Правка разделов «Statement», «Solution», «Answer»

naz правка от 9 июн. 2026 г., 11:31

правка #18466 предыдущая #18461 ← раньше

en/9.3.6.md +22 −5

		@@ -1,20 +1,37 @@
			### Statement

		−	$9.3.6.$ ~~[Insert the problem statement]~~
		+	$9.3.6.$
		+	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:
			$$\frac{\Delta}{a}=\frac{P}{E}$$
			$$\Delta=\frac{\mu_0aI^2}{8Eb^2}$$
		−	б)$$P=\frac{B^2}{2\mu_0}$$
		+	b)$$P=\frac{B^2}{2\mu_0}$$
			$$B=\sqrt{2\mu_0P}$$
		−	$$B_1\approx10Тл$$
		−	$$B_2\approx35Тл$$
		+	$$B_1\approx10Т$$
		+	$$B_2\approx35Т$$

			#### 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

	$9.3.6.$ ~~[Insert the problem statement]~~		$9.3.6.$
			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:
	$$\frac{\Delta}{a}=\frac{P}{E}$$		$$\frac{\Delta}{a}=\frac{P}{E}$$
	$$\Delta=\frac{\mu_0aI^2}{8Eb^2}$$		$$\Delta=\frac{\mu_0aI^2}{8Eb^2}$$
	б)$$P=\frac{B^2}{2\mu_0}$$		b)$$P=\frac{B^2}{2\mu_0}$$
	$$B=\sqrt{2\mu_0P}$$		$$B=\sqrt{2\mu_0P}$$
	$$B_1\approx10Тл$$		$$B_1\approx10Т$$
	$$B_2\approx35Тл$$		$$B_2\approx35Т$$

	#### Answer		#### Answer

	[Insert a concise answer or boxed result]		$$
			\Delta=\mu_{0}aI^{2}/(8Eb^{2})
			$$

			$$
			\quad B_{1}\approx10\textrm{ T.}
			$$

			$$
			B_{2}\approx35\textrm{ T}.
			$$