing through any closed surface $S$ is zero. Up to now, we have only considered stationary charges and steady currents. It is clear that if all charges are stationary and all currents are steady then there can be no net current flowing through a closed surface $S$, since this would imply a build up of charge in the volume $V$ enclosed by $S$. In other words, as long as we restrict our investigation to stationary charges, and steady currents, then we expect $\nabla\cdot{\bf j} = 0$, and Ampère’s circuital law makes sense. However, suppose that we now relax this restriction. Suppose that some of the charges in a volume $V$ decide to move outside $V$. Clearly, there will be a non-zero net flux of electric current through the bounding surface $S$ whilst this is happening. This implies from Gauss’ theorem that $\nabla\cdot{\bf j} \neq 0$. Under these circumstances Ampère’s circuital law collapses in a heap. We shall see later that we can rescue Ampère’s circuital law by adding an extra term involving a time derivative to the right-hand side of the field equation (274). For steady-state situations (i.e., $\partial/\partial t=0$), this extra term can be neglected. Thus, the field equation $\nabla\times{\bf B} = \mu_0 {\bf j}$ is, in fact, only two-thirds of Maxwell’s third equation: there is a term missing on the right-hand side.

We have now derived two field equations involving magnetic fields (actually, we have only derived one and two-thirds):

$\displaystyle \nabla\cdot{\bf B}$ $\textstyle =$ $\displaystyle 0,$ (277)

$\displaystyle \nabla\times{\bf B}$ $\textstyle =$ $\displaystyle \mu_0 {\bf j}.$ (278)

We obtained these equations by looking at the fields generated by infinitely long, straight, steady currents. This, of course, is a rather special class of currents. We should now go back and repeat the process for general currents. In fact, if we did this we would find that the above field equations still hold (provided that the currents are steady). Unfortunately, this demonstration is rather messy and extremely tedious. There is a better approach. Let us assume that the above field equations are valid for any set of steady currents. We can then, with relatively little effort, use these equations to generate the correct formula for the magnetic field induced by a general set of steady currents, thus proving that our assumption is correct. More of this later.

**PUERTOGALERA**#5

The calculation of the magnetic field of a current distribution can, in principle, be carried out using Ampere’s law which relates the path integral of the magnetic field around a closed path to the current intercepted by an arbitrary surface that spans this path:

(35.1)

Ampere’s law is independent of the shape of the surface chosen as long as the current flows along a continuous, unbroken circuit. However, consider the case in which the current wire is broken and connected to a parallel-plate capacitor (see Figure 35.1). A current will flow through the wire during the charging process of the capacitor. This current will generate a magnetic field and if we are far away from the capacitor, this field should be very similar to the magnetic field produced by an infinitely long, continuous, wire. However, the current intercepted by an arbitrary surface now depends on the surface chosen. For example, the surface shown in Figure 35.1 does not intercept any current. Clearly, Ampere’s law can not be applied in this case to find the magnetic field generated by the current.

Figure 35.1. Ampere’s law in a capacitor circuit.

Although the surface shown in Figure 35.1 does not intercept any current, it intercepts electric flux. Suppose the capacitor is an ideal capacitor, with a homogeneous electric field E between the plates and no electric field outside the plates. At a certain time t the charge on the capacitor plates is Q. If the plates have a surface area A then the electric field between the plates is equal to

(35.2)

The electric field outside the capacitor is equal to zero. The electric flux, [Phi]E, intercepted by the surface shown in Figure 35.1 is equal to

(35.3)

If a current I is flowing through the wire, then the charge on the capacitor plates will be time dependent. The electric flux will therefore also be time dependent, and the rate of change of electric flux is equal to

(35.4)

The magnetic field around the wire can now be found by modifying Ampere’s law

(35.5)

where [Phi]E is the electric flux through the surface indicated in Figure 35.1 In the most general case, the surface spanned by the integration path of the magnetic field can intercept current and electric flux. In such a case, the effects of the electric flux and the electric current must be combined, and Ampere’s law becomes

