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en/9.3.21.md
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| + | ### Statement | ||
| + | |||
| + | $9.3.21.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | In previous problems, we derived a formula for the induction at a distance from the axis inside a cylindrical wire: | ||
| + | $$B = \frac{\mu_0 j x}{2}$$ | ||
| + | However, we did not mention one remarkable property of this formula: since the induction is directed tangentially to the circular contours along the axis, mathematically this can be rewritten in vector form: | ||
| + | $$\hat{B} = \frac{\mu_0[\hat{j}\times \hat{x}]}{2}$$ | ||
| + | Let's use an equally remarkable property of the vector product: | ||
| + | $$(\hat{a}+\hat{b})\times \hat{c} = \hat{a}\times \hat{c} + \hat{b} \times \hat{c}$$ | ||
| + | Let's take a point at a distance x from the cavity axis; in vector form from the wire axis, it is located on the vector $\hat{r}=\hat{d}+\hat{x}$. We represent the field from the wire with the cavity as the difference between the fields of the small and large cylinders in vector form. Taking into account the properties of the vector product, we obtain: | ||
| + | $$\hat{B} = \hat{B} = \frac{\mu_0[\hat{j}\times \hat{d}]}{2}+\frac{\mu_0[\hat{j}\times \hat{x}]}{2} - \frac{\mu_0[\hat{j}\times \hat{x}]}{2}$$ | ||
| + | $$B = \frac{\mu_0 j d}{2}$$ | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $9.3.21.$ [Insert the problem statement] | |||
| ### Solution | |||
| In previous problems, we derived a formula for the induction at a distance from the axis inside a cylindrical wire: | |||
| $$B = \frac{\mu_0 j x}{2}$$ | |||
| However, we did not mention one remarkable property of this formula: since the induction is directed tangentially to the circular contours along the axis, mathematically this can be rewritten in vector form: | |||
| $$\hat{B} = \frac{\mu_0[\hat{j}\times \hat{x}]}{2}$$ | |||
| Let's use an equally remarkable property of the vector product: | |||
| $$(\hat{a}+\hat{b})\times \hat{c} = \hat{a}\times \hat{c} + \hat{b} \times \hat{c}$$ | |||
| Let's take a point at a distance x from the cavity axis; in vector form from the wire axis, it is located on the vector $\hat{r}=\hat{d}+\hat{x}$. We represent the field from the wire with the cavity as the difference between the fields of the small and large cylinders in vector form. Taking into account the properties of the vector product, we obtain: | |||
| $$\hat{B} = \hat{B} = \frac{\mu_0[\hat{j}\times \hat{d}]}{2}+\frac{\mu_0[\hat{j}\times \hat{x}]}{2} - \frac{\mu_0[\hat{j}\times \hat{x}]}{2}$$ | |||
| $$B = \frac{\mu_0 j d}{2}$$ | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||