$9.3.11.$ a. A solid cylinder is cut from iron magnetized to saturation so that its axis coincides with the direction
+
of magnetization. Prove that the magnetic field of this cylinder is equivalent to the field of a transverse
+
current flowing over its surface, the linear density of which is equal to the magnetic moment of a unit
+
volume of iron.
+
b. A cube was cut from a long rod magnetized to saturation along the axis so that one of the edges of
+
the cube was directed along the direction of magnetization. How many times will the induction of the
+
magnetic field in the center of the cube be less than the induction in the rod?
+
+
c. Determine the magnetic field induction in the center of a cylinder of length $l$ and radius $r$. The
+
magnetic moment of a unit volume of iron is equal to $M$. What is this induction equal to for $r ≪ l$ ? for
+
$r ≫ l$?
+
+
d. Solve the previous problem if a small-radius hole is drilled along the axis of the cylinder
+
### Solution
$а)\$Let us prove that the current in a cylinder magnetized to saturation occurs only on its outer surface:
$$j=\Delta\times M$$$$i=\hat{n}\times M$$ где $\hat{n}$ -
normal vector to the surface.
Because the conductor is magnetized to saturation the rotor of $M=0$ .
At the ends the vector product vanishes, so there is no current there either. $\Rightarrow$
the current is only at the ends and according to the vector product rule its linear density is equal to $i=M$
$b) \$
The difference in induction arises from the difference in solid angles at which current-carrying surfaces are visible. Similar to point a, currents arise only on the outer surface and are absent at the ends and inside the conductors. In the case of an infinite stub, the solid angle is $\Omega = 4\pi$.
In the case of a square $\Omega' = \frac{8\pi}{3}$ therefore the induction is $n=1.5$ times less.
$c)\$The solid angle at which the outer surface along which the current flows is visible is equal to $$\Omega = 4\pi - 2\cdot2\pi (1-cos\alpha)=4\pi cos\alpha$$ where $cos\alpha = \frac{1}{\sqrt{4\frac{r^2}{l^2}+1}}$
So:
$$B = \frac{\mu_0M}{\sqrt{1+4\frac{r^2}{l^2}}}$$
$r/l \to 0$:
$$B \to\mu_0 M$$
$l/r \to 0$:
$$B \to\frac{\mu_0 Ml}{2r}$$
$d)\$
Let us imagine the field of the past cylinder as a superposition of the fields of the thin cylinder in the center and the rest of it. $$\\hat{B}_{thin}+\hat{B}_{rest}=\hat{B}_{full}$$
$a) \ $ The magnetic field of the cylinder is composed of the magnetic fields of the thin disks
+
of thickness $∆$, into which the cylinder can be divided. The magnetic field of each disk coincides with the magnetic field of a current flowing with a linear density $M$ ($M$ is the magnetic moment
+
per unit volume of iron); along the outer surface of the disk (see the solution to problem 9.2.19∗).
+
+
$b) \ $The direction of the magnetic field induction at the center of the cube coincides with the direction
+
of magnetization. The magnitude of this vector will be smaller than the magnitude of the magnetic field induction inside the rod by as many times as $8π/3$ (the solid angle subtended by the lateral
+
faces of cube 1–4) is less than $4π$, i.e., $n = 1.5$ times.
$9.3.11.$ a. A solid cylinder is cut from iron magnetized to saturation so that its axis coincides with the direction
of magnetization. Prove that the magnetic field of this cylinder is equivalent to the field of a transverse
current flowing over its surface, the linear density of which is equal to the magnetic moment of a unit
volume of iron.
b. A cube was cut from a long rod magnetized to saturation along the axis so that one of the edges of
the cube was directed along the direction of magnetization. How many times will the induction of the
magnetic field in the center of the cube be less than the induction in the rod?
c. Determine the magnetic field induction in the center of a cylinder of length $l$ and radius $r$. The
magnetic moment of a unit volume of iron is equal to $M$. What is this induction equal to for $r ≪ l$ ? for
$r ≫ l$?
d. Solve the previous problem if a small-radius hole is drilled along the axis of the cylinder
### Solution
### Solution
$а)\$Let us prove that the current in a cylinder magnetized to saturation occurs only on its outer surface:
$а)\$Let us prove that the current in a cylinder magnetized to saturation occurs only on its outer surface:
$$j=\Delta\times M$$$$i=\hat{n}\times M$$ где $\hat{n}$ -
$$j=\Delta\times M$$$$i=\hat{n}\times M$$ где $\hat{n}$ -
normal vector to the surface.
normal vector to the surface.
