Edits to “Statement”, “Solution”, “Answer”
en/9.3.6.md
+22 −5
| @@ -1,20 +1,37 @@ | |||
| ### Statement | |||
| − | $9.3.6.$ | ||
| + | $9.3.6.$ | ||
| + | а. | ||
| + | Through a rectangular plate $a \times b$ $(a \ll b)$ passed the current $I$. Modulus of longitudinal elasticity of the plate $E$. | ||
| + | Determine by how much the size of $a$ will decrease under the action of magnetic forces. | ||
| + | b. | ||
| + | Soft copper "flows" under pressure $4 \cdot 10^7 \, Pa$, | ||
| + | and steel - under pressure $5 \cdot 10^8 \, Pa.$ | ||
| + | Estimate the minimum magnetic field induction that will cause copper and steel to “flow”. | ||
| ### Solution | |||
| а)Energy density (magnetic pressure) is calculated using the formula $$P=w=\frac{B^2}{2\mu_0}$$ | |||
| Considering the induction to be constant throughout the entire cross-section and equal to $B=\frac{\mu_0I}{2b}$, | |||
| we get a formula for pressure:$$P=\frac{\mu_0I^2}{8b^2}$$ | |||
| Hooke's law: | |||
| $$\frac{\Delta}{a}=\frac{P}{E}$$ | |||
| $$\Delta=\frac{\mu_0aI^2}{8Eb^2}$$ | |||
| − | |||
| + | b)$$P=\frac{B^2}{2\mu_0}$$ | ||
| $$B=\sqrt{2\mu_0P}$$ | |||
| − | $$B_1\approx10Т | ||
| − | $$B_2\approx35Т | ||
| + | $$B_1\approx10Т$$ | ||
| + | $$B_2\approx35Т$$ | ||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $$ | ||
| + | \Delta=\mu_{0}aI^{2}/(8Eb^{2}) | ||
| + | $$ | ||
| + | |||
| + | $$ | ||
| + | \quad B_{1}\approx10\textrm{ T.} | ||
| + | $$ | ||
| + | |||
| + | $$ | ||
| + | B_{2}\approx35\textrm{ T}. | ||
| + | $$ | ||
| @@ -1,20 +1,37 @@ | |||
| ### Statement | ### Statement | ||
| $9.3.6.$ |
$9.3.6.$ | ||
| а. | |||
| Through a rectangular plate $a \times b$ $(a \ll b)$ passed the current $I$. Modulus of longitudinal elasticity of the plate $E$. | |||
| Determine by how much the size of $a$ will decrease under the action of magnetic forces. | |||
| b. | |||
| Soft copper "flows" under pressure $4 \cdot 10^7 \, Pa$, | |||
| and steel - under pressure $5 \cdot 10^8 \, Pa.$ | |||
| Estimate the minimum magnetic field induction that will cause copper and steel to “flow”. | |||
| ### Solution | ### Solution | ||
| а)Energy density (magnetic pressure) is calculated using the formula $$P=w=\frac{B^2}{2\mu_0}$$ | а)Energy density (magnetic pressure) is calculated using the formula $$P=w=\frac{B^2}{2\mu_0}$$ | ||
| Considering the induction to be constant throughout the entire cross-section and equal to $B=\frac{\mu_0I}{2b}$, | Considering the induction to be constant throughout the entire cross-section and equal to $B=\frac{\mu_0I}{2b}$, | ||
| we get a formula for pressure:$$P=\frac{\mu_0I^2}{8b^2}$$ | we get a formula for pressure:$$P=\frac{\mu_0I^2}{8b^2}$$ | ||
| Hooke's law: | Hooke's law: | ||
| $$\frac{\Delta}{a}=\frac{P}{E}$$ | $$\frac{\Delta}{a}=\frac{P}{E}$$ | ||
| $$\Delta=\frac{\mu_0aI^2}{8Eb^2}$$ | $$\Delta=\frac{\mu_0aI^2}{8Eb^2}$$ | ||
| b)$$P=\frac{B^2}{2\mu_0}$$ | |||
| $$B=\sqrt{2\mu_0P}$$ | $$B=\sqrt{2\mu_0P}$$ | ||
| $$B_1\approx10Т |
$$B_1\approx10Т$$ | ||
| $$B_2\approx35Т |
$$B_2\approx35Т$$ | ||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $$ | ||
| \Delta=\mu_{0}aI^{2}/(8Eb^{2}) | |||
| $$ | |||
| $$ | |||
| \quad B_{1}\approx10\textrm{ T.} | |||
| $$ | |||
| $$ | |||
| B_{2}\approx35\textrm{ T}. | |||
| $$ | |||