 1.
The Coulomb field (5.3) is a special case of a radial field of the form \( \overrightarrow{E}\left(\overrightarrow{r}\right)=E(r)\;\hat{r} \), where \( r=\mid \overrightarrow{r}\mid \) and \( \hat{r}=\overrightarrow{r}/r \) (the position vector \( \overrightarrow{r} \) is taken relative to the location of the point charge that produces the field; by \( \hat{r} \) we denote the unit vector in the direction of \( \overrightarrow{r} \)). Show that no radial magnetic field of the form \( \overrightarrow{B}\kern0.1em \left(\overrightarrow{r}\right)=B\kern0.1em (r)\;\hat{r} \) may exist. What is the physical significance of this?

Solution: Consider a spherical surface S of radius r, centered at some fixed reference point Ο (Fig. 7.5). Let \( \overrightarrow{B} \) be the magnetic field vector at some point P of S. The position vector of P with respect to Ο is \( \overrightarrow{r}=\overrightarrow{OP} \). By Gauss’ law for Magnetism, the total magnetic flux through S is zero:
$$ {\oint}_S\overrightarrow{B}\cdot \overrightarrow{da}=0\kern0.48em \Rightarrow \kern0.48em {\oint}_S\left[B(r)\;\hat{r}\right]\cdot \left[(da)\;\hat{r}\right]=0\kern0.48em \Rightarrow \kern0.48em {\oint}_SB(r)\; da=0 $$
or, since
B(
r) has a constant value on
S,
$$ B(r)\kern0.24em {\oint}_S da=0\kern0.48em \Rightarrow \kern0.48em B(r)\;\left(4\pi {r}^2\right)=0\kern0.48em \Rightarrow \kern0.48em B(r)=0\kern0.48em \Rightarrow \kern0.48em \overrightarrow{B}=0 $$
Therefore the assumption
\( \overrightarrow{B}\kern0.1em \left(\overrightarrow{r}\right)=B\kern0.1em (r)\;\hat{r} \) cannot hold for
\( \overrightarrow{B}\ne 0 \). Indeed, if a magnetic field of this form existed, the magnetic flux through
S would not vanish, which would indicate the presence of an isolated magnetic “charge” at the point
Ο at which the radial magnetic field lines begin. But, as we know, no such magnetic poles exist!
 2.
Show that a nonzero total current passes through any (closed) field line of a static magnetic field.
Solution: Let C be a field line of \( \overrightarrow{B} \). At each point of C the vector \( \overrightarrow{B} \) is tangential, having the direction of the element \( \overrightarrow{d\kern0em l} \) of C. Hence,
$$ {\oint}_C\overrightarrow{B}\cdot \overrightarrow{d\kern0em l}={\oint}_C\mid \overrightarrow{B}\mid \mid \overrightarrow{d\kern0em l}\mid \kern0.36em >0 $$
(1)
On the other hand, by Ampère’s law,
$$ \kern0.22em {\oint}_C\overrightarrow{B}\cdot \overrightarrow{d\kern0em l}={\mu}_0\kern0.2em {I}_{in}\kern0.22em $$
(2)
From (
1) and (
2) it follows that
I_{in} ≠ 0 .
 3.
A long, thin, hollow metal cylinder of radius R carries a constant current Ι that runs parallel to the cylinder’s axis. Determine the magnetic field both inside and outside the cylinder.
Solution: We will first take a look at a simpler and more fundamental problem: the magnetic field produced by a long rectilinear current Ι (Fig. 7.6). As found by the BiotSavart law, the magnetic field lines are circular, each circle having its center on the axis of the current and belonging to a plane normal to that axis. The magnetic field \( \overrightarrow{B} \) is tangential at every point of a field line, its direction being determined by the righthand rule; that is, if we rotate the fingers of our right hand in the direction of \( \overrightarrow{B} \), our extended thumb will point in the direction of Ι.
By symmetry, the magnitude
B of
\( \overrightarrow{B} \) is constant along a field line. If
r is the radius of this line, it can be shown that
$$ \boxed{\kern0.22em B\kern0.2em (r)=\frac{\mu_0\kern0.1em I}{2\pi r}\kern0.22em } $$
(1)
By integrating
\( \overrightarrow{B} \) on a field line
C we thus recover Ampère’s law:
$$ \kern0.22em {\oint}_C\overrightarrow{B}\cdot \overrightarrow{dl}={\oint}_C\mid \overrightarrow{B}\mid \kern0.3em \mid \overrightarrow{dl}\mid ={\oint}_CB\; dl=B\kern0.24em {\oint}_C dl=B\;\left(2\pi r\right)={\mu}_0\kern0.2em I={\mu}_0\kern0.2em {I}_{in}\kern0.22em $$
where we have used the fact that
B is constant on
C, equal to
B(
r).
Let us now return to our original problem. Consider a normal crosssection of the cylinder, as seen in Fig.
