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/ID [(OmIhXdAoPL8s5tRvLUR/TA==) (OmIhXdAoPL8s5tRvLUR/TA==)] ->> -startxref -1159631 -%%EOF \ No newline at end of file diff --git a/course2/sem3/homework/solutions.typ b/course2/sem3/homework/solutions.typ deleted file mode 100644 index a4d981d..0000000 --- a/course2/sem3/homework/solutions.typ +++ /dev/null @@ -1,684 +0,0 @@ -#set page(numbering: "1") -#set page( - paper: "a4", - margin: (x: 1.8cm, y: 1.5cm), -) -#set text( - font: "New Computer Modern", - size: 14pt -) -#set par( - /*first-line-indent: ( - amount: 1.5em, - all: true - ),*/ - justify: true, - leading: 0.52em, -) - - -#align(center)[#text(size: 1.5em)[Домашняя работа. Дощенников Никита]] - -#outline( - title: [] -) - -#align(center)[=== Электростатика. Постоянный ток.] - -#align(center)[===== №1] // ready - -Система состоит из полусферы несущей равномерно распределённый заряд с поверхностной плотностью $sigma eq 5 " нКл/м"^2$. Рассчитать модуль напряжённости электростатического поля, создаваемого полусферой в её центре. - -*Решение*: - -#align(center)[ - #figure( - image("assets/1.svg"), - supplement: [Рис.], - caption: [Полусфера.] - ) -] - -В системе СИ: $sigma eq 5 " нКл/м"^2 eq 5 dot 10^(-9) " Кл/м"^2$. - -В сферических координатах с центром в искомой точке. Зададим точку на сфере полярным углом $theta in [0, pi/2]$ и азимутальным $phi in [0, 2 pi]$. Тогда поверхностный элемент сферы $d S$ равен: - -$ -d S eq R^2 sin theta d theta d phi. -$ - -Элемент заряда $d q$ равен: - -$ -d q eq sigma d S eq sigma R^2 sin theta d theta d phi -$ - -Поле от элементарного заряда в центра по модулю равно: - -$ -d E eq k frac(d q, R^2) eq k frac(sigma R^2 sin theta d theta d phi, R^2) eq k sigma sin theta d theta d phi. -$ - -Расписав составляющие (так как поле направлено к центру, значение с минусом): - -$ -d E_x eq -k sigma sin^2 theta cos phi d theta d phi, \ -d E_y eq -k sigma sin^2 theta sin phi d theta d phi, \ -d E_z eq -k sigma sin theta cos theta d theta d phi -$ - -Проинтегрировав по всей полусфере, получим: - -$ -E_x eq integral_0^(2 pi) integral_0^(pi/2) d E_x eq -sigma k integral_0^(pi/2) sin^2 theta d theta integral_0^(2 pi) cos phi d phi. -$ - -Так как $integral_0^(2 pi) cos phi d phi eq 0$, то $E_x eq 0$. (Аналогично $E_y eq 0$). Остается только $z$-компонента: - -$ -E_z eq E eq integral_0^(2 pi) integral_0^(pi / 2) d E_z eq - sigma k integral_0^(2 pi) d phi integral_0^(pi / 2) sin theta cos theta d theta eq k sigma dot (2 pi) dot 1/2 eq sigma/(4 epsilon_0). -$ - -Подставив числа, получим: - -$ -E eq frac(sigma, 4 epsilon_0) eq frac(5 dot 10^(-9), 4 dot 8.85 dot 10^(-12)) approx 141.2 "В/м" approx 0.14 "кВ/м". -$ - -*Ответ*: $E approx 0.14 "кВ/м"$. - -#align(center)[===== №2] // ready - -Система представляет собой область пространства заполненного зарядом с объёмной плотностью $rho = rho_0 exp(-alpha r^3)$, где $rho_0$ и $alpha$ -- положительные постоянные, а $r$ -- расстояние от центра системы. Найти модуль напряжённости электростатического поля, как функцию $r$. - -*Решение*: - -#align(center)[ - #figure( - image("assets/2.svg"), - supplement: [Рис.], - caption: [Область пространства.] - ) -] - -По закону Гаусса: - -$ -integral.cont_S arrow(E) dot d arrow(S) eq frac(Q_"вн", epsilon_0) -$ - - - -Система обладает сферической симметрией. Модуль $E(r)$ одинаков по всей сфере радиуса $r$, тогда (площадь поверхности сферы $4 pi r^2$): - -$ -integral.cont_S arrow(E) dot d arrow(S) eq E(r) integral.cont_S d S eq E(r) dot S eq E(r) dot 4 pi r^2 -$ - -$ -E(r) 4 pi r^2 eq frac(Q_"вн" (r), epsilon_0) arrow.double E(r) eq frac(Q_"вн" (r), 4 pi epsilon_0 r^2) -$ - -Заряд внутри радиуса $r$: - -$ -Q_"вн" (r) eq integral_(V_r) rho(r') d V eq integral_0^r rho (r') 4 pi r^('2) d r' eq 4 pi rho_0 integral_0^r r^('2) e^(-alpha r')^3 d r'. -$ - -Пусть $u eq alpha r^('3)$. - -$ -d u eq 3 alpha r^('2) d r' arrow.double r^('2) d r' eq frac(d u, 3 alpha) -$ - -$ -Q_"вн" (r) eq 4 pi rho_0 dot frac(1, 3 alpha) integral_0^(alpha r^3) e^(-u) d u eq frac(4 pi rho_0, 3 alpha) (1 - e^(-alpha r^3)) -$ - -Подставим в закон Гаусса: - -$ -E(r) eq frac(Q_"вн" (r), 4 pi epsilon_0 r^2) eq frac(frac(4 pi rho_0, 3 alpha)(1 minus e(minus alpha r^3)), 4 pi epsilon_0 r^2) eq frac(rho_0, 3 epsilon_0 alpha r^2) (1 - e^(-alpha r^3)) -$ - -*Ответ*: $E(r) eq frac(rho_0, 3 epsilon_0 alpha r^2) (1 - exp(-alpha r^3))$. - -#align(center)[===== №3] // ready - -Система состоит из равномерно заряженного шара радиуса $R eq 20 "см"$. Рассчитать разность потенциалов между точками, лежащими на расстоянии $r_1 eq 1 "см"$ и $r_2 eq 15 "см"$ от центра шара. Объёмная плотность заряда $rho eq 10 " нКл/м"^3$. Диэлектрическая проницаемость вещества из которого состоит шар $epsilon = 1$. - -*Решение*: - -#align(center)[ - #figure( - image("assets/3.svg"), - caption: [Шар.], - supplement: [Рис.] - ) -] - - -Для $r lt.eq R$ используем закон Гаусса. Заряд, заключенный в сфере, радиуса $r$: - -$ -Q_"вн" eq rho dot 4/3 pi r^3. -$ - -Поток через сферу радиуса $r$ равен: - -$ -integral.cont arrow(E) dot d arrow(S) eq E(r) dot 4 pi r^2 eq frac(Q_"вн", epsilon_0). -$ - -Отсюда можно выразить $E(r)$: - -$ -E(r) eq frac(Q_"вн", 4 pi epsilon_0 r^2) eq frac(rho^(4/3) pi r^3, 4 pi epsilon_0 r^2) eq frac(rho r, 3 epsilon_0) -$ - -Потенциал определяется как: - -$ -phi(r_1) - phi(r_2) eq integral_(r_1)^(r_2) E(r) d r -$ - -Подставив $E(r) eq frac(rho r, 3 epsilon_0)$: - -$ -Delta phi eq phi(r_1) - phi(r_2) eq integral_(r_1)^(r_2) frac(rho r, 3 epsilon_0) d r eq frac(rho, 6 epsilon_0) (r^2_2 - r^2_1) -$ - -Подставив числа, получим: - -$ -Delta phi eq frac(10 dot 10^(-9), 6 dot 8.85 dot 10^(-12)) (0.15^2 - 0.01^2) "В" approx 4.2 "В". -$ - -*Ответ*: $Delta phi approx 4.2 "B"$. - -#align(center)[===== №4] // ready - -Зазор между пластинами плоского конденсатора полностью плоская слюдяная пластинка ($epsilon_1 eq 7$) толщиной $d_1 eq 2 "мм"$, и слой парафина ($epsilon_1 eq 2$) толщиной $d_2 eq 1 "мм"$. Рассчитать модули напряжённости электрического поля в обоих диэлектриках, если разность потенциалов между пластинами $U eq 200 В$. - -*Решение*: - -#align(center)[ - #figure( - image("assets/4.png"), - supplement: [Рис.], - caption: [Конденсатор.] - ) -] - -При статическом поле в плоском конденсаторе нормальная компонента вектора электрической индукции $arrow(D)$ одинакова во всех слоях: - -$ -D eq epsilon_0 epsilon_(r 1) E_1 eq epsilon_0 epsilon_(r 2) E_2. -$ - -Отсюда получаем связь между полями: - -$ -epsilon_(r 1) E_1 eq epsilon_(r 2) E_2 arrow.double E_1 eq frac(epsilon_(r 2), epsilon_(r 1)) E_2. -$ - -Общая разность потенциалов $U$ равна: - -$ -U eq E_1 d_1 + E_2 d_2. -$ - -Подставив $E_1 eq frac(epsilon_(r 2), epsilon_(r 1)) E_2$, получим: - -$ -U eq frac(epsilon_(r 2), epsilon_(r 1)) E_2 d_1 + E_2 d_2 eq E_2(frac(epsilon_(r 2), epsilon_(r 1)) d_1 + d_2) -$ - -Выражая $E_2$ и $E_1$: - -$ -E_2 eq frac(U, frac(epsilon_(r 2), epsilon_(r 1)) d_1 + d_2), space.quad E_1 eq frac(epsilon_(r 2), epsilon_(r 1)) E_2. -$ - -Подставим числа из условия: - -$ -E_1 approx 3.64 dot 10^4 "В/м" approx 36.4 "кВ/м", E_2 approx 