Because the conductor is magnetized to saturation the rotor of $M=0$ .
Because the conductor is magnetized to saturation the rotor of $M=0$ .
At the ends the vector product vanishes, so there is no current there either. $\Rightarrow$
At the ends the vector product vanishes, so there is no current there either. $\Rightarrow$
the current is only at the ends and according to the vector product rule its linear density is equal to $i=M$
the current is only at the ends and according to the vector product rule its linear density is equal to $i=M$
$b) \$
$b) \$
The difference in induction arises from the difference in solid angles at which current-carrying surfaces are visible. Similar to point a, currents arise only on the outer surface and are absent at the ends and inside the conductors. In the case of an infinite stub, the solid angle is $\Omega = 4\pi$.
The difference in induction arises from the difference in solid angles at which current-carrying surfaces are visible. Similar to point a, currents arise only on the outer surface and are absent at the ends and inside the conductors. In the case of an infinite stub, the solid angle is $\Omega = 4\pi$.
In the case of a square $\Omega' = \frac{8\pi}{3}$ therefore the induction is $n=1.5$ times less.
In the case of a square $\Omega' = \frac{8\pi}{3}$ therefore the induction is $n=1.5$ times less.
$c)\$The solid angle at which the outer surface along which the current flows is visible is equal to $$\Omega = 4\pi - 2\cdot2\pi (1-cos\alpha)=4\pi cos\alpha$$ where $cos\alpha = \frac{1}{\sqrt{4\frac{r^2}{l^2}+1}}$
$c)\$The solid angle at which the outer surface along which the current flows is visible is equal to $$\Omega = 4\pi - 2\cdot2\pi (1-cos\alpha)=4\pi cos\alpha$$ where $cos\alpha = \frac{1}{\sqrt{4\frac{r^2}{l^2}+1}}$
So:
So:
$$B = \frac{\mu_0M}{\sqrt{1+4\frac{r^2}{l^2}}}$$
$$B = \frac{\mu_0M}{\sqrt{1+4\frac{r^2}{l^2}}}$$
$r/l \to 0$:
$r/l \to 0$:
$$B \to\mu_0 M$$
$$B \to\mu_0 M$$
$l/r \to 0$:
$l/r \to 0$:
$$B \to\frac{\mu_0 Ml}{2r}$$
$$B \to\frac{\mu_0 Ml}{2r}$$
$d)\$
$d)\$
Let us imagine the field of the past cylinder as a superposition of the fields of the thin cylinder in the center and the rest of it. $$\\hat{B}_{thin}+\hat{B}_{rest}=\hat{B}_{full}$$
Let us imagine the field of the past cylinder as a superposition of the fields of the thin cylinder in the center and the rest of it. $$\\hat{B}_{thin}+\hat{B}_{rest}=\hat{B}_{full}$$
$a) \ $ The magnetic field of the cylinder is composed of the magnetic fields of the thin disks
of thickness $∆$, into which the cylinder can be divided. The magnetic field of each disk coincides with the magnetic field of a current flowing with a linear density $M$ ($M$ is the magnetic moment
per unit volume of iron); along the outer surface of the disk (see the solution to problem 9.2.19∗).
$b) \ $The direction of the magnetic field induction at the center of the cube coincides with the direction
of magnetization. The magnitude of this vector will be smaller than the magnitude of the magnetic field induction inside the rod by as many times as $8π/3$ (the solid angle subtended by the lateral
faces of cube 1–4) is less than $4π$, i.e., $n = 1.5$ times.