7.7. Imagine that the axis of the cylinder is normal to the page and that the current
Ι is directed
outward, i.e., toward the reader. Since the problem has the same symmetry as the problem of the long rectilinear current (
cylindrical symmetry), the field lines of the magnetic field
\( \overrightarrow{B} \) both inside and outside the cylinder will be circles normal to the cylinder and centered on its axis, while the magnitude
B of
\( \overrightarrow{B} \) will be constant along a field line. Let
C be a field line of radius
r > R, i.e., a line external to the cylinder. By applying Ampère’s law on the closed path
C and by noticing that
I_{in} = I, we have:
$$ \kern0.22em {\oint}_C\overrightarrow{B}\cdot \overrightarrow{dl}={\mu}_0\kern0.2em {I}_{in}\kern0.48em \Rightarrow \kern0.48em {\mu}_0\kern0.2em I={\oint}_C\mid \overrightarrow{B}\mid \kern0.3em \mid \overrightarrow{dl}\mid ={\oint}_CB\; dl=B\kern0.24em {\oint}_C dl=B\;\left(2\pi r\right)\kern0.48em \Rightarrow $$
$$ \kern0.22em B\kern0.2em (r)=\frac{\mu_0\kern0.1em I}{2\pi r}\kern0.46em ,\kern0.6em r>R $$
(2)
By comparing (
2) with (
1) we notice that the magnetic field in the
exterior of the cylinder is the same as that of a hypothetical rectilinear current
Ι flowing along the cylinder’s axis! Now, to find the magnetic field in the
interior of the cylinder, we apply Ampère’s law on a loop
C΄ with
r < R, noticing that
I_{in} = 0 in this case:
$$ \kern0.22em {\oint}_{C\prime}\overrightarrow{B}\cdot \overrightarrow{dl}={\mu}_0\kern0.2em {I}_{in}\kern0.48em \Rightarrow \kern0.48em 0={\oint}_{C\prime}\mid \overrightarrow{B}\mid \kern0.3em \mid \overrightarrow{dl}\mid ={\oint}_{C\prime }B\; dl=B\kern0.24em {\oint}_{C\prime } dl=B\;\left(2\pi r\right)\kern0.48em \Rightarrow $$
$$ B(r)=0,\kern0.5em r<R $$
That is, there is no magnetic field inside the cylinder.
 4.
By using Gauss’ law for the magnetic field, show that an equal magnetic flux passes through any two open surfaces S_{1}, S_{2} having a common border C.
Solution: The magnetic flux through a surface S is defined as the surface integral of the magnetic field \( \overrightarrow{B} \) over S. According to Gauss’ law, the field \( \overrightarrow{B} \) is solenoidal. Now, as discussed in Sect. 4.3, the value of the surface integral of a solenoidal field, over an open surface S bordered by a closed curve C, depends only on the border C of S and is the same for any two open surfaces sharing a common border C. This is exactly what we needed to prove.
It is instructive, however, to view an alternative proof of the above property. Let
S be a
closed surface (Fig.
7.8). According to Gauss’ law,
$$ {\oint}_S\overrightarrow{B}\cdot \overrightarrow{da}=0 $$
(1)
The surface element
\( \overrightarrow{da} \) in the above integral is always directed
outward relative to
S, at all points of this surface. Imagine now that we partition
S into two
open surfaces
S_{1} and
S_{2} by drawing a closed curve
C on
S. Clearly, these two surfaces have a common border
C.
Equation (
1) is written
$$ {\oint}_S\overrightarrow{B}\cdot \overrightarrow{da}={\int}_{S_1}\overrightarrow{B}\cdot \overrightarrow{da}+{\int}_{S_2}\overrightarrow{B}\cdot \overrightarrow{d{a}^{\prime }}=0 $$
(2)
where both
\( \overrightarrow{da} \) and
\( \overrightarrow{d{a}^{\prime }} \) point
outward with respect to
S. Now, if we assign a positive direction of traversing the closed curve
C, only one of the above surface elements, say
\( \overrightarrow{da} \), will be oriented consistently with the direction of
C according to the righthand rule, while the orientation of
\( \overrightarrow{d{a}^{\prime }} \) will be inconsistent with that of
C. The element
\( \overrightarrow{d{a}^{\prime }} \), however, will be properly oriented. By setting
\( \overrightarrow{da_1}\equiv \overrightarrow{da} \) and
\( \overrightarrow{da_2}\equiv \overrightarrow{d{a}^{\prime }} \), relation (
2) is finally rewritten as.
$$ {\int}_{S_1}\overrightarrow{B}\cdot \overrightarrow{da_1}{\int}_{S_2}\overrightarrow{B}\cdot \overrightarrow{da_2}=0\kern0.48em \iff \kern0.48em {\int}_{S_1}\overrightarrow{B}\cdot \overrightarrow{da_1}={\int}_{S_2}\overrightarrow{B}\cdot \overrightarrow{da_2};\kern0.5em \mathrm{Q}.\mathrm{E}.\mathrm{D}. $$
 5.
Consider a region of space in which the distribution of charge is static. Let C be a closed curve in this region and let S be any open surface bordered by C. We define the total current through C as the surface integral of the current density over S:
$$ {I}_{in}={\int}_S\overrightarrow{J}\cdot \overrightarrow{da} $$
Show that, for a given
C, the quantity
I_{in} has a welldefined value independent of the particular choice of
S.
Solution: Since the charge density ρ inside the region is static (∂ρ/∂t = 0), the equation of continuity, Eq. (6.9), reduces to \( \overrightarrow{\nabla}\cdot \overrightarrow{J}=0 \). This means that, within the considered region of space, the current density has the properties of a solenoidal field. In particular, the value of the surface integral of \( \overrightarrow{J} \) over an open surface S depends only on the border C of S and is the same for all surfaces sharing a common border C.
Exercise: Suggest an alternative proof, in the spirit of Prob. 4.