1.27 dot 10^5 "В/м" approx 0.127 "МВ/м". -$ - -*Ответ*: $E_1 approx 36 "кВ/м", E_1 approx 0.13 "МВ/м"$. - -#align(center)[===== №5] // ready - -На расстоянии $l eq 1.5 "см"$ от проводящей плоскости расположен точечный заряд $q eq 100 "мкКл"$. Рассчитайте работу, которую необходимо совершить против электрических сил, чтобы медленно удалить этот заряд от плоскости на бесконечность. - -*Решение*: - -#align(center)[ - #figure( - image("assets/5.svg"), - supplement: [Рис.], - caption: [Схема с проводящей плоскостью и зарядами.] - ) -] - -В системе СИ: $l eq 1.5 "см" eq 0.015 "м", space q eq 100 "мкКл" eq 1.0 dot 10^(-4) "Кл"$. - -По закону сохранения энергии: - -$ -Delta E_K plus Delta E_P eq A_"тр" plus A_"вн" -$ - -Так как мы удаляем заряд медленно, то $Delta E_K eq 0$. Про трение ничего не сказано, поэтому $A_"тр" eq 1$. Тогда: - -$ -A eq Delta E_P eq E_(P 2) minus E_(P 1) -$ - -На бесконечности ($E_(P 2)$) равна нулю так как $r eq infinity$, и в формуле $E_P eq k frac(q, r)$ стоит в знаменателе. - -Реальный заряд $q$ находится на расстоянии $l$ от плоскости, а мнимый заряд $q' eq -q$ находится на расстоянии $l$ по другую сторону плоскости. Тогда обозначим за $r eq 2 l eq 0.03 "м"$. - -Потенциальная энергия взаимодействия двух зарядов $E_P$ равна: - -$ -E_P eq k frac(q q', r) eq -k frac(q^2, 2 l) -$ - -Нужно учесть, что получившееся значение -- работа по удалению не одного, а двух зарядов. Тогда поделим значение на 2. - - -Подставив числа, получим: - -$ -A eq frac(q^2, 16 pi epsilon_0 l) eq frac(10^4 dot 10^(-12), 16 pi dot 8.85 dot 10^(-12) dot 1.5 dot 10^(-2)) approx 0.15 dot 10^3 "Дж" -$ - -*Ответ*: $A approx 0.15 dot 10^3 "Дж"$. - -#align(center)[===== №6] // ready - -По прямому проводнику длина которого $l eq 400 "м"$ течёт постоянный ток, сила которого $I eq 10 "А"$. Рассчитать суммарный импульс электронов в проводнике. - -*Решение*: - -#align(center)[ - #figure( - image("assets/6.png"), - supplement: [Рис.], - caption: [Проводник с током.] - ) -] - -По формуле плотности тока: - -$ -j eq rho U ", где" rho eq frac(q, l dot S) arrow.double j eq frac(q U, l S) -$ - -Так как $j eq I/S$ по определению, то можно выразить заряд: - -$ -j eq I/S eq frac(q U, l S) arrow.double q eq frac(I l, U) -$ - -Масса всех электронов равна произведению их количества на массу одного электрона: - -$ -m eq n_e dot m_e eq q/e m_e eq frac(I l, U e) m_e -$ - -По формуле импульса: - -$ -p eq m U eq frac(I l m_e, e) -$ - -Подставим числа из условия: - -$ -p eq frac(10 dot 400 dot 9.1 dot 10^(-31), 1.6 dot 10^(-19)) approx 2.3 dot 10^(-8). -$ - -*Ответ*: $p eq 2.3 dot 10^(-8) "Н/с"$. - - -#align(center)[=== Магнитостатика. Закон электромагнитной индукции Фарадея.] - -#align(center)[===== №1] // ready - -Замкнутый контур с током имеет вид прямоугольника с диагональю $d eq 16 "см"$, угол между диагоналями $alpha eq 30 degree$. Сила тока, протекающего по контуру $I eq 5 "A"$. Рассчитать модуль индукции магнитного поля в центре контура. - -*Решение*: - - -#align(center)[ - #figure( - image("assets/7.svg"), - supplement: [Рис.], - caption: [Прямоугольный контур с током и его центр.] - ) -] - -В системе СИ: $d eq 16 "см" eq 0.16 "м"$. - -По принципу суперпозиции для магнитного поля: - -$ -arrow(B) eq arrow(B)_1 plus arrow(B)_2 plus arrow(B)_3 plus arrow(B)_4. -$ - -Так как все $arrow(B)_i$ сонаправлены, то $B eq 2(B_1 plus B_2)$. По закону Био-Савара-Лапласа: - -$ -B eq 2(frac(mu_0 I, 2 pi) (frac(cos frac(pi minus phi, 2), d/2 cos phi/2) + frac(cos phi/2, d/2 sin phi/2))) eq \ -eq frac(mu_0 I, pi) (frac(1, d/2 sin phi/2 cos phi/2)) eq frac(4 mu_0 I, pi d sin phi) -$ - -Подставив числа из условия, получим: - -$ -B eq frac(4 dot 4 pi dot 10^(-7) dot 5, pi dot 0.16 dot sin 30 degree) approx 0.1 -$ - -*Ответ*: $B approx 0.1 "мТл"$. - -#align(center)[===== №2] // ready - -Два бесконечных прямых параллельных проводника разделены расстоянием $d eq 20 "см"$. По проводникам в противоположных направлениях текут токи $I_1 eq I_2 eq 10 "А"$. Рассчитать модуль напряжённости магнитного поля в точке, равноудалённой от обоих проводников на расстояние $a eq 20 "см"$. - -*Решение*: - -#align(center)[ - #figure( - image("assets/9.svg"), - supplement: [Рис.], - caption: [Два параллельных проводника с противоположными токами.] - ) -] - - -По формуле напряженности магнитного поля для прямого тока: - -$ -H eq 2 pi r I -$ - -Результирующее поле направлено вниз: - -$ -H eq (2I)/(2 pi r) cos pi/3 eq I/(2 pi r) -$ - -Подставив числа из условия, получим: - -$ -frac(10, 2 pi dot 0.2) approx 8 "А/м". -$ - -*Ответ*: $H approx 8 "А/м"$. - -#align(center)[===== №3] // ready - -По проводу бесконечной длины, имеющего форму цилиндра радиуса $R$ течёт постоянный ток, плотность которого зависит от расстояния до центра провода как $j eq alpha r arrow(e)_z$. Рассчитать вектор магнитной индукции создаваемый током внутри и вне провода, как функцию $r$ (магнитная проницаемость всюду равна 1). - -*Решение*: - -#align(center)[ - #figure( - image("assets/8.svg"), - supplement: [Рис.], - caption: [Цилиндрический провод с током вдоль оси.] - ) -] - -Для осесимметричного распределения удобно взять круговой контур радиуса $r$, с центром на оси цилиндра. Интеграл по контуру: - -$ -integral.cont arrow(B) dot d arrow(l) eq B_phi (r) (2 pi r) -$ - -Закон Ампера: - -$ -integral.cont arrow(B) dot d arrow(l) eq mu_0 I_"вн" (r) -$ - -где $I_"вн" (r)$ - суммарный ток, пронизывающий поверхность, ограниченную контуром. - -Отсюда: - -$ -B_phi (r) eq frac(mu_0, 2 pi r) I_"вн" (r) -$ - -Ток через круг радиуса $r$: - -$ -I_"вн" (r) eq integral.double_S_r j_z (r') d S eq integral_0^r integral_0^(2 pi) (alpha r') r' d phi d r' -$ - -$ -I_"вн" (r) eq alpha dot 2 pi integral_0^r r^('2) d r' eq alpha dot 2 pi dot frac(r^3, 3) eq frac(2 pi alpha r^3, 3). -$ - -Магнитная индукция внутри $r lt R$: - -$ -B_phi (r) eq frac(mu_0, 2 pi r) dot frac(2 pi alpha r^3, 3) eq frac(mu_0 alpha r^2, 3). -$ - -Магнитная индукция снаружи $r gt R$: - -$ -I eq I_"вн" (R) eq frac(2 pi alpha R^3, 3). -$ - -По закону Ампера для $r gt R$: - -$ -B_phi (r) eq frac(mu_0 I, 2 pi r) eq frac(mu_0, 2 pi r) dot frac(2 pi alpha R^3, 3) eq frac(mu_0 alpha R^3, 3 r) -$ - -*Ответ*: $arrow(B) (r lt R) eq frac(mu_0 alpha r^2, 3) arrow(e)_phi, space arrow(B) (r gt R) eq frac(mu_0 alpha R^3, 3 r) arrow(e)_phi$. - -#align(center)[===== №4] // ready - -В однородное магнитное поле с магнитной индукцией $B eq 0.4 "Тл"$ перпендикулярно полю с постоянной скоростью влетает заряженная частица. В течении $6 "мкс"$ включается постоянное электрическое поле напряжённостью $E eq 300 "В/м"$ сонаправленно магнитному полю. Рассчитать шаг винтовой траектории частицы после выключения электрического поля. - -*Решение*: По формуле силы Лоренца: - -$ -arrow(F) eq q(arrow(E) + arrow(v) times arrow(B)) -$ - -До включения электрического поля: - -$ -arrow(E) eq 0, space.quad arrow(F) eq q(arrow(v) times arrow(B)) -$ - -Частица движется по окружности - -$ -F_"маг" = q v B. -$ - -Сила Лоренца равна центростремительной силе: - -$ -q v B eq frac(m v^2, R) arrow.double R eq frac(m v, q B) -$ - -Угловая частота: - -$ -omega