2018年IOAA理论第7题-太阳黑子 - 版本历史

2022年4月4日 (一) 03:26 某九流辣鸡民科

2022-04-04T03:26:09Z

2022-04-04T03:25:52Z

‎中文解答

2022-04-04T03:17:45Z

2022-04-04T03:17:30Z

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2022-04-04T03:16:02Z

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2022-04-04T03:15:09Z

‎中文解答

2022-04-04T03:14:58Z

‎中文解答

2022-04-04T03:13:58Z

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2022-04-04T03:13:09Z

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2022-04-04T03:12:30Z

‎英文解答

←上一版本		2022年4月4日 (一) 03:25的版本
第50行：		第50行：

	==中文解答==		==中文解答==
		+	(a)压力是来自各个方向的。太阳黑子内外的等离子体的压强必须相等，使黑子边界处保持平衡。

		+	太阳黑子内部的动力学压强为<br/>
		+
		+	$$p_i=n_i k_B T_i$$<br/>
		+
		+	其中 n<sub>i</sub> 是黑子内的数密度 <br/>
		+
		+	同样有,<br/>
		+
		+	$$p_e=n_e k_B T_e$$<br/>
		+
		+	(<math>p_e V=N_e k_B T_e</math> 也是正确的。但是最好写成压强和数密度的关系)<br/>
		+
		+	由假设知,n<sub>i</sub>=n<sub>e</sub>=n。平衡要求:<br/>
		+
		+	$$n k_{B} T_{i}+\frac{B_{i}^{2}}{2 \mu_{0}}=n k_{B} T_{e}+\frac{B_{e}^{2}}{2 \mu_{0}}$$<br/>
		+	于是有<br/>
		+
		+	$$n=\frac{1}{2 \mu_{0} k_{B}} \frac{\left(B_{i}^{2}-B_{e}^{2}\right)}{\left(T_{e}-T_{i}\right)}=1.43 \times 10^{23} \mathrm{~m}^{-3}$$<br/>
		+
		+	(b)在地球上， <math>p_{E}=n_{E} k_{B} T_{E}</math>，其中<math>P_{E} \approx 10^{5} \mathrm{~Pa}</math> ， <math>T_{E} \approx 300 \mathrm{~K}</math><br/>
		+
		+	于是有, <math>n_{E}=\frac{p_{E}}{k_{B} T_{E}}=2.4 \times 10^{25} m^{-3}</math><br/>
		+
		+	这意味着地球表面附近大气中粒子的数密度至少是太阳大气中等离子数密度的100倍（当然质量密度会更高）。
	[[分类:磁场]]		[[分类:磁场]]

@@ 第45行： / 第45行： @@
 (b)On the Earth, <math>p_{E}=n_{E} k_{B} T_{E}</math> where <math>P_{E} \approx 10^{5} \mathrm{~Pa}</math> and <math>T_{E} \approx 300 \mathrm{~K}</math><br/>
-Thus, <math>n_{E}=\frac{p_{E}}{k_{B} T_{E}}=2.4 \times 10^{25} m^{-3}<math><br/>
+Thus, <math>n_{E}=\frac{p_{E}}{k_{B} T_{E}}=2.4 \times 10^{25} m^{-3}</math><br/>
 This means that the number density of the atmospheric particles at the surface of the earth is at least 100 times larger that the number density of particles in the solar photosphere (of course the mass density is even higher).

@@ 第25行： / 第25行： @@
 (a)The pressure is from all directions. The plasma inside the sunspot must have the same total pressure as the plasma outside to maintain equilibrium on the border.
 The kinetic pressure inside the sunspot is <br/>
 $$p_i=n_i k_B T_i$$<br/>
 where n<sub>i</sub> is the number density inside the sunspot. <br/>
 And also,<br/>
 $$p_e=n_e k_B T_e$$<br/>
 (<math>p_e V=N_e k_B T_e</math> is also correct. But it’s better to write the pressure in terms of number density)<br/>
 From the assumption,n<sub>i</sub>=n<sub>e</sub>=n. The equilibrium requires:<br/>
@@ 第37行： / 第43行： @@
 $$n=\frac{1}{2 \mu_{0} k_{B}} \frac{\left(B_{i}^{2}-B_{e}^{2}\right)}{\left(T_{e}-T_{i}\right)}=1.43 \times 10^{23} \mathrm{~m}^{-3}$$<br/>
-(b)On the Earth, <math>p_{E}=n_{E} k_{B} T_{E}</math> where $P_{E} \approx 10^{5} \mathrm{~Pa}$ and $T_{E} \approx 300 \mathrm{~K}$<br/>
+(b)On the Earth, <math>p_{E}=n_{E} k_{B} T_{E}</math> where <math>P_{E} \approx 10^{5} \mathrm{~Pa}</math> and <math>T_{E} \approx 300 \mathrm{~K}</math><br/>
-Thus, $n_{E}=\frac{p_{E}}{k_{B} T_{E}}=2.4 \times 10^{25} m^{-3}$<br/>
 This means that the number density of the atmospheric particles at the surface of the earth is at least 100 times larger that the number density of particles in the solar photosphere (of course the mass density is even higher).

@@ 第25行： / 第25行： @@
 (a)The pressure is from all directions. The plasma inside the sunspot must have the same total pressure as the plasma outside to maintain equilibrium on the border.
 The kinetic pressure inside the sunspot is <br/>
-$$p_{i}=n_{i} k_{B} T_{i}$$<br/>
+$$p_i=n_i k_B T_i$$<br/>
 where n<sub>i</sub> is the number density inside the sunspot. <br/>
 And also,<br/>
-$$p_{e}=n_{e} k_{B} T_{e}$$<br/>
+$$p_e=n_e k_B T_e$$<br/>
-(<math>p_{e} V=N_{e} k_{B} T_{e}</math> is also correct. But it’s better to write the pressure in terms of number density)<br/>
+(<math>p_e V=N_e k_B T_e</math> is also correct. But it’s better to write the pressure in terms of number density)<br/>
 From the assumption,n<sub>i</sub>=n<sub>e</sub>=n. The equilibrium requires:<br/>

@@ 第42行： / 第42行： @@
 ==中文解答==
-:<math>\int_0^\infty e^{-x^2}\,dx</math>
 [[分类:磁场]]

@@ 第31行： / 第31行： @@
 (<math>p_{e} V=N_{e} k_{B} T_{e}</math> is also correct. But it’s better to write the pressure in terms of number density)<br/>
 From the assumption,n<sub>i</sub>=n<sub>e</sub>=n. The equilibrium requires:<br/>
 $$n k_{B} T_{i}+\frac{B_{i}^{2}}{2 \mu_{0}}=n k_{B} T_{e}+\frac{B_{e}^{2}}{2 \mu_{0}}$$<br/>
 Then<br/>
 $$n=\frac{1}{2 \mu_{0} k_{B}} \frac{\left(B_{i}^{2}-B_{e}^{2}\right)}{\left(T_{e}-T_{i}\right)}=1.43 \times 10^{23} \mathrm{~m}^{-3}$$<br/>
 (b)On the Earth, $p_{E}=n_{E} k_{B} T_{E}$ where $P_{E} \approx 10^{5} \mathrm{~Pa}$ and $T_{E} \approx 300 \mathrm{~K}$<br/>