Edits to “Statement”, “Solution”, “Answer”
en/9.2.19.md
+14 −4
| @@ -1,12 +1,20 @@ | |||
| ### Statement | |||
| − | |||
| + | $9.2.19.$ a. Inside a large square circuit with current, many square micro-circuits with current are evenly distributed. The magnetic moment of each micro-circuit is $M_0$. Prove that at a distance much greater than the distance between the micro-contours, the induction of their magnetic field coincides with the induction of the magnetic field of a large contour, the magnetic moment of which is $nM_0$, where n is the | ||
| + | number of micro-contours inside the large contour. | ||
| + | b. A thin square plate with dimensions $a × a × h (h ≪ a)$ is magnetized in a direction perpendicular to | ||
| + | its plane. Magnetic field induction in the center of plate $B$. Determine the magnetic moment of the unit | ||
| + | volume of the plate substance. | ||
| + | |||
| ### Solution | |||
|  | |||
| − | a) let current flow in each micro circuit | ||
| − | This means that at large distances we can assume that we are dealing with a frame with the same current, the magnetic moment of such a frame (assuming the distribution of contours as in the problem $9.2.18$ | ||
| + | a) let current flow in each micro circuit $I = \frac{M_0}{\Delta S} $ | ||
| + | This means that at large distances we can assume that we are dealing with a frame with the same current, the magnetic moment of such a frame (assuming the distribution of contours as in the problem $9.2.18$): | ||
| + | |||
| + | $$ M = n \Delta S = nM_0$$ | ||
| + | |||
| b)Find the equivalent current flowing along the edges of the plate: | |||
| $$MhS=IS$$ | |||
| $$I = Mh$$ | |||
| Here, the field in the center is represented as a superposition of four fields from the frame contour. A field from one frame: | |||
| $$dB_1 = \frac{\mu_0 Idlsin\varphi}{4\pi r^2}$$ | |||
| $$dB_1 = \frac{\mu_0 I\frac{a/2d\varphi}{sin^2\varphi} sin\varphi}{4\pi \frac{(a/2)^2}{sin^2\varphi}}=\frac{\mu_0 Isin\varphi d\varphi}{4\pi a/2}$$ | |||
| $$B_1 = 2\int_{0}^{\pi/4}\frac{\mu_0 Isin\varphi d\varphi}{4\pi a/2}$$ | |||
| $$B_1 = \frac{\mu_0 I}{4\pi a/2}\cdot 2 cos(\pi/4)$$ | |||
| From total 4: | |||
| $$B = 4B_1 = \frac{2\sqrt{2}\mu_0I}{\pi a}$$ | |||
| Substitute the expression for the current and get the answer: | |||
| $$M = \frac{B\pi a}{\sqrt{8}\mu_0 h}$$ | |||
| @@ -22,4 +30,6 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | а)Q.E.D | ||
| + | |||
| + | б)$$M = \frac{B\pi a}{\sqrt{8}\mu_0 h}$$ | ||
| @@ -1,12 +1,20 @@ | |||
| ### Statement | ### Statement | ||
| $9.2.19.$ a. Inside a large square circuit with current, many square micro-circuits with current are evenly distributed. The magnetic moment of each micro-circuit is $M_0$. Prove that at a distance much greater than the distance between the micro-contours, the induction of their magnetic field coincides with the induction of the magnetic field of a large contour, the magnetic moment of which is $nM_0$, where n is the | |||
| number of micro-contours inside the large contour. | |||
| b. A thin square plate with dimensions $a × a × h (h ≪ a)$ is magnetized in a direction perpendicular to | |||
| its plane. Magnetic field induction in the center of plate $B$. Determine the magnetic moment of the unit | |||
| volume of the plate substance. | |||
| ### Solution | ### Solution | ||
|  |  | ||
| a) let current flow in each micro circuit |
a) let current flow in each micro circuit $I = \frac{M_0}{\Delta S} $ | ||
| This means that at large distances we can assume that we are dealing with a frame with the same current, the magnetic moment of such a frame (assuming the distribution of contours as in the problem $9.2.18$ |
This means that at large distances we can assume that we are dealing with a frame with the same current, the magnetic moment of such a frame (assuming the distribution of contours as in the problem $9.2.18$): | ||
| $$ M = n \Delta S = nM_0$$ | |||
| b)Find the equivalent current flowing along the edges of the plate: | b)Find the equivalent current flowing along the edges of the plate: | ||
| $$MhS=IS$$ | $$MhS=IS$$ | ||
| $$I = Mh$$ | $$I = Mh$$ | ||
| Here, the field in the center is represented as a superposition of four fields from the frame contour. A field from one frame: | Here, the field in the center is represented as a superposition of four fields from the frame contour. A field from one frame: | ||
| $$dB_1 = \frac{\mu_0 Idlsin\varphi}{4\pi r^2}$$ | $$dB_1 = \frac{\mu_0 Idlsin\varphi}{4\pi r^2}$$ | ||
| $$dB_1 = \frac{\mu_0 I\frac{a/2d\varphi}{sin^2\varphi} sin\varphi}{4\pi \frac{(a/2)^2}{sin^2\varphi}}=\frac{\mu_0 Isin\varphi d\varphi}{4\pi a/2}$$ | $$dB_1 = \frac{\mu_0 I\frac{a/2d\varphi}{sin^2\varphi} sin\varphi}{4\pi \frac{(a/2)^2}{sin^2\varphi}}=\frac{\mu_0 Isin\varphi d\varphi}{4\pi a/2}$$ | ||
| $$B_1 = 2\int_{0}^{\pi/4}\frac{\mu_0 Isin\varphi d\varphi}{4\pi a/2}$$ | $$B_1 = 2\int_{0}^{\pi/4}\frac{\mu_0 Isin\varphi d\varphi}{4\pi a/2}$$ | ||
| $$B_1 = \frac{\mu_0 I}{4\pi a/2}\cdot 2 cos(\pi/4)$$ | $$B_1 = \frac{\mu_0 I}{4\pi a/2}\cdot 2 cos(\pi/4)$$ | ||
| From total 4: | From total 4: | ||
| $$B = 4B_1 = \frac{2\sqrt{2}\mu_0I}{\pi a}$$ | $$B = 4B_1 = \frac{2\sqrt{2}\mu_0I}{\pi a}$$ | ||
| Substitute the expression for the current and get the answer: | Substitute the expression for the current and get the answer: | ||
| $$M = \frac{B\pi a}{\sqrt{8}\mu_0 h}$$ | $$M = \frac{B\pi a}{\sqrt{8}\mu_0 h}$$ | ||
| @@ -22,4 +30,6 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | а)Q.E.D | ||
| б)$$M = \frac{B\pi a}{\sqrt{8}\mu_0 h}$$ | |||