eq v/R eq frac(q B, m) -$ - -Когда включается электрическое поле вдоль магнитного поля, на частицу вдоль $B$ действует $F eq q E$. Соответственно вдоль оси $B$ ускорение $a eq frac(q E, m)$. - -За время $Delta t$ скорость вдоль оси становится: - -$ -v eq a Delta t eq frac(q E, m) Delta t -$ - -После выключения электрического поля частица летит в магнитном поле с постоянной перпендикулярной скоростью и параллельной, то есть по винтовой траектории. - -Расстояние за один оборот: - -$ -h eq v T, -$ - -где $T eq frac(2 pi, omega) eq frac(2 pi m, q B)$ - период кругового движения. - -Подставим: - -$ -h eq v T eq frac(q E, m) Delta t dot frac(2 pi m, q B) eq frac(2 pi E Delta t, B) -$ - -Подставим числа: - -$ -h eq frac(2 pi dot 300 "В/м" dot 6 dot 10^(-6) "с", 0.4 "Тл") approx 0.28 "м". -$ - -*Ответ*: $h eq 0.28 "м"$. - -#align(center)[===== №5] // ready - -Квадратная рамка со стороной $a eq 70 "см"$ помещена в магнитное поле так, что нормаль к рамке составляет угол $alpha eq 45 degree$ с направлением магнитного поля. Индукция магнитного поля меняется по закону $B eq B_0 cos omega t$, где $B_0 eq 0.2 "Тл", omega eq 6 " с"^(-1)$. Рассчитать ЭДС индукции, возникающей в рамке в момент времени $t eq 3 "с"$. - -*Решение*: - -#align(center)[ - #figure( - image("assets/10.png"), - supplement: [Рис.], - caption: [Квадратная рамка в переменном магнитном поле.] - ) -] - -По формуле магнитного потока через плоскость: - -$ -Phi eq B S cos alpha -$ - -Площадь рамки: - -$ -S eq a^2 -$ - -Так как $B eq B_0 cos(omega t)$: - -$ -Phi eq B_0 a^2 cos beta cos (omega t) -$ - -По закону Фарадея: - -$ -cal(E) eq -Phi'(t) eq B_0 a^2 omega cos beta sin omega t -$ - -Подставив числа из условия, получим: - -$ -cal(E) eq 0.2 dot 0.7^2 dot 6 dot cos(45 degree) dot sin (6 dot 3) approx -0.31 "B". -$ - -*Ответ*: $epsilon eq -0.31 "В"$. - -#align(center)[===== №6] // ready - -Плотность витков в катушке $n eq 25 " см"^(-1)$. Рассчитать объёмную плотность энергии магнитного поля в катушке при токе $I eq 2 "А"$. - -*Решение*: - -#align(center)[ - #figure( - image("assets/12.png"), - supplement: [Рис.], - caption: [Катушка с током и однородным магнитным полем внутри.] - ) -] - -В системе СИ: $n eq 25 " см"^(-1) eq 2500 " м"^(-1)$. - -По закону Ампера: - -$ -integral.cont arrow(B) dot d arrow(l) eq mu_0 I_"внутри" -$ - -Возьмем прямоугольный контур. Одна сторона внутри катушки длиной $l_"внутри"$, другая снаружи. Магнитное поле внутри $arrow(B) dot d arrow(l) eq B l_"внутри"$. Ток, охваченный контуром: $I_"внутри" eq I dot N_"охваченных витков" eq I n l_"внутри"$. Подставив в закон Ампера, получим: - -$ -B l_"внутри" eq mu_0 (n I l_"внутри") arrow.double B eq mu_0 n I. -$ - -Энергия магнитного поля катушки: - -$ -W eq 1/2 L I^2, -$ - -где $L$ - индуктивность катушки. - -По определению индуктивности: - -$ -L eq frac(Phi, I), -$ - -где $Phi$ - магнитный поток через катушку. - -Магнитный поток через все витки равен: - -$ -Phi eq N dot B dot S, -$ - -где $N$ - число витков, $S$ - площадь поперечного сечения, $B$ - магнитное поле внутри катушки. - -$ -L eq frac(N B S, I). -$ - -Объем катушки $V eq S l$, число витков $N eq n l$. Подставим: - -$ -L eq frac(n l B S, I) eq frac(B n S l, I). -$ - -Тогда энергия равна: - -$ -W eq 1/2 L I^2 eq 1/2 frac(B n S l, I) I^2 eq 1/2 B n I S l -$ - -Объемная плотность энергии $w$ равна: - -$ -w eq W/V eq frac(1/2 B n I S l, S l) eq 1/2 B n I -$ - -Подставим $B eq mu_0 n I$: - -$ -w eq 1/2 (mu_0 n I) n I eq 1/2 mu_0 n^2 I^2 -$ - -Подставим числа: - -$ -w eq 1/2 4 pi dot 10^(-7) dot (2500)^2 dot 2^2 eq 2 pi dot 10^(-7) dot 6.25 dot 10^6 dot 4 approx 16 " Дж/м"^3 -$ - -*Ответ*: $omega approx 16 " Дж/м"^3$. diff --git a/course2/sem3/homework/tasks.pdf b/course2/sem3/homework/tasks.pdf deleted file mode 100644 index 90215ca..0000000 Binary files a/course2/sem3/homework/tasks.pdf and /dev/null differ diff --git a/course2/sem3/practice/assets/1.png b/course2/sem3/practice/archive/assets/1.png similarity index 100% rename from course2/sem3/practice/assets/1.png rename to course2/sem3/practice/archive/assets/1.png diff --git a/course2/sem3/practice/archive/solutions.pdf b/course2/sem3/practice/archive/solutions.pdf new file mode 100644 index 0000000..88864c1 Binary files /dev/null and b/course2/sem3/practice/archive/solutions.pdf differ diff --git a/course2/sem3/practice/archive/solutions.typ b/course2/sem3/practice/archive/solutions.typ new file mode 100644 index 0000000..85e81cb --- /dev/null +++ b/course2/sem3/practice/archive/solutions.typ @@ -0,0 +1,523 @@ +#set page(numbering: "- 1 -") +#set text(size: 1.3em) + +#align(center)[= _Задачи_] + +#align(center)[=== _Закон Кулона. Принцип суперпозиции_] + +*1*. _На шелковой нити подвешен шар массы $m$, заряд которого $q_1^+$. Рассчитать на какое расстояние необходимо поднести положительно заряженный шар, с зарядом $q_2^+$, чтобы сила натяжения нити уменьшилась вдвое._ + +*Решение*: + +На шар действуют следующие силы: + +- Сила тяжести: $F_g = m g$ (действует вниз) +- Сила натяжения нити: $T$ (действует вдоль нити) +- Сила электрического отталкивания $F_e = k frac(q_1 q_2, l^2)$ (действует горизонтально, если шар $q_2$ подносят сбоку) + +Теперь нить отклонилась на угол $theta$ с вертикалью (под действием электрической силы). + +Найдем компоненты: + +- Вертикальная: $T cos theta = m g$ +- Горизонтальная: $T sin theta = F_e = k frac(q_1 q_2, l^2)$ + +Так как натяжение нити уменьшилось вдвое: $T = frac(m g, 2)$. + +Подставим в вертикальную компоненту: + +$ +T cos theta = m g arrow.double frac(m g, 2) cos theta = m g arrow.double cos theta = 2 +$ + +Так как $cos theta$ не может быть больше $1$, мы понимаем, что что-то не так... + +Используем теорему Пифагора для сил: + +$ +T = sqrt((m g)^2 + F^2_e) +$ + +- Изначально $F_e = 0 arrow.double T_0 = m g$ +- Теперь $T = frac(m g, 2)$ + +$ +T = sqrt((m g)^2 + F_e^2) = frac(m g, 2) arrow.double F_e^2 + (m g)^2 = (frac(m g, 2))^2 \ +F_e^2 = (frac(m g, 2))^2 - (m g)^2 = frac(m^2 g^2, 4) - m^2 g^2 = -frac(3m^2 g^2, 4) +$ + +Получилось отрицательное число... + +То есть: + +$ +F_e = T_0 - T = m g - frac(m g, 2) = frac(m g, 2) +$ + +$ +F_e = k frac(q_1 q_2, l^2) = frac(m g, 2) +$ + +Отсюда: + +$ +l^2 = frac(2 k q_1 q_2, m g) arrow.double l = sqrt(frac(2 k q_1 q_2, m g)) +$ + +*Ответ*: $l = sqrt(frac(2 k q_1^+ q_2^+, m g))$ + +#line(length: 100%) + +*2*. _К потолку в одной точке на шелковых нитях длины $l$ подвешены два одинаковых шара обладающих одинаковым зарядом $q$ и массой $m$. Раастояние между шарами $x lt.double l$. Рассчитать скорость утечки зарядов $frac(d q, d t)$ с каждого шара, если скорость их сближения, как функция от $x$ имеет виды: $v(x) = frac(alpha, sqrt(x))$ ($alpha$ - некоторая постоянная)._ + +*Решение*: + +Для одного шарика вертикальная и горизонтальная составляющие сил дают: + +- вертикальная: $T cos theta = m g$ +- горизонтальная: $T sin theta = F_e$ + +гдe $F_e$ - кулоновская сила отталкивания между шариками: + +$ +F_e = k frac(q^2, x^2) +$ + +При малом отклонении $sin theta approx tan theta approx frac(x/2, l) = frac(x, 2 l)$ (смещение одного шарика по горизонтали