(35.6)

The current I is the current intercepted by whatever surface is used in the calculation, and is not necessarily the same as the current in the wires. Equation (35.6) is frequently written as

(35.7)

where Id is called the displacement current and is defined as

(35.8)

Example: Problem 35.8

A parallel-plate capacitor has circular plates of area A separated by a distance d. A thin straight wire of length d lies along the axis of the capacitor and connects the two plates. This wire has a resistance R. The exterior terminals of the plates are connected to a source of alternating emf with a voltage V = V0 sin([omega] t).

a) What is the current in the thin wire ?

b) What is the displacement current through the capacitor ?

c) What is the current arriving at the outside terminals of the capacitor ?

d) What is the magnetic field between the capacitor plates at a distance r from the axis ? Assume that r is less than the radius of the plates.

a) The setup can be regarded as a parallel circuit of a resistor with resistance R and a capacitor with capacitance C (see Figure 35.2). The current in the thin wire can be obtained using Ohm’s law

(35.9)

Figure 35.2. Circuit Problem 35.8.

b) The voltage across the capacitor is equal to the external emf. The electric field between the capacitor plates is therefore equal to

(35.10)

The electric flux through the capacitor is therefore equal to

(35.11)

The displacement current Id can be obtained by substituting eq.(35.11) into eq.(35.8)

(35.12)

The current at the outside terminals of the capacitor is the sum of the current used to charge the capacitor and the current through the resistor. The charge on the capacitor is equal to

(35.13)

The charging current is thus equal to

(35.14)

The total current is therefore equal to

(35.15)

d) The magnetic field lines inside the capacitor will form concentric circles, centered around the resistor (see Figure 35.3). The path integral of the magnetic field around a circle of radius r is equal to

(35.16)

Figure 35.3. Amperian loop used to determine the magnetic field inside a capacitor.

The surface to be used to determine the current and electric flux intercepted is the disk of radius r shown Figure 35.3. The electric flux through this disk is equal to

(35.17)

The displacement current intercepted by this surface is equal to

(35.18)

The current intercepted by the surface is equal to the current through the resistor (eq.(35.9)). Ampere’s law thus requires

(35.19)

The strength of the magnetic field is thus equal to

(35.20)

35.2. Maxwells Equations

The fundamental equations describing the behavior of electric and magnetic fields are known as the Maxwell equations. They are

(35.21)

(35.22)

(35.23)

(35.24)

Maxwell’s equations provide a complete description of the interactions among charges, currents, electric fields, and magnetic fields. All the properties of the fields can be obtained by mathematical manipulations of these equations. If the distribution of charges and currents is given, than these equations uniquely determine the corresponding fields.

Example: Problem 35.10

Prove that Maxwell’s equations mathematically imply the conservation of electric charge; that is, prove that if no electric current flows into or out a given volume, then the electric charge within this volume remains constant.

Equation (35.21) shows that the enclosed charge Q is related to the electric flux [Phi]E:

(35.25)

The rate of change of the enclosed charge can be determined by differentiating eq.(35.25) with respect to time

(35.26)

The closed surface used in eq.(35.24) to determine the flux and current intercepted can be replaced by a bag whose mouth has shrunk to zero. The path integral of the magnetic field along the mouth is therefore equal to zero, and eq.(35.24) can be written as

(35.27)

Using eq.(35.26) we can rewrite eq.(35.27) as

(35.28)

In other words, if no current flows in or out of the enclosed volume (I = 0) then the electric charge within this volume will remain constant. This implies conservation of charge.

35.3. Cavity Oscillations

The electric field between the plates of a parallel-plate capacitor is determined by the external emf. If the distance between the plates is d (see Figure 35.4) then the electric field between the plates is equal to

(35.29)

This time-dependent electric field will induce a magnetic field with a strength that can be obtained via Ampere’s law. Consider a circular Amperian loop of radius r. The path integral of the magnetic field around this loop is equal to

(35.30)

The electric flux though the surface spanned by this path is equal to

(35.31)

Figure 35.4. The oscillating parallel-plate capacitor.

