Правка разделов «Statement», «Solution», «Answer»
en/9.2.20.md
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| ### Statement | |||
| − | $9.2.20.$ [Insert the problem statement] | ||
| − | ### Solution | ||
| − | |||
| − | ### Problem | ||
| − | |||
| $9.2.20.$ A flat, thin disk of radius $R$ and thickness $h$ was cut from magnetized iron. The plane of the disk is perpendicular to the direction of magnetization. The magnetic moment per unit volume of the iron is $M$. Determine the magnetic field induction on the axis of the disk at a distance $l$ from its center. | |||
| ### Solution | |||
| The magnetic moment per unit volume of a magnet $M$ is called its magnetization. | |||
| Generally, the magnetic properties of magnets are determined by the so-called magnetization currents. $M = const$, which means magnetization currents flow only on the surface of the disk (all molecular currents inside the volume cancel each other out). | |||
| To find these currents, we use the circulation theorem for the vector $\vec{M}$: | |||
| $$\oint \vec{M} d\vec{l} = I$$ | |||
| where $I$ is the algebraic sum of the magnetization currents enclosed by a closed contour of length $l$. | |||
| Consider a contour enclosing the end surface of the cylinder. In this case, $\vec{M} \perp \vec{l}$, | |||
| $$I = \oint \vec{M} d\vec{l} = \vec{M} \cdot \vec{l} = 0 \quad (1)$$ | |||
| Now consider a contour enclosing the lateral surface of the cylinder. In this case, $\vec{M} \parallel \vec{l}$, | |||
| $$I = \oint \vec{M} d\vec{h} = \vec{M} \cdot \vec{h} = Mh \quad (2)$$ | |||
| @@ -28,14 +23,10 @@Solution | |||
| Thus, the magnetic field induction of our disk is equivalent to the field on the axis of a ring of radius $R$ carrying a current $I$, which is equal to | |||
| $$B(l) = \frac{\mu_0 I R^2}{2(R^2 + l^2)^{3/2}} \quad (3)$$ | |||
| − | This formula was already obtained in problem 9.2.10. | ||
| + | This formula was already obtained in problem [9.2.10](https://savchenkosolutions.com/en/9.2.10). | ||
| Substituting (2) into (3), we get | |||
| $$B = \frac{\mu_0 M R^2 h}{2(R^2 + l^2)^{3/2}}$$ | |||
| ### Answer | |||
| $$B = \frac{\mu_0 M R^2 h}{2(R^2 + l^2)^{3/2}}$$ | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,11 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $9.2.20.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Problem | |||
| $9.2.20.$ A flat, thin disk of radius $R$ and thickness $h$ was cut from magnetized iron. The plane of the disk is perpendicular to the direction of magnetization. The magnetic moment per unit volume of the iron is $M$. Determine the magnetic field induction on the axis of the disk at a distance $l$ from its center. | $9.2.20.$ A flat, thin disk of radius $R$ and thickness $h$ was cut from magnetized iron. The plane of the disk is perpendicular to the direction of magnetization. The magnetic moment per unit volume of the iron is $M$. Determine the magnetic field induction on the axis of the disk at a distance $l$ from its center. | ||
| ### Solution | ### Solution | ||
| The magnetic moment per unit volume of a magnet $M$ is called its magnetization. | The magnetic moment per unit volume of a magnet $M$ is called its magnetization. | ||
| Generally, the magnetic properties of magnets are determined by the so-called magnetization currents. $M = const$, which means magnetization currents flow only on the surface of the disk (all molecular currents inside the volume cancel each other out). | Generally, the magnetic properties of magnets are determined by the so-called magnetization currents. $M = const$, which means magnetization currents flow only on the surface of the disk (all molecular currents inside the volume cancel each other out). | ||
| To find these currents, we use the circulation theorem for the vector $\vec{M}$: | To find these currents, we use the circulation theorem for the vector $\vec{M}$: | ||
| $$\oint \vec{M} d\vec{l} = I$$ | $$\oint \vec{M} d\vec{l} = I$$ | ||
| where $I$ is the algebraic sum of the magnetization currents enclosed by a closed contour of length $l$. | where $I$ is the algebraic sum of the magnetization currents enclosed by a closed contour of length $l$. | ||
| Consider a contour enclosing the end surface of the cylinder. In this case, $\vec{M} \perp \vec{l}$, | Consider a contour enclosing the end surface of the cylinder. In this case, $\vec{M} \perp \vec{l}$, | ||
| $$I = \oint \vec{M} d\vec{l} = \vec{M} \cdot \vec{l} = 0 \quad (1)$$ | $$I = \oint \vec{M} d\vec{l} = \vec{M} \cdot \vec{l} = 0 \quad (1)$$ | ||
| Now consider a contour enclosing the lateral surface of the cylinder. In this case, $\vec{M} \parallel \vec{l}$, | Now consider a contour enclosing the lateral surface of the cylinder. In this case, $\vec{M} \parallel \vec{l}$, | ||
| $$I = \oint \vec{M} d\vec{h} = \vec{M} \cdot \vec{h} = Mh \quad (2)$$ | $$I = \oint \vec{M} d\vec{h} = \vec{M} \cdot \vec{h} = Mh \quad (2)$$ | ||
| @@ -28,14 +23,10 @@Solution | |||
| Thus, the magnetic field induction of our disk is equivalent to the field on the axis of a ring of radius $R$ carrying a current $I$, which is equal to | Thus, the magnetic field induction of our disk is equivalent to the field on the axis of a ring of radius $R$ carrying a current $I$, which is equal to | ||
| $$B(l) = \frac{\mu_0 I R^2}{2(R^2 + l^2)^{3/2}} \quad (3)$$ | $$B(l) = \frac{\mu_0 I R^2}{2(R^2 + l^2)^{3/2}} \quad (3)$$ | ||
| This formula was already obtained in problem 9.2.10. | This formula was already obtained in problem [9.2.10](https://savchenkosolutions.com/en/9.2.10). | ||
| Substituting (2) into (3), we get | Substituting (2) into (3), we get | ||
| $$B = \frac{\mu_0 M R^2 h}{2(R^2 + l^2)^{3/2}}$$ | $$B = \frac{\mu_0 M R^2 h}{2(R^2 + l^2)^{3/2}}$$ | ||
| ### Answer | ### Answer | ||
| $$B = \frac{\mu_0 M R^2 h}{2(R^2 + l^2)^{3/2}}$$ | $$B = \frac{\mu_0 M R^2 h}{2(R^2 + l^2)^{3/2}}$$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||