равно $x/2$). Подставим $T approx m g$ (поскольку $cos theta approx 1$) в горизонтальное уравнение: + +$ +m g dot frac(x, 2 l) approx F_e = k frac(q^2, x^2) +$ + +Отсюда найдем зависимость $q$ от $x$: + +$ +k frac(q^2, x^2) = frac(m g x, 2 l) arrow.double q^2 = frac(m g, 2 k l) x^3 +$ + +Значит + +$ +q(x) = sqrt(frac(m g, 2 k l)) x^(3/2) +$ + +По цепному правилу: + +$ +frac(d q, d t) = frac(d q, d x) dot frac(d x, d t). +$ + +Вычислим производную по $x$: + +$ +frac(d q, d x) = sqrt(frac(m g, 2 k l)) dot 3/2 x^(1/2). +$ + +Теперь используем заданную скорость. Поскольку $v(x)$ дана как модуль скорости сближения, скорость изменения расстояния $x$ равна + +$ +frac(d x, d t) = -frac(alpha, sqrt(x)) +$ + +Подставляем: + +$ +frac(d q, d t) = 3/2 sqrt(frac(m g, 2 k l)) x^(1/2) dot (-frac(alpha, sqrt(x))) = -frac(3, 2) alpha sqrt(frac(m g, 2 k l)) +$ + + +*Ответ*: $frac(d q, d t) = frac(3 alpha, 2)sqrt(frac(m g, 2 k l))$ + +#line(length: 100%) +*3*. _Радиус векторы двух положительных зарядов $q_1$ и $q_2$ соответственно $arrow(r_1)$ и $arrow(r_2)$. Рассчитать отрицательный заряд $q_3$ и его радиус-вектор $arrow(r_3)$ точки в которую его надо поместить, чтобы сила, действующая на каждый из зарядов была равна $0$._ + +*Решение*: + +Чтобы сила на $q_1$ была нуль, векторная сумма сил от $q_2$ и $q_3$ на $q_1$ должна быть нулём. Это значит, что силы от $q_2$ и $q_3$ действуют вдоль одной линии и противоположны по направлению. Отсюда следует, что $arrow(r)_3$ лежит на прямой, проходящей через $arrow(r)_1$ и $arrow(r)_2$. Аналогично для равновесия $q_2$. Поэтому $arrow(r)_3$ лежит на отрезке между $arrow(r)_1$ и $arrow(r)_2$. + +Пусть $L = |arrow(r)_2 - arrow(r)_1|$ — расстояние между первыми двумя зарядами. Обозначим + +$ +d_(13) = |arrow(r)_3 - arrow(r)_1|, space.quad d_(23) = |arrow(r)_3 - arrow(r)_2|. +$ + +Тогда $d_(13) + d_(23) = L$ + +Уравнения равновесия + +Сила Кулона по модулю между точками $i$ и $j$ (в масштабе $k$): +$ +F_(i j) = k frac(|q_i q_j|, d_(i j)^2). +$ + +Для заряда $q_1$: силы от $q_2$ и $q_3$ должны компенсировать друг друга, значит по модулю +$ +k frac(q_1 q_2, L^2) = k frac(q_1 |q_3|, d_(13)^2). +$ +Отсюда (сокращая $k$ и $q_1$, $q_1 > 0$): +$ +frac(q_2, L^2) = frac(|q_3|, d_(13)^2). +$ +Помня, что $q_3$ отрицателен, можно записать +$ +q_3 = -q_2 frac(d_(13)^2, L^2). space.quad space.quad (1) +$ + +Аналогично для заряда $q_2$: +$ +k frac(q_1 q_2, L^2) = k frac(q_2 |q_3|, d_(23)^2), +$ +откуда (сократив $k$ и $q_2$): +$ +frac(q_1, L^2) = frac(|q_3|, d_(23)^2) space.quad arrow.double space.quad q_3 = -q_1 frac(d_(23)^2, L^2). space.quad space.quad (2) +$ + +Приравняем правые части (1) и (2) — обе равны $q_3$: +$ +-q_2 frac(d_(13)^2, L^2) = -q_1 frac(d_(23)^2, L^2) space.quad arrow.double space.quad q_2 d_(13)^2 = q_1 d_(23)^2. +$ + +Возвращаемся к параметризации расстояний: положим $d_(13) = t$. Тогда $d_(23) = L - t$. Подставим: +$ +q_2 t^2 = q_1 (L-t)^2. +$ + +Возьмём корни (положительные, так как $t gt 0$, $L - t > 0$): +$ +sqrt(q_2), t = sqrt(q_1) (L-t). +$ + +Решаем относительно $t$: +$ +t(sqrt(q_2) + sqrt(q_1)) = sqrt(q_1) L space.quad arrow.double space.quad t = frac(sqrt(q_1), sqrt(q_1) + sqrt(q_2)) L. +$ + +Теперь вернёмся к векторной форме: точка $arrow(r)_3$ находится на отрезке $arrow(r)_1 arrow arrow(r)_2$ на расстоянии $t$ от $arrow(r)_1$. Значит +$ +arrow(r)_3 = arrow(r)_1 + frac(t, L)(arrow(r)_2 - arrow(r)_1) = arrow(r)_1 + frac(sqrt(q_1), sqrt(q_1) + sqrt(q_2))(arrow(r)_2 - arrow(r)_1). +$ + +Разворачивая: + +$ +arrow(r)_3 = frac(sqrt(q_1) arrow(r)_2 + sqrt(q_2) arrow(r)_1 , sqrt(q_1) + sqrt(q_2)). +$ + +Наконец, подставим $t/L = frac(sqrt(q_1), sqrt(q_1) + sqrt(q_2))$ в (1) для $q_3$: +$ +q_3 = -q_2 (t/L)^2 = -q_2(frac(sqrt(q_1), sqrt(q_1) + sqrt(q_2)))^2 = -frac(q_1 q_2, (sqrt(q_1) + sqrt(q_2))^2). +$ + +*Ответ*: $q_3 = -frac(q_1 q_2, (sqrt(q_1) + sqrt(q_2))^2), arrow(r)_3 = frac(sqrt(q_1) arrow(r)_2 + sqrt(q_2) arrow(r)_1, sqrt(q_1) + sqrt(q_2))$ + +#line(length: 100%) + +*4*. _Точечный заряд $q = 50$ мкКл расположен в точке с радиус-вектором $arrow(r)_0 = 2 arrow(i) + 3 arrow(j)$. Найти напряженность $arrow(E)$ электрического поля и ее модуль в точке с радиус-вектором $arrow(r) = 8 arrow(i) - 5 arrow(j)$. Координаты векторов заданы в метрах._ + +*Решение*: + +Найдём вектор $arrow(R)$ — радиус-вектор от заряда до точки наблюдения: +$ +arrow(R) = arrow(r) - arrow(r)_0 = (8-2, -5-3) = (6, -8) "м". +$ + +Длина вектора +$ +R=|arrow(R)| = sqrt(6^2+(-8)^2) = sqrt(36+64) = sqrt(100) = 10 "м". +$ + +Постоянная Кулона (в СИ): +$ +k = frac(1, 4 pi epsilon_0) approx 8.9875517923 times 10^9 frac("Н" dot "м"^2, "Кл"^2). +$ + +Поле точечного заряда: +$ +arrow(E)(arrow(r)) = k q frac(arrow(R), R^3). +$ + +Вычислим по шагам. + +1. $k q = 8.9875517923 times 10^9 dot 50 times 10^(-6) = 449377.589615$. + +2. $R^3 = 10^3 = 1000$. + +3. множитель перед вектором $arrow(R)$: +$ +frac(k q, R^3) = frac(449377.589615, 1000) = 449.377589615. +$ + +4. Компоненты поля: +$ +E_x = 449.377589615 dot 6 = 2696.26553769 "В/м", +$ +$ +E_y = 449.377589615 dot (-8) = -3595.02071692 "В/м". +$ + +5. Модуль поля: +$ +|arrow(E)| = sqrt(E_x^2 + E_y^2) = sqrt(2696.2655^2 + (-3595.0207)^2) = 4493.77589615 "В/м". +$ + +*Ответ*: $E = 4.5 "кВ/м"; arrow(E) = 2.7 arrow(i) - 3.6 arrow(j)$ + +#line(length: 100%) +*5*. _Точечные заряды $q^((+))$ и $q^((-))$ расположены по углам квадрата, диагональ которого равна $2 l$. Найти модуль напряженности электрического поля в точке, отстоящей на расстояние $x$ от плоскости квадрата, симметрично относительно его вершин._ + +#align(center)[#image("assets/1.png")] + +*Решение*: + +Положение вершин и параметры + +Диагональ квадрата = $2 l$. Пусть центр квадрата — начало координат, стороны параллельны осям $x, y$. Тогда координаты вершин можно взять как +$ +(plus.minus a, plus.minus a, 0), +$ +где +$ +a = frac(l, sqrt(2)) +$ + +Пусть заряды: + +- в верхних вершинах ($ -a, +a, 0$) и ($+a, +a, 0$) — по $+q$; +- в нижних вершинах ($ -a, -a, 0$) и ($+a, -a, 0$) — по $-q$. + +Точка наблюдения (вершина «пирамиды») находится на оси, проходящей через центр и перпендикулярно плоскости квадрата: +$ +P(0,0,x). +$ + +Расстояние от любой вершины до точки $P$: +$ +R = sqrt(a^2+a^2+x^2) = sqrt(2a^2+x^2) = sqrt(l^2+x^2). +$ + +Поле от одной вершины — векторная форма + +Поле точечного заряда $q_i$ в точке $P$ равно +$ +arrow(E)_i = k frac(q_i, R^3) arrow(R)_i, +$ +где $arrow(R)_i$ — вектор от вершины к точке $P$. + +Возьмём, например, вершину ($+a,+a,0$) с зарядом $+q$. Тогда +$ +arrow(R) = (0 - a, 0 - a, x - 0) = (-a, -a, x). +$ +Компоненты поля от этой вершины: +$ +E_x^((1)) = k frac(q(-a), R^3), space.quad +E_y^((1)) = k frac(q(-a), R^3), space.quad +E_z^((1)) = k frac(q x, R^3). +$ + +Аналогично для остальных вершин — запишем вклады по компонентам и просуммируем, учитывая знаки зарядов. + +Суммирование вкладов — симметрия + +Из симметрии видно: + +- $x$ - компоненты от вершин попарно отменяются (пары ($+a,+a$) и ($+a,-a$) дают противоположные $x$-компоненты с одинаковыми коэффициентами и одинаковыми зарядами по модулю — суммарно ноль). +- вертикальные ($z$) компоненты: верхние вершины дают вклад $+ k frac(q x, R^3)$ каждая, нижние — дают вклад $-q$ каждое, то есть вклад нижних равен $-k frac(q x, R^3)$ для каждой; суммарно $E_z = k frac(q x, R^3) + k frac(q x, R^3) - k frac(q x, R^3) - k frac(q x, R^3) = 0$. Иначе говоря, вертикальные компоненты компенсируются, потому что суммарный заряд равен нулю (две + и две −). +- $y$ - компоненты не компенсируются, а складываются с одинаковым знаком. Посчитаем их. + +Возьмём по очереди все четыре угла. Для вершины ($+a,+a$) с $+q$ её $y$ - компонента равна +$ +E_y^((+a,+a)) = k frac(q(-a), R^3). +$ +Для вершины ($-a,+a$) с $+q$: $E_y^((-a,+a)) = k frac(q(-a), R^3)$. Их сумма даёт $2 dot k frac(q(-a), R^3)$. + +Для нижних вершин с зарядом $-q$: + +- вершина ($+a,-a$): вектор $arrow(R) = (-a, +a, x)$ (заметим: её $y$ - компонента равна $+a$), и заряд $-q$ даёт вклад $E_y^((+a,-a)) = (-q) dot k frac(+a, R^3) = -k frac(q a, R^3)$. +- вершина ($-a,-a$): аналогично даёт $-k frac(q a, R^3)$. + +Сумма вкладов от нижних вершин: $-2 k frac(q a, R^3)$. Но обратим внимание: при записи выше знаки «минус» в координатах и знак заряда дают плюс в результате (следует внимательно проследить направление векторов). Если аккуратно пройти по всем четырём, то итоговая сумма $y$-компонент равна +$ +E_y = 4 k frac(q a, R^3). +$ +(Короткая проверка знаков: для верхних вершин $y$-вклад направлен в отрицательную сторону $y$ (т.к. вектор от вершины к точке имеет $y$-компонент $-a$), а для нижних вершин заряд отрицательный, и отрицательный заряд умноженный на положительную геометрическую $y$-компонент даёт тоже отрицательный вклад; все четыре вклада ориентированы в одну сторону — поэтому складываются.) + +Компоненты $x$ суммарно 0, $z$ суммарно 0, остаётся единственная ненулевая компонента $y$. + +Подставляем $a = frac(l, sqrt(2))$ и $R = sqrt(l^2+x^2)$ + +$ +E = |E_y| = 4 k frac(q a, R^3) = 4 k frac(q, R^3) dot frac(l, sqrt(2)) = k frac(4, sqrt(2)) frac(q l, (l^2 + x^2)^(3/2)). +$ +Упростим коэффициент: +$ +frac(4, sqrt(2)) = 2 sqrt(2). +$ + +*Ответ*: $E = k frac(2 sqrt(2) q l, (l^2 + x^2)^(3/2))$ + +#line(length: 100%) +/* +**. __ + +*Решение*: + +*Ответ*: + +#line(length: 100%) +**. __ + +*Решение*: + +*Ответ*: + +#line(length: 100%) +**. __ + +*Решение*: + +*Ответ*: + +#line(length: 100%) +**. __ + +*Решение*: + +*Ответ*: + +#line(length: 100%) +**. __ + +*Решение*: + +*Ответ*: + +#line(length: 100%) +**. __ + +*Решение*: + +*Ответ*: + +#line(length: 100%) +**. __ + +*Решение*: + +*Ответ*: + +#line(length: 100%) +**. __ + +*Решение*: + +*Ответ*: + +#line(length: 100%) +**. __ + +*Решение*: + +*Ответ*: + +#line(length: 100%) +**. __ + +*Решение*: + +*Ответ*: + +#line(length: 100%) +**. __ + +*Решение*: + +*Ответ*: + +#line(length: 100%) +**. __ + +*Решение*: + +*Ответ*: + +#line(length: 100%) +**. __ + +*Решение*: + +*Ответ*: + +#line(length: 100%) +*/ + +#align(center)[=== _Расчет напряженности непрерывного распределения заряда на основе теоремы Гаусса_] + +*1*. _Напряженность электрического поля, как функция координат имеет вид: $arrow(E) = frac(alpha x arrow(i) + alpha y arrow(j), x^2 + y^2)$, где $alpha = "const"$, a $arrow(i), arrow(j)$ - орты координатных осей $O X$ и $O Y$ соответственно. Найти поток вектора $arrow(E)$ через сферу радиуса $R$ с центром в начале координат._ + +*Решение*: Преобразуем $arrow(E)$: + +$ +arrow(E) = frac(alpha x arrow(i) + alpha y arrow(j), x^2 + y^2) = frac(alpha (x arrow(i) + y arrow(j)), x^2 + y^2) = frac(alpha arrow(r), r^2) +$ + +Будем использовать сферические координаты: + +$ +x = R sin theta cos phi \ +y = R sin theta sin phi \ +z = R cos theta +$ + + +*Ответ*: $P = 4 pi alpha R$. + +#line(length: 100%) + +*2*. _Объемная плотность положительно заряженного шара радиуса $R$ зависит только от расстояния до центра шара: $rho(r) = rho_0 (1 - r/R)$, где $rho_0 = "const"$. Найти:_ + +- _модуль напряженности электрического поля внутри и вне шара, как функцию $r$;_ +- _максимальное значение модуля напряженности $E_max$ и соответствующее ему значение $r_max$._ + +_Диэлектрическая проницаемость всюду $epsilon = 1$._ + +*Ответ*: $E_r (r lt.eq R) = frac(rho_0 r, 3 epsilon_0) (1 - frac(3 r, 4 R)), space.quad E_r (r gt.eq R) = frac(rho_0 R^3, 12 epsilon_0 r^2), space.quad r_max = 2/3 R, space.quad E_r (r_max) = frac(rho_0 R, 9 epsilon_0)$. + +*3*. _Система состоит из равномерно заряженного шара радиуса $R = 0.2 " м"$, объемная плотность которого $rho = 20 " нКл/м"^3$. Рассчитать модуль напряженности электрического поля:_ + +- _на расстоянии $r = 0.1 " м"$ от центра шара;_ +- _на поверхности шара;_ +- _на расстоянии $r = 0.25 " м"$ от центра шара;_ + +_Диэлектрическая проницаемость материала из которого состоит шар $epsilon = 5$._ + +*Ответ*: $E(0.1) approx 15 " В/м"$, $E(0.2) approx 30 " В/м" (r lt.eq R)$, $E(0.25) approx 96 " В/м"$, $E(0.2) approx 151 " В/м" (r gt.eq R)$. + +*4*. _Шар радиуса $R$ заряженный равномерно помещен в некоторую среду диэлектрическая проницаемость которой $epsilon = 1$. Среда заполнена зарядом, объемная плотность которого $rho = alpha/r$, где $alpha$ - постоянная, а $r$ - расстояние от центра шара. Рассчитать заряд шара при котором модуль напряженности электрического поля вне шара не зависит от $r$._ + +*Ответ*: $q = 2 pi alpha R^2$. + +*5*. _Система представлена областью пространства. По пространству распределен заряд, плотность которого зависит от расстояния до центра по закону $rho = rho_0 "exp"(-alpha r^3)$, где $alpha$ некоторая постоянная. Найти модуль напряженности, как функцию $r$._ + +*Ответ*: $E_r = frac(rho_0, 3 epsilon_0 alpha r^2) (1 - "exp"(-alpha r^3))$. + +*6*. _Рассчитать напряженность электрического поля бесконечной плоскости, заряженной равномерно. Поверхностная плотность заряда - $sigma$. Расчет произвести 2-мя способами:_ + +- _с использованием закона Кулона;_ +- _с использованием теоремы Гаусса._ + +*Ответ*: $arrow(E) = frac(sigma, 2 epsilon_0) arrow(n)$. + +*7*. _Рассчитать напряженность электростатического поля создаваемого бесконечно длинной нитью, заряженной равномерно. Поверхностная плотность заряда - $lambda$. Расчет произвести 2-мя способами:_ + +- _с использованием закона Кулона;_ +- _с использованием теоремы Гаусса._ + +*Ответ*: $arrow(E) = frac(lambda, 2 pi epsilon_0 r) arrow(n)$. + +*8*. _Рассчитать вектор напряженности электростатического поля в области пересечения двух шаров, равномерно заполненными разноименными зарядами с объемной плотностью $rho$ и $-rho$. Расстояния между центрами шаров характеризуется вектором $a$._ + +*Ответ*: $arrow(E) = frac(rho, 3 epsilon_0) arrow(a)$. + +*9*. _Напряжённость аксиально симметричного электростатическое поля зависит от расстояния до источника по закону $arrow(E) = frac(alpha, r^2) arrow(r)$ ($alpha$ - постоянная). Рассчитать заряд внутри сферы радиуса $R$, центр которой расположен на источнике._ + + diff --git a/course2/sem3/practice/archive/tasks.pdf b/course2/sem3/practice/archive/tasks.pdf new file mode 100644 index 0000000..cbf0060 Binary files /dev/null and b/course2/sem3/practice/archive/tasks.pdf differ diff --git a/course2/sem3/practice/solutions.pdf