The displacement current is thus equal to

(35.32)

Using Ampere’s law we obtain for the magnetic field

(35.33)

This time-dependent magnetic field will induce an electric field. The total electric field inside the capacitor will therefore be the sum of the constant electric field generated by the source of emf and the induced electric field, generated by the time-dependent magnetic field. The strength of the induced electric field can be calculated using Faraday’s law of induction. Consider the closed path indicated in Figure 35.4. We take the induced electric field on the capacitor axis equal to zero. The path integral of the induced electric field along the path indicated is then equal to

(35.34)

where Eind is reckoned to be positive if it is directed upwards. The magnetic flux through the surface spanned by the loop indicated in Figure 35.4 is equal to

(35.35)

Thus

(35.36)

The induced electric field, Eind, can be obtained from Faraday’s law of induction (eq.(35.23)) and is equal to

(35.37)

The total electric field is therefore equal to

(35.38)

But the addition of the induced field implies that a correction needs to be made to the magnetic field calculated before (eq.(35.33)). This in turn will modify the induced current and this process will go on forever. If we neglect the additional corrections, then eq.(35.38) shows that the electric field vanishes at a radius R if

(35.39)

or

(35.40)

If we create a cavity by enclosing the capacitor with a conducting cylinder of radius R then eq.(35.40) can be used to determine the frequency of the driving emf that will produce a standing wave. This frequency is called the resonance frequency and it is equal to

(35.41)

For a cavity with R = 0.5 m the resonance frequency is 1.2 GHz. Electromagnetic radiation with a frequency in this range is called microwave radiation, and the cavity is called a microwave oven.

35.4. The Electric Field of an Accelerated Charge

The electric field produced by a stationary charge is stationary. When the charge accelerates, it produces extra electric and magnetic fields that travel outward from the position of the charge. These radiation fields are called electromagnetic waves. They travel with the speed of light (in vacuum) and carry energy and momentum away from the charge. Their properties are determined by the properties of the accelerated charge, and in this manner provide a means to transmit information at the speed of light over long distances.

Consider a charge q initially at rest (for t < 0). Between t = 0 and t = [tau], the charge accelerates with an acceleration a. After t = [tau] the charge moves with a constant velocity (v = a[tau]). We will assume that the final velocity of the charge is small compared to the speed of light (v << c) and that the time period [tau] during which the charge accelerates is short. The electric field generated by the charge at a time t > [tau] consists of three separate regions (see Figure 35.5). In the region r > ct the field lines will be that of a point charge at rest at the origin (electromagnetic waves travel with the speed of light, and the region with r > ct can not know yet that the charge has moved away from the origin). In the spherical region with radius r < c(t - [tau]), centered at x = vt, the electric field will that of a uniformly moving charge. The disturbance produced by the accelerated charge is confined to the region between these two spheres and the effect of the acceleration is a kink in the field lines. The electric field in this region has two components: a radial component and a transverse component.

The radial component is determined by Gauss’ law. Consider a spherical Gaussian surface located between the two spheres shown in Figure 35.5. The charge enclosed by this surface is equal to q. The electric flux through this surface depends only on the radial component of the field. Applying Gauss’ law we conclude that the radial component of the electric field is simply the usual Coulomb field

(35.42)

Figure 35.5. Electric field lines generated by an accelerated charge.

The relation between the radial component of the electric field and the transverse component of the electric field can be determined by carefully examining one field line (see Figure 35.6). The field line shown in Figure 35.6 makes an angle [theta] with the direction of the moving charge. The ratio between the magnitude of the transverse electric field and the radial electric field is equal to

(35.43)

Since the radial field is known, eq.(35.42), we can use eq.(35.43) to determine the transverse component of the electric field:

(35.44)

The distance r at which the kink occurs is related to the time t at which we look at the field:

(35.45)

Eliminating the dependence on t in eq.(35.44) we obtain the following expression for the transverse component of the electric field:

(35.46)

Figure 35.6. Calculation of transverse electric field.

