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Консультация № 189158
11.04.2016, 13:47
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Уважаемые эксперты! Пожалуйста, ответьте на вопрос:
цепь состоит из соленоида с сердечником, магнитная проницаемость которого равна 1000, и источника тока .соленоид длиной 15 см и диаметром 4 см имеет
плотную намотку из двух слоев медного провода диаметром 0,2 мм. По соленоиду течет ток 1 А. Определить ЭДС самоиндукции в соленоиде в тот момент времени после отключения его от источника тока, когда сила тока уменьшилась в два раза. Сопротивлением источника тока и проводящих проводов пренебречь.
цепь состоит из соленоида с сердечником, магнитная проницаемость которого равна 1000, и источника тока .соленоид длиной 15 см и диаметром 4 см имеет
плотную намотку из двух слоев медного провода диаметром 0,2 мм. По соленоиду течет ток 1 А. Определить ЭДС самоиндукции в соленоиде в тот момент времени после отключения его от источника тока, когда сила тока уменьшилась в два раза. Сопротивлением источника тока и проводящих проводов пренебречь.
Обсуждение
11.04.2016, 14:46
общий
это ответ
Здравствуйте, amelya.kuro!
Во-первых, в условии очень плохо оговорён сам процесс отключения источника тока. Видимо, следует предположить, что контакты соленоида при этом замыкаются проводом с пренебрежительно малым сопротивлением.
Найдём сопротивление самого соленоида:
имеем N0=2 слоя, в каждом N=15см/0,2мм=750 витков длиной [$8467$]0=[$960$]D=[$960$][$183$]4см (разницей диаметров обмоток можно пренебречь)
итого, длина медной проволоки в солениоде: [$8467$]=N0N[$8467$]0=2[$183$]750[$183$]4[$960$] см=1,885[$183$]105 см=188,5 м
Площадь сечения проволоки S=[$960$](d/2)2=[$960$][$183$](0,2мм/2)2=[$960$][$183$](10-4м)2=3,14[$183$]10-8 м
удельное сопротивление меди [$961$]=1,75[$183$]10-8 Ом[$183$]м
сопротивление обмоток R=[$961$][$8467$]/S=105 Ом
После отключения источника тока в любой момент времени ЭДС самоиндукции должна быть равна падению напряжения на сопротивлении обмоток
[$8496$]=IR
То есть в момент, когда I=I0/2=0,5 А, ЭДС равна
[$8496$]=0,5А[$183$]105[$937$]=52,5 В
Как видите, индуктивность соленоида нам не понадобилась. Она будет нужна для нахождения, за сколько времени сила тока уменьшится до заданного значения и сколько тепловой энергии при этом выделится.
Во-первых, в условии очень плохо оговорён сам процесс отключения источника тока. Видимо, следует предположить, что контакты соленоида при этом замыкаются проводом с пренебрежительно малым сопротивлением.
Найдём сопротивление самого соленоида:
имеем N0=2 слоя, в каждом N=15см/0,2мм=750 витков длиной [$8467$]0=[$960$]D=[$960$][$183$]4см (разницей диаметров обмоток можно пренебречь)
итого, длина медной проволоки в солениоде: [$8467$]=N0N[$8467$]0=2[$183$]750[$183$]4[$960$] см=1,885[$183$]105 см=188,5 м
Площадь сечения проволоки S=[$960$](d/2)2=[$960$][$183$](0,2мм/2)2=[$960$][$183$](10-4м)2=3,14[$183$]10-8 м
удельное сопротивление меди [$961$]=1,75[$183$]10-8 Ом[$183$]м
сопротивление обмоток R=[$961$][$8467$]/S=105 Ом
После отключения источника тока в любой момент времени ЭДС самоиндукции должна быть равна падению напряжения на сопротивлении обмоток
[$8496$]=IR
То есть в момент, когда I=I0/2=0,5 А, ЭДС равна
[$8496$]=0,5А[$183$]105[$937$]=52,5 В
Как видите, индуктивность соленоида нам не понадобилась. Она будет нужна для нахождения, за сколько времени сила тока уменьшится до заданного значения и сколько тепловой энергии при этом выделится.
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