1 525.5 м/с | 0.50 |
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1 $a = \frac{\mu_s}{m} \frac{dB}{dx}$ | 0.40 |
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2 $\delta_1 = \frac{1}{2} \frac{\mu_s}{m} \frac{dB}{dx} \frac{l_2^2}{v_z^2}$ | 0.60 |
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3 $\delta_2 = l_3 l_2 \frac{\mu_s}{m v_z^2} \frac{dB}{dx}$ | 0.40 |
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4 $\Delta x = 2 \frac{\mu_s}{m} \frac{dB}{dx} \frac{l_2}{v_z^2} \left( \frac{l_2}{2}+l_3 \right) $ | 0.60 |
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5 Штраф за потерю 2 | -0.30 |
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1 Поле одного провода: $\vec{B}_1(x,y)=\frac{\mu I_0}{2 \pi} \frac{\hat{k} \times (x \hat{i} + (y-a)\hat{j})}{r_1^2} $ | 0.40 |
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2 Поле другого провода: $\vec{B}_2(x,y)=- \frac{\mu I_0}{2 \pi} \frac{\hat{k} \times (x \hat{i} + (y+a)\hat{j})}{r_2^2} $ | 0.40 |
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3 Ответ: $\vec{B}(x,y)=\frac{\mu I_0 a}{\pi r_1^2 r_2^2} \left[ 2xy \hat{j} + (x^2-y^2+a^2)\hat{i} \right]$ | 0.70 |
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4 Если не собраны все слагаемые по компонентам | -0.10 |
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5 Если ошибка при упрощении ответа | -0.10 |
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1 $\vec{B}(x,0) \propto \hat{i} $ | 0.20 |
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2 $\vec{B}(x_c , (x_c^2+a^2)^{1/2}) \propto \hat{j} $ | 0.30 |
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1 $\vec{B} = \frac{\mu I_0 a}{\pi (x^2 + a^2)}\hat{i}$ | 0.50 |
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1 $F_x = \mu_s \frac{\partial B_x}{\partial x} = \frac{\mu_s \mu I_0}{\pi} \times \frac{2ax}{(x^2+a^2)^2}$ | 0.50 |
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1 $x_P = 1.624 \times 10^{-2}$ м | 0.50 |
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2 $B_x(x_P,0) = 0.16$ Тл | 1.00 |
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3 $\left( \frac{\partial B_x}{ \partial x} \right)_ {x_P} = 17.34$ Тл/м | 0.50 |
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1 $\mu_s = \frac{m \Delta x}{2 \left( \frac{\partial B_x}{ \partial x} \right)_ {x_P}} \times \frac{1}{\left[ \frac{l_2}{v_z^2} \left( \frac{l_2}{2} + l_3 \right) \right]}$ | 0.50 |
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2 $\mu_s = 1.04 \times 10^{-23}$ Дж/Тл | 1.00 |
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1 $\delta (\Delta x/2) = 2 (\Delta x/2) \frac{\delta v_z}{v_z} = 0.04$ см | 0.30 |
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2 Hence the spread in the line from the centre is 0.1 - 0.04 = 0.06 cm to 0.1 + 0.04 = 0.14 cm. 1. Credit will also be given if 20% is interpreted as 10% on each side 2. Answer reported in terms of percentages receive full credit | 0.20 |
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1 Полный разброс $0.84 \times 10^{-23}$ Дж/Тл | 0.30 |
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2 $\mu_s = (1.04 \pm 0.42) \times 10^{-23}$ Дж/Тл | 0.20 |
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