b/course2/sem3/practice/solutions.pdf index 88864c1..128f018 100644 Binary files a/course2/sem3/practice/solutions.pdf and b/course2/sem3/practice/solutions.pdf differ diff --git a/course2/sem3/practice/solutions.typ b/course2/sem3/practice/solutions.typ index 85e81cb..5ee9fa4 100644 --- a/course2/sem3/practice/solutions.typ +++ b/course2/sem3/practice/solutions.typ @@ -1,523 +1,383 @@ -#set page(numbering: "- 1 -") -#set text(size: 1.3em) +#set text( + font: "New Computer Modern", + size: 14pt +) -#align(center)[= _Задачи_] +#set page( + paper: "a4", + numbering: "1" +) -#align(center)[=== _Закон Кулона. Принцип суперпозиции_] +#set par( + justify: true, + leading: 0.52em +) -*1*. _На шелковой нити подвешен шар массы $m$, заряд которого $q_1^+$. Рассчитать на какое расстояние необходимо поднести положительно заряженный шар, с зарядом $q_2^+$, чтобы сила натяжения нити уменьшилась вдвое._ +#outline( + title: "Содержание" +) -*Решение*: +#pagebreak() -На шар действуют следующие силы: +#align(center)[= Постоянное магнитное поле] -- Сила тяжести: $F_g = m g$ (действует вниз) -- Сила натяжения нити: $T$ (действует вдоль нити) -- Сила электрического отталкивания $F_e = k frac(q_1 q_2, l^2)$ (действует горизонтально, если шар $q_2$ подносят сбоку) +#align(center)[=== Индукция магнитного поля. Закон Био-Савара] -Теперь нить отклонилась на угол $theta$ с вертикалью (под действием электрической силы). +#align(center)[===== №1] -Найдем компоненты: - -- Вертикальная: $T cos theta = m g$ -- Горизонтальная: $T sin theta = F_e = k frac(q_1 q_2, l^2)$ - -Так как натяжение нити уменьшилось вдвое: $T = frac(m g, 2)$. - -Подставим в вертикальную компоненту: - -$ -T cos theta = m g arrow.double frac(m g, 2) cos theta = m g arrow.double cos theta = 2 -$ - -Так как $cos theta$ не может быть больше $1$, мы понимаем, что что-то не так... - -Используем теорему Пифагора для сил: - -$ -T = sqrt((m g)^2 + F^2_e) -$ - -- Изначально $F_e = 0 arrow.double T_0 = m g$ -- Теперь $T = frac(m g, 2)$ - -$ -T = sqrt((m g)^2 + F_e^2) = frac(m g, 2) arrow.double F_e^2 + (m g)^2 = (frac(m g, 2))^2 \ -F_e^2 = (frac(m g, 2))^2 - (m g)^2 = frac(m^2 g^2, 4) - m^2 g^2 = -frac(3m^2 g^2, 4) -$ - -Получилось отрицательное число... - -То есть: - -$ -F_e = T_0 - T = m g - frac(m g, 2) = frac(m g, 2) -$ - -$ -F_e = k frac(q_1 q_2, l^2) = frac(m g, 2) -$ - -Отсюда: - -$ -l^2 = frac(2 k q_1 q_2, m g) arrow.double l = sqrt(frac(2 k q_1 q_2, m g)) -$ - -*Ответ*: $l = sqrt(frac(2 k q_1^+ q_2^+, m g))$ - -#line(length: 100%) - -*2*. _К потолку в одной точке на шелковых нитях длины $l$ подвешены два одинаковых шара обладающих одинаковым зарядом $q$ и массой $m$. Раастояние между шарами $x lt.double l$. Рассчитать скорость утечки зарядов $frac(d q, d t)$ с каждого шара, если скорость их сближения, как функция от $x$ имеет виды: $v(x) = frac(alpha, sqrt(x))$ ($alpha$ - некоторая постоянная)._ - -*Решение*: - -Для одного шарика вертикальная и горизонтальная составляющие сил дают: - -- вертикальная: $T cos theta = m g$ -- горизонтальная: $T sin theta = F_e$ - -гдe $F_e$ - кулоновская сила отталкивания между шариками: - -$ -F_e = k frac(q^2, x^2) -$ - -При малом отклонении $sin theta approx tan theta approx frac(x/2, l) = frac(x, 2 l)$ (смещение одного шарика по горизонтали равно $x/2$). Подставим $T approx m g$ (поскольку $cos theta approx 1$) в горизонтальное уравнение: - -$ -m g dot frac(x, 2 l) approx F_e = k frac(q^2, x^2) -$ - -Отсюда найдем зависимость $q$ от $x$: - -$ -k frac(q^2, x^2) = frac(m g x, 2 l) arrow.double q^2 = frac(m g, 2 k l) x^3 -$ - -Значит - -$ -q(x) = sqrt(frac(m g, 2 k l)) x^(3/2) -$ - -По цепному правилу: - -$ -frac(d q, d t) = frac(d q, d x) dot frac(d x, d t). -$ - -Вычислим производную по $x$: - -$ -frac(d q, d x) = sqrt(frac(m g, 2 k l)) dot 3/2 x^(1/2). -$ - -Теперь используем заданную скорость. Поскольку $v(x)$ дана как модуль скорости сближения, скорость изменения расстояния $x$ равна - -$ -frac(d x, d t) = -frac(alpha, sqrt(x)) -$ - -Подставляем: - -$ -frac(d q, d t) = 3/2 sqrt(frac(m g, 2 k l)) x^(1/2) dot (-frac(alpha, sqrt(x))) = -frac(3, 2) alpha sqrt(frac(m g, 2 k l)) -$ - - -*Ответ*: $frac(d q, d t) = frac(3 alpha, 2)sqrt(frac(m g, 2 k l))$ - -#line(length: 100%) -*3*. _Радиус векторы двух положительных зарядов $q_1$ и $q_2$ соответственно $arrow(r_1)$ и $arrow(r_2)$. Рассчитать отрицательный заряд $q_3$ и его радиус-вектор $arrow(r_3)$ точки в которую его надо поместить, чтобы сила, действующая на каждый из зарядов была равна $0$._ +*Условие*: Заряженная элементарная частица движется со скоростью, модуль которой $v eq 900 "м/c"$. В некоторый момент в точке наблюдения $P$ модуль напряжённости электрического поля этой частицы $E eq 600 "В/м"$, а угол между векторами скорости и напряжённости $alpha eq 30 degree$. Рассчитать индукцию магнитного поля данной частицы. *Решение*: -Чтобы сила на $q_1$ была нуль, векторная сумма сил от $q_2$ и $q_3$ на $q_1$ должна быть нулём. Это значит, что силы от $q_2$ и $q_3$ действуют вдоль одной линии и противоположны по направлению. Отсюда следует, что $arrow(r)_3$ лежит на прямой, проходящей через $arrow(r)_1$ и $arrow(r)_2$. Аналогично для равновесия $q_2$. Поэтому $arrow(r)_3$ лежит на отрезке между $arrow(r)_1$ и $arrow(r)_2$. - -Пусть $L = |arrow(r)_2 - arrow(r)_1|$ — расстояние между первыми двумя зарядами. Обозначим - -$ -d_(13) = |arrow(r)_3 - arrow(r)_1|, space.quad d_(23) = |arrow(r)_3 - arrow(r)_2|. -$ - -Тогда $d_(13) + d_(23) = L$ +*Ответ*: $B eq 3 "пТл"$. -Уравнения равновесия - -Сила Кулона по модулю между точками $i$ и $j$ (в масштабе $k$): -$ -F_(i j) = k frac(|q_i q_j|, d_(i j)^2). -$ - -Для заряда $q_1$: силы от $q_2$ и $q_3$ должны компенсировать друг друга, значит по модулю -$ -k frac(q_1 q_2, L^2) = k frac(q_1 |q_3|, d_(13)^2). -$ -Отсюда (сокращая $k$ и $q_1$, $q_1 > 0$): -$ -frac(q_2, L^2) = frac(|q_3|, d_(13)^2). -$ -Помня, что $q_3$ отрицателен, можно записать -$ -q_3 = -q_2 frac(d_(13)^2, L^2). space.quad space.quad (1) -$ - -Аналогично для заряда $q_2$: -$ -k frac(q_1 q_2, L^2) = k frac(q_2 |q_3|, d_(23)^2), -$ -откуда (сократив $k$ и $q_2$): -$ -frac(q_1, L^2) = frac(|q_3|, d_(23)^2) space.quad arrow.double space.quad q_3 = -q_1 frac(d_(23)^2, L^2). space.quad space.quad (2) -$ +#align(center)[===== №2] -Приравняем правые части (1) и (2) — обе равны $q_3$: -$ --q_2 frac(d_(13)^2, L^2) = -q_1 frac(d_(23)^2, L^2) space.quad arrow.double space.quad q_2 d_(13)^2 = q_1 d_(23)^2. -$ - -Возвращаемся к параметризации расстояний: положим $d_(13) = t$. Тогда $d_(23) = L - t$. Подставим: -$ -q_2 t^2 = q_1 (L-t)^2. -$ +*Условие*: Используя закон Био-Савара, получить формулу для рассчёта модуля вектора индукции магнитного поля, создаваемого током $I$, протекающем в линейном бесконечном проводнике в точке, расположенной на расстоянии $r_0$ от проводника. -Возьмём корни (положительные, так как $t gt 0$, $L - t > 0$): -$ -sqrt(q_2), t = sqrt(q_1) (L-t). -$ +*Решение*: -Решаем относительно $t$: -$ -t(sqrt(q_2) + sqrt(q_1)) = sqrt(q_1) L space.quad arrow.double space.quad t = frac(sqrt(q_1), sqrt(q_1) + sqrt(q_2)) L. -$ +*Ответ*: $B eq frac(mu_0 I, 2 pi r_0)$. -Теперь вернёмся к векторной форме: точка $arrow(r)_3$ находится на отрезке $arrow(r)_1 arrow arrow(r)_2$ на расстоянии $t$ от $arrow(r)_1$. Значит -$ -arrow(r)_3 = arrow(r)_1 + frac(t, L)(arrow(r)_2 - arrow(r)_1) = arrow(r)_1 + frac(sqrt(q_1), sqrt(q_1) + sqrt(q_2))(arrow(r)_2 - arrow(r)_1). -$ -Разворачивая: +#align(center)[===== №3] -$ -arrow(r)_3 = frac(sqrt(q_1) arrow(r)_2 + sqrt(q_2) arrow(r)_1 , sqrt(q_1) + sqrt(q_2)). -$ +*Условие*: Рассчитать модуль вектора индукции магнитного поля, создаваемого конечным прямолинейным участком проводника, длины $l$, по которому протекает ток $I$, в точке отстоящей на произвольном расстоянии $r_0$ от оси проводника. -Наконец, подставим $t/L = frac(sqrt(q_1), sqrt(q_1) + sqrt(q_2))$ в (1) для $q_3$: -$ -q_3 = -q_2 (t/L)^2 = -q_2(frac(sqrt(q_1), sqrt(q_1) + sqrt(q_2)))^2 = -frac(q_1 q_2, (sqrt(q_1) + sqrt(q_2))^2). -$ - -*Ответ*: $q_3 = -frac(q_1 q_2, (sqrt(q_1) + sqrt(q_2))^2), arrow(r)_3 = frac(sqrt(q_1) arrow(r)_2 + sqrt(q_2) arrow(r)_1, sqrt(q_1) + sqrt(q_2))$ - -#line(length: 100%) - -*4*. _Точечный заряд $q = 50$ мкКл расположен в точке с радиус-вектором $arrow(r)_0 = 2 arrow(i) + 3 arrow(j)$. Найти напряженность $arrow(E)$ электрического поля и ее модуль в точке с радиус-вектором $arrow(r) = 8 arrow(i) - 5 arrow(j)$. Координаты векторов заданы в метрах._ - -*Решение*: - -Найдём вектор $arrow(R)$ — радиус-вектор от заряда до точки наблюдения: -$ -arrow(R) = arrow(r) - arrow(r)_0 = (8-2, -5-3) = (6, -8) "м". -$ - -Длина вектора -$ -R=|arrow(R)| = sqrt(6^2+(-8)^2) = sqrt(36+64) = sqrt(100) = 10 "м". -$ - -Постоянная Кулона (в СИ): -$ -k = frac(1, 4 pi epsilon_0) approx 8.9875517923 times 10^9 frac("Н" dot "м"^2, "Кл"^2). -$ - -Поле точечного заряда: -$ -arrow(E)(arrow(r)) = k q frac(arrow(R), R^3). -$ - -Вычислим по шагам. - -1. $k q = 8.9875517923 times 10^9 dot 50 times 10^(-6) = 449377.589615$. - -2. $R^3 = 10^3 = 1000$. - -3. множитель перед вектором $arrow(R)$: -$ -frac(k q, R^3) = frac(449377.589615, 1000) = 449.377589615. -$ - -4. Компоненты поля: -$ -E_x = 449.377589615 dot 6 = 2696.26553769 "В/м", -$ -$ -E_y = 449.377589615 dot (-8) = -3595.02071692 "В/м". -$ - -5. Модуль поля: -$ -|arrow(E)| = sqrt(E_x^2 + E_y^2) = sqrt(2696.2655^2 + (-3595.0207)^2) = 4493.77589615 "В/м". -$ +*Решение*: -*Ответ*: $E = 4.5 "кВ/м"; arrow(E) = 2.7 arrow(i) - 3.6 arrow(j)$ +*Ответ*: $B eq frac(mu_0 I, 4 pi r_0) (cos alpha_1 plus cos alpha_2)$. -#line(length: 100%) -*5*. _Точечные заряды $q^((+))$ и $q^((-))$ расположены по углам квадрата, диагональ которого равна $2 l$. Найти модуль напряженности электрического поля в точке, отстоящей на расстояние $x$ от плоскости квадрата, симметрично относительно его вершин._ -#align(center)[#image("assets/1.png")] +#align(center)[===== №4] -*Решение*: +*Условие*: Замкнутый контур с током имеет вид прямоугольника с диагональю $d eq 16 "см"$, угол между диагоналями $alpha eq 30 degree$. Сила тока, протекающего по контуру $I eq 5 "A"$. Рассчитать модуль индукции магнитного поля в центре контура. -Положение вершин и параметры +*Решение*: -Диагональ квадрата = $2 l$. Пусть центр квадрата — начало координат, стороны параллельны осям $x, y$. Тогда координаты вершин можно взять как -$ -(plus.minus a, plus.minus a, 0), -$ -где -$ -a = frac(l, sqrt(2)) -$ +*Ответ*: $B eq 0.1 "мТл"$. -Пусть заряды: -- в верхних вершинах ($ -a, +a, 0$) и ($+a, +a, 0$) — по $+q$; -- в нижних вершинах ($ -a, -a, 0$) и ($+a, -a, 0$) — по $-q$. +#align(center)[===== №5] -Точка наблюдения (вершина «пирамиды») находится на оси, проходящей через центр и перпендикулярно плоскости квадрата: -$ -P(0,0,x). -$ +*Условие*: -Расстояние от любой вершины до точки $P$: -$ -R = sqrt(a^2+a^2+x^2) = sqrt(2a^2+x^2) = sqrt(l^2+x^2). -$ +*Решение*: -Поле от одной вершины — векторная форма +*Ответ*: -Поле точечного заряда $q_i$ в точке $P$ равно -$ -arrow(E)_i = k frac(q_i, R^3) arrow(R)_i, -$ -где $arrow(R)_i$ — вектор от вершины к точке $P$. +#align(center)[===== №6] -Возьмём, например, вершину ($+a,+a,0$) с зарядом $+q$. Тогда -$ -arrow(R) = (0 - a, 0 - a, x - 0) = (-a, -a, x). -$ -Компоненты поля от этой вершины: -$ -E_x^((1)) = k frac(q(-a), R^3), space.quad -E_y^((1)) = k frac(q(-a), R^3), space.quad -E_z^((1)) = k frac(q x, R^3). -$ +*Условие*: -Аналогично для остальных вершин — запишем вклады по компонентам и просуммируем, учитывая знаки зарядов. +*Решение*: -Суммирование вкладов — симметрия +*Ответ*: -Из симметрии видно: -- $x$ - компоненты от вершин попарно отменяются (пары ($+a,+a$) и ($+a,-a$) дают противоположные $x$-компоненты с одинаковыми коэффициентами и одинаковыми зарядами по модулю — суммарно ноль). -- вертикальные ($z$) компоненты: верхние вершины дают вклад $+ k frac(q x, R^3)$ каждая, нижние — дают вклад $-q$ каждое, то есть вклад нижних равен $-k frac(q x, R^3)$ для каждой; суммарно $E_z = k frac(q x, R^3) + k frac(q x, R^3) - k frac(q x, R^3) - k frac(q x, R^3) = 0$. Иначе говоря, вертикальные компоненты компенсируются, потому что суммарный заряд равен нулю (две + и две −). -- $y$ - компоненты не компенсируются, а складываются с одинаковым знаком. Посчитаем их. +#align(center)[===== №7] -Возьмём по очереди все четыре угла. Для вершины ($+a,+a$) с $+q$ её $y$ - компонента равна -$ -E_y^((+a,+a)) = k frac(q(-a), R^3). -$ -Для вершины ($-a,+a$) с $+q$: $E_y^((-a,+a)) = k frac(q(-a), R^3)$. Их сумма даёт $2 dot k frac(q(-a), R^3)$. +*Условие*: -Для нижних вершин с зарядом $-q$: +*Решение*: -- вершина ($+a,-a$): вектор $arrow(R) = (-a, +a, x)$ (заметим: её $y$ - компонента равна $+a$), и заряд $-q$ даёт вклад $E_y^((+a,-a)) = (-q) dot k frac(+a, R^3) = -k frac(q a, R^3)$. -- вершина ($-a,-a$): аналогично даёт $-k frac(q a, R^3)$. +*Ответ*: -Сумма вкладов от нижних вершин: $-2 k frac(q a, R^3)$. Но обратим внимание: при записи выше знаки «минус» в координатах и знак заряда дают плюс в результате (следует внимательно проследить направление векторов). Если аккуратно пройти по всем четырём, то итоговая сумма $y$-компонент равна -$ -E_y = 4 k frac(q a, R^3). -$ -(Короткая проверка знаков: для верхних вершин $y$-вклад направлен в отрицательную сторону $y$ (т.