Правка разделов «Statement», «Answer»
en/9.3.20.md
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| ### Statement | |||
| − | $9.3.20.$ | ||
| + | $9.3.20.$ In an infinite plate of thickness $h$, a cylindrical cavity of radius $h/2$ was cut out, the axis of which is parallel | ||
| + | to the surfaces of the plate. In the entire volume of the plate, with the exception of the cavity, a current | ||
| + | flows along the axis of the cavity. Find the distribution of magnetic field induction along a straight line | ||
| + | $OA$ that passes through the cavity axis and is perpendicular to the plate surfaces. Current density $j$. | ||
| ### Solution | |||
| In problem $9.3.15$, we obtained formulas for the field inside and outside a cylindrical rod and plate. Replacing $I=j \pi (h/2)^2$ in the formulas yields the formulas for induction, respectively: | |||
| $$B^{in}_{full} = \mu_0 j x$$ | |||
| $$B^{out}_{full}=\frac{\mu_0 j h}{2}$$ | |||
| $$B^{in}_{cyl} = \frac{\mu_0 j x}{2}$$ | |||
| $$B^{out}_{cyl}=\frac{\mu_0 j h^2}{8x}$$ | |||
| Now, we represent the desired field as the difference between the field of the plate and the cylindrical tube. We obtain the answers: | |||
| For $0<x<h/2$: | |||
| $$B = \frac{\mu_0 j x}{2}$$ | |||
| For $x > h/2$: | |||
| $$B = \frac{\mu_0 j h}{2}(1-\frac{h}{4x})$$ | |||
| @@ -20,4 +23,8 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | For $0<x<h/2$: | ||
| + | $$B = \frac{\mu_0 j x}{2}$$ | ||
| + | |||
| + | For $x > h/2$: | ||
| + | $$B = \frac{\mu_0 j h}{2}(1-\frac{h}{4x})$$ | ||
| @@ -1,6 +1,9 @@ | |||
| ### Statement | ### Statement | ||
| $9.3.20.$ |
$9.3.20.$ In an infinite plate of thickness $h$, a cylindrical cavity of radius $h/2$ was cut out, the axis of which is parallel | ||
| to the surfaces of the plate. In the entire volume of the plate, with the exception of the cavity, a current | |||
| flows along the axis of the cavity. Find the distribution of magnetic field induction along a straight line | |||
| $OA$ that passes through the cavity axis and is perpendicular to the plate surfaces. Current density $j$. | |||
| ### Solution | ### Solution | ||
| In problem $9.3.15$, we obtained formulas for the field inside and outside a cylindrical rod and plate. Replacing $I=j \pi (h/2)^2$ in the formulas yields the formulas for induction, respectively: | In problem $9.3.15$, we obtained formulas for the field inside and outside a cylindrical rod and plate. Replacing $I=j \pi (h/2)^2$ in the formulas yields the formulas for induction, respectively: | ||
| $$B^{in}_{full} = \mu_0 j x$$ | $$B^{in}_{full} = \mu_0 j x$$ | ||
| $$B^{out}_{full}=\frac{\mu_0 j h}{2}$$ | $$B^{out}_{full}=\frac{\mu_0 j h}{2}$$ | ||
| $$B^{in}_{cyl} = \frac{\mu_0 j x}{2}$$ | $$B^{in}_{cyl} = \frac{\mu_0 j x}{2}$$ | ||
| $$B^{out}_{cyl}=\frac{\mu_0 j h^2}{8x}$$ | $$B^{out}_{cyl}=\frac{\mu_0 j h^2}{8x}$$ | ||
| Now, we represent the desired field as the difference between the field of the plate and the cylindrical tube. We obtain the answers: | Now, we represent the desired field as the difference between the field of the plate and the cylindrical tube. We obtain the answers: | ||
| For $0<x<h/2$: | For $0<x<h/2$: | ||
| $$B = \frac{\mu_0 j x}{2}$$ | $$B = \frac{\mu_0 j x}{2}$$ | ||
| For $x > h/2$: | For $x > h/2$: | ||
| $$B = \frac{\mu_0 j h}{2}(1-\frac{h}{4x})$$ | $$B = \frac{\mu_0 j h}{2}(1-\frac{h}{4x})$$ | ||
| @@ -20,4 +23,8 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | For $0<x<h/2$: | ||
| $$B = \frac{\mu_0 j x}{2}$$ | |||
| For $x > h/2$: | |||
| $$B = \frac{\mu_0 j h}{2}(1-\frac{h}{4x})$$ | |||