Equation (35.46) shows that the transverse electric field is directly proportional to the acceleration a and inversely proportional to the distance r. The Coulomb component of the field falls of as 1/r2. This shows that the transverse component, also called the radiation field, remains significant at distances where the Coulomb field practically disappears.

The equation for the transverse electric field (eq.(35.46)) is valid in general, even if the acceleration is not constant. If the charge oscillates back and forth with a simple harmonic motion of frequency [omega], then the acceleration at time t will be equal to

(35.47)

In order to determine the radiation field at a time t and a distance r, we have to realize that the acceleration a used in eq.(35.46) should be the acceleration at time t - r/c, where r/c is the time required for a signal to travel over a distance r. The radiation field for the oscillating charge is therefore equal to

(35.48)

Example: Problem 35.25

On a radio antenna (a straight piece of wire), electrons move back and forth in unison. Suppose that the velocity of the electrons is v = v0 cos([omega] t), where v0 = 8.0 x 10 -3 m/s and [omega] = 6.0 x 10 6 rad/s.

a) What is the maximum acceleration of the electrons ?

b) Corresponding to this maximum acceleration, what is the strength of the transverse electric field produced by one electron at a distance of 1.0 km fro the antenna in a direction perpendicular to the antenna ? What is the time delay (or retardation) between the instant of maximum acceleration and the instant at which the corresponding electric field reaches a distance of 1.0 km ?

c) There are 2.0 x 10 24 electrons on the antenna. What is the collective electric field produced by all electrons acting together ? Assume the antenna is sufficiently small so that all electrons contribute just about the same electric field at a distance of 1.0 km.

a) The acceleration of the electrons can be obtained by differentiating their velocity with respect to time:

(35.49)

The maximum acceleration is thus equal to

(35.50)

b) The maximum transverse electric field at a distance of 1.0 km (= 1000 m), in a direction perpendicular to the antenna ([theta] = 90deg.), can be obtained using eq.(35.48):

(35.51)

Since the propagation speed of the radiation field is equal to the speed of light, c, the maxima in the radiation field will occur a time period [Delta]t after the maxima in the acceleration of the electron. The length of this period, [Delta]t, is equal to

(35.52)

c) Assuming that all the electrons are in phase, then the maximum total transverse electric field at a distance of 1.0 km is equal to the number of electrons times the maximum transverse electric field produced by a single electron. Thus

(35.53)

35.5. The magnetic field of an accelerated charge

When a charge accelerates from rest it will produce a magnetic field. Initially, the magnetic field will be equal to zero (charge at rest). As a result of the acceleration a disturbance will move outward and change the magnetic field from its initial value (B = 0 T) to its final value, in much the same way as we observed for the electric field. The magnetic field can be obtained from the electric flux via the Maxwell-Ampere law which states that

(35.54)

Note that the current I does not appear in eq. (35.54). Since we are looking in the region away from the moving charge, the current intercepted by a surface spanned by the path used to evaluate the path integral of B is equal to zero. The induced magnetic field will be time-dependent and, therefore, will induce an electric field via Faraday’s law. This induced electric field will again be time dependent and induce another magnetic field, and this process continues. The combined electric and magnetic radiation fields produced by the accelerating charge are called electromagnetic waves. They are self-supporting; the electric field induces a magnetic field, and the induced magnetic field induces an electric field. Because the electric and magnetic fields naturally support each other, the electromagnetic wave does not require a medium for its propagation, and it readily propagates in vacuum. The Maxwell equations can be used to show that the product of u0 and [epsilon]0 is equal to 1/c2. Or,

(35.55)

This equation was one of the great and early triumphs of Maxwell’s electromagnetic theory of light. It shows that electricity and magnetism are two different aspects of the same phenomena.