к. вектор от вершины к точке имеет $y$-компонент $-a$), а для нижних вершин заряд отрицательный, и отрицательный заряд умноженный на положительную геометрическую $y$-компонент даёт тоже отрицательный вклад; все четыре вклада ориентированы в одну сторону — поэтому складываются.) -Компоненты $x$ суммарно 0, $z$ суммарно 0, остаётся единственная ненулевая компонента $y$. +#align(center)[===== №8] -Подставляем $a = frac(l, sqrt(2))$ и $R = sqrt(l^2+x^2)$ -$ -E = |E_y| = 4 k frac(q a, R^3) = 4 k frac(q, R^3) dot frac(l, sqrt(2)) = k frac(4, sqrt(2)) frac(q l, (l^2 + x^2)^(3/2)). -$ -Упростим коэффициент: -$ -frac(4, sqrt(2)) = 2 sqrt(2). -$ +*Условие*: -*Ответ*: $E = k frac(2 sqrt(2) q l, (l^2 + x^2)^(3/2))$ +*Решение*: -#line(length: 100%) -/* -**. __ +*Ответ*: -*Решение*: +#align(center)[===== №9] -*Ответ*: +*Условие*: -#line(length: 100%) -**. __ +*Решение*: -*Решение*: +*Ответ*: -*Ответ*: -#line(length: 100%) -**. __ +#align(center)[===== №10] -*Решение*: -*Ответ*: +*Условие*: -#line(length: 100%) -**. __ +*Решение*: -*Решение*: +*Ответ*: -*Ответ*: -#line(length: 100%) -**. __ +#align(center)[=== Закон полного тока] -*Решение*: +#align(center)[===== №1] -*Ответ*: +*Условие*: -#line(length: 100%) -**. __ +*Решение*: -*Решение*: +*Ответ*: -*Ответ*: -#line(length: 100%) -**. __ +#align(center)[===== №2] -*Решение*: +*Условие*: -*Ответ*: +*Решение*: -#line(length: 100%) -**. __ +*Ответ*: -*Решение*: -*Ответ*: +#align(center)[===== №3] -#line(length: 100%) -**. __ +*Условие*: -*Решение*: +*Решение*: -*Ответ*: +*Ответ*: -#line(length: 100%) -**. __ -*Решение*: +#align(center)[===== №4] -*Ответ*: +*Условие*: -#line(length: 100%) -**. __ +*Решение*: -*Решение*: +*Ответ*: -*Ответ*: +#align(center)[===== №5] -#line(length: 100%) -**. __ +*Условие*: -*Решение*: +*Решение*: -*Ответ*: +*Ответ*: -#line(length: 100%) -**. __ -*Решение*: +#align(center)[===== №6] -*Ответ*: +*Условие*: -#line(length: 100%) -*/ +*Решение*: -#align(center)[=== _Расчет напряженности непрерывного распределения заряда на основе теоремы Гаусса_] +*Ответ*: -*1*. _Напряженность электрического поля, как функция координат имеет вид: $arrow(E) = frac(alpha x arrow(i) + alpha y arrow(j), x^2 + y^2)$, где $alpha = "const"$, a $arrow(i), arrow(j)$ - орты координатных осей $O X$ и $O Y$ соответственно. Найти поток вектора $arrow(E)$ через сферу радиуса $R$ с центром в начале координат._ +#align(center)[===== №7] -*Решение*: Преобразуем $arrow(E)$: +*Условие*: -$ -arrow(E) = frac(alpha x arrow(i) + alpha y arrow(j), x^2 + y^2) = frac(alpha (x arrow(i) + y arrow(j)), x^2 + y^2) = frac(alpha arrow(r), r^2) -$ +*Решение*: -Будем использовать сферические координаты: +*Ответ*: -$ -x = R sin theta cos phi \ -y = R sin theta sin phi \ -z = R cos theta -$ +#align(center)[===== №8] +*Условие*: -*Ответ*: $P = 4 pi alpha R$. +*Решение*: -#line(length: 100%) +*Ответ*: -*2*. _Объемная плотность положительно заряженного шара радиуса $R$ зависит только от расстояния до центра шара: $rho(r) = rho_0 (1 - r/R)$, где $rho_0 = "const"$. Найти:_ +#align(center)[===== №9] -- _модуль напряженности электрического поля внутри и вне шара, как функцию $r$;_ -- _максимальное значение модуля напряженности $E_max$ и соответствующее ему значение $r_max$._ +*Условие*: -_Диэлектрическая проницаемость всюду $epsilon = 1$._ +*Решение*: -*Ответ*: $E_r (r lt.eq R) = frac(rho_0 r, 3 epsilon_0) (1 - frac(3 r, 4 R)), space.quad E_r (r gt.eq R) = frac(rho_0 R^3, 12 epsilon_0 r^2), space.quad r_max = 2/3 R, space.quad E_r (r_max) = frac(rho_0 R, 9 epsilon_0)$. +*Ответ*: -*3*. _Система состоит из равномерно заряженного шара радиуса $R = 0.2 " м"$, объемная плотность которого $rho = 20 " нКл/м"^3$. Рассчитать модуль напряженности электрического поля:_ -- _на расстоянии $r = 0.1 " м"$ от центра шара;_ -- _на поверхности шара;_ -- _на расстоянии $r = 0.25 " м"$ от центра шара;_ +#align(center)[=== Магнитное поле при наличии Магнетиков. Магнитный момент.] -_Диэлектрическая проницаемость материала из которого состоит шар $epsilon = 5$._ +#align(center)[===== №1] -*Ответ*: $E(0.1) approx 15 " В/м"$, $E(0.2) approx 30 " В/м" (r lt.eq R)$, $E(0.25) approx 96 " В/м"$, $E(0.2) approx 151 " В/м" (r gt.eq R)$. +*Условие*: -*4*. _Шар радиуса $R$ заряженный равномерно помещен в некоторую среду диэлектрическая проницаемость которой $epsilon = 1$. Среда заполнена зарядом, объемная плотность которого $rho = alpha/r$, где $alpha$ - постоянная, а $r$ - расстояние от центра шара. Рассчитать заряд шара при котором модуль напряженности электрического поля вне шара не зависит от $r$._ +*Решение*: -*Ответ*: $q = 2 pi alpha R^2$. +*Ответ*: -*5*. _Система представлена областью пространства. По пространству распределен заряд, плотность которого зависит от расстояния до центра по закону $rho = rho_0 "exp"(-alpha r^3)$, где $alpha$ некоторая постоянная. Найти модуль напряженности, как функцию $r$._ +#align(center)[===== №2] -*Ответ*: $E_r = frac(rho_0, 3 epsilon_0 alpha r^2) (1 - "exp"(-alpha r^3))$. +*Условие*: -*6*. _Рассчитать напряженность электрического поля бесконечной плоскости, заряженной равномерно. Поверхностная плотность заряда - $sigma$. Расчет произвести 2-мя способами:_ +*Решение*: -- _с использованием закона Кулона;_ -- _с использованием теоремы Гаусса._ +*Ответ*: -*Ответ*: $arrow(E) = frac(sigma, 2 epsilon_0) arrow(n)$. +#align(center)[===== №3] -*7*. _Рассчитать напряженность электростатического поля создаваемого бесконечно длинной нитью, заряженной равномерно. Поверхностная плотность заряда - $lambda$. Расчет произвести 2-мя способами:_ +*Условие*: -- _с использованием закона Кулона;_ -- _с использованием теоремы Гаусса._ +*Решение*: -*Ответ*: $arrow(E) = frac(lambda, 2 pi epsilon_0 r) arrow(n)$. +*Ответ*: -*8*. _Рассчитать вектор напряженности электростатического поля в области пересечения двух шаров, равномерно заполненными разноименными зарядами с объемной плотностью $rho$ и $-rho$. Расстояния между центрами шаров характеризуется вектором $a$._ +#align(center)[===== №4] -*Ответ*: $arrow(E) = frac(rho, 3 epsilon_0) arrow(a)$. +*Условие*: -*9*. _Напряжённость аксиально симметричного электростатическое поля зависит от расстояния до источника по закону $arrow(E) = frac(alpha, r^2) arrow(r)$ ($alpha$ - постоянная). Рассчитать заряд внутри сферы радиуса $R$, центр которой расположен на источнике._ +*Решение*: + +*Ответ*: + +#align(center)[===== №5] + +*Условие*: + +*Решение*: + +*Ответ*: + +#align(center)[===== №6] + +*Условие*: + +*Решение*: + +*Ответ*: + +#align(center)[===== №7] + +*Условие*: + +*Решение*: + +*Ответ*: + +#align(center)[===== №8] + +*Условие*: + +*Решение*: + +*Ответ*: + +#align(center)[===== №9] + +*Условие*: + +*Решение*: + +*Ответ*: + + +#align(center)[=== Частица в магнитном поле] + +#align(center)[===== №1] + +*Условие*: + +*Решение*: + +*Ответ*: + +#align(center)[===== №2] + +*Условие*: + +*Решение*: + +*Ответ*: + + +#align(center)[=== Работа по перемещению проводника с током в магнитном поле] + +#align(center)[===== №1] + +*Условие*: + +*Решение*: + +*Ответ*: + +#align(center)[===== №2] + +*Условие*: + +*Решение*: + +*Ответ*: + + +#align(center)[= Электромагнитная индукция] + +#align(center)[=== Индукция токов. Закон электромагнитной индукции Фарадея] + +#align(center)[===== №1] + +*Условие*: + +*Решение*: + +*Ответ*: + +#align(center)[===== №2] + +*Условие*: + +*Решение*: + +*Ответ*: + +#align(center)[===== №3] + +*Условие*: + +*Решение*: + +*Ответ*: + +#align(center)[===== №4] + +*Условие*: + +*Решение*: + +*Ответ*: + +#align(center)[===== №5] + +*Условие*: + +*Решение*: + +*Ответ*: + +#align(center)[===== №6] + +*Условие*: + +*Решение*: + +*Ответ*: + +#align(center)[===== №7] + +*Условие*: + +*Решение*: + +*Ответ*: + +#align(center)[===== №8] + +*Условие*: + +*Решение*: + +*Ответ*: + +#align(center)[===== №9] + +*Условие*: + +*Решение*: + +*Ответ*: diff --git a/course2/sem3/practice/tasks.pdf b/course2/sem3/practice/tasks.pdf index cbf0060..5b55dc0 100644 Binary files a/course2/sem3/practice/tasks.pdf and b/course2/sem3/practice/tasks.pdf differ