Правка разделов «Statement», «Answer»
en/9.3.22.md
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| ### Statement | |||
| − | $9.3.22.$ | ||
| + | $9.3.22.$ a. Two cylinders of radius $R$, whose axes are at a distance $a$ from each other, intersect, as shown in the | ||
| + | figure. Currents with a density of $±j$ flow through the shaded areas along the axes in opposite directions. | ||
| + | Find the magnetic field induction in the region lying between the shaded areas. | ||
| + | b. Using the result of the previous problem and applying the limit transition method, find for $a → 0$, | ||
| + | $j → ∞$ the distribution of the linear current density on the surface of a cylinder of radius $R$, which gives | ||
| + | a homogeneous magnetic field of induction $B_0$ inside the cylinder. How is the maximum linear current | ||
| + | density related to the field induction $B_0$? | ||
| + | |||
| + | |||
| ### Solution | |||
| a) As in the previous problem, we will use the property of the vector product and the formula for the field inside a cylindrical wire. Consider a point at distances $r_1$ and $r_2$ from the centers of the cylinders: | |||
| $$\hat{B} = \frac{\mu_0 [\hat{r}_1 \times \hat{j}]}{2} - \frac{\mu_0 [\hat{r}_2 \times \hat{j}]}{2} = \frac{\mu_0 [\hat{a} \times \hat{j}]}{2}$$ | |||
| $$B = \frac{\mu_0 a j}{2}$$ | |||
| b) The detailed solution to this point repeats in all its main points the solution to problem 6.2.14, which I recommend you read. However, it should be mentioned that here it is more convenient to express the answer in terms of a sine, since the induction lines inside are perpendicular to the vector $\hat{a}$. | |||
| $$i = j a sin \varphi$$ | |||
| $$j = \frac{2B_0}{\mu_0 a}$$ | |||
| $$i = \frac{2B_0}{\mu_0}sin\varphi$$ | |||
| $$i_{max} = \frac{2B_0}{\mu_0}$$ | |||
| @@ -15,4 +23,6 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | а)$$B = \frac{\mu_0 a j}{2}$$ | ||
| + | b)$$i = \frac{2B_0}{\mu_0}sin\varphi$$ | ||
| + | $$i_{max} = \frac{2B_0}{\mu_0}$$ | ||
| @@ -1,7 +1,15 @@ | |||
| ### Statement | ### Statement | ||
| $9.3.22.$ |
$9.3.22.$ a. Two cylinders of radius $R$, whose axes are at a distance $a$ from each other, intersect, as shown in the | ||
| figure. Currents with a density of $±j$ flow through the shaded areas along the axes in opposite directions. | |||
| Find the magnetic field induction in the region lying between the shaded areas. | |||
| b. Using the result of the previous problem and applying the limit transition method, find for $a → 0$, | |||
| $j → ∞$ the distribution of the linear current density on the surface of a cylinder of radius $R$, which gives | |||
| a homogeneous magnetic field of induction $B_0$ inside the cylinder. How is the maximum linear current | |||
| density related to the field induction $B_0$? | |||
| ### Solution | ### Solution | ||
| a) As in the previous problem, we will use the property of the vector product and the formula for the field inside a cylindrical wire. Consider a point at distances $r_1$ and $r_2$ from the centers of the cylinders: | a) As in the previous problem, we will use the property of the vector product and the formula for the field inside a cylindrical wire. Consider a point at distances $r_1$ and $r_2$ from the centers of the cylinders: | ||
| $$\hat{B} = \frac{\mu_0 [\hat{r}_1 \times \hat{j}]}{2} - \frac{\mu_0 [\hat{r}_2 \times \hat{j}]}{2} = \frac{\mu_0 [\hat{a} \times \hat{j}]}{2}$$ | $$\hat{B} = \frac{\mu_0 [\hat{r}_1 \times \hat{j}]}{2} - \frac{\mu_0 [\hat{r}_2 \times \hat{j}]}{2} = \frac{\mu_0 [\hat{a} \times \hat{j}]}{2}$$ | ||
| $$B = \frac{\mu_0 a j}{2}$$ | $$B = \frac{\mu_0 a j}{2}$$ | ||
| b) The detailed solution to this point repeats in all its main points the solution to problem 6.2.14, which I recommend you read. However, it should be mentioned that here it is more convenient to express the answer in terms of a sine, since the induction lines inside are perpendicular to the vector $\hat{a}$. | b) The detailed solution to this point repeats in all its main points the solution to problem 6.2.14, which I recommend you read. However, it should be mentioned that here it is more convenient to express the answer in terms of a sine, since the induction lines inside are perpendicular to the vector $\hat{a}$. | ||
| $$i = j a sin \varphi$$ | $$i = j a sin \varphi$$ | ||
| $$j = \frac{2B_0}{\mu_0 a}$$ | $$j = \frac{2B_0}{\mu_0 a}$$ | ||
| $$i = \frac{2B_0}{\mu_0}sin\varphi$$ | $$i = \frac{2B_0}{\mu_0}sin\varphi$$ | ||
| $$i_{max} = \frac{2B_0}{\mu_0}$$ | $$i_{max} = \frac{2B_0}{\mu_0}$$ | ||
| @@ -15,4 +23,6 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | а)$$B = \frac{\mu_0 a j}{2}$$ | ||
| b)$$i = \frac{2B_0}{\mu_0}sin\varphi$$ | |||
| $$i_{max} = \frac{2B_0}{\mu_0}$$ | |||