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EN_B00000_S03252_W000196 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S03252_W000197 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S03252_W000198 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S03252_W000199 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S03252_W000200 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " Which is the same as that of an electron but the opposite sign. It's 1.6 times 10 to the negative 19 coulombs. So if we type these numbers into the calculator,",
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EN_B00000_S03252_W000201 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S03252_W000202 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S03252_W000203 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S03252_W000204 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S03252_W000205 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"speaker": "EN_B00000_S03252",
"text": " So this is equal to 2.09 times 10 to negative 14 joules. Now, once you have the energy in joules, you can convert it to electron volts. Electron volts is basically another uni... | EN_B00000_S03252_W000205.mp3 | [
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EN_B00000_S03252_W000206 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " Is equal to 1.6 times 10 to the negative 19 joules. We'll talk about why later. But that's what it's equal to.",
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EN_B00000_S03252_W000207 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"speaker": "EN_B00000_S03252",
"text": " And so you just gotta convert it to electron volts. And this will give you the answer.",
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EN_B00000_S03252_W000208 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " So why is it that 1 electron volt is equal to 1.6 times 10 to the negative 19 joules? Why is that the case?",
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EN_B00000_S03252_W000209 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " Electric potential, which is measured in volts, is equal to the electric potential energy",
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EN_B00000_S03252_W000210 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " Which is measured in joules divided by the charge.",
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EN_B00000_S03252_W000211 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " So the unit volt, 1 volt, is equal to 1 joule per 1 coulomb.",
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EN_B00000_S03252_W000212 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " An electron has a charge of 1.6 x 10-19 C. It's negative, but let's ignore the negative sign. That's the charge of an electron. And an electron that has 1 volt",
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EN_B00000_S03252_W000213 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " Or 1EV, that's 1 election vote and election with 1 vote.",
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EN_B00000_S03252_W000214 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S03252_W000215 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"duration": 10.153,
"id": "EN_B00000_S03252_W000215",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " Volt is basically the ratio between joules and coulombs. So if you have a charged particle that has one joule and one coulomb, its voltage is one volt.",
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EN_B00000_S03252_W000216 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"language": "en",
"speaker": "EN_B00000_S03252",
"text": " Or the electric potential is one volt. Voltage is really work per unit charge. Electric potential is energy per unit charge.",
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EN_B00000_S03252_W000217 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"duration": 18.455,
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"language": "en",
"speaker": "EN_B00000_S03252",
"text": " Now, if we have a charged particle that has an energy of 1.6 times 10th negative 19 joules, and a charge of that many coulombs, then these two will cancel, and the electric po... | EN_B00000_S03252_W000217.mp3 | [
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EN_B00000_S03252_W000218 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.3675,
"duration": 4.143,
"id": "EN_B00000_S03252_W000218",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " Has a charge of 1.6 times 10 to the power of 19 Coulombs.",
"wav": "EN_B00000/EN_B00000_S03252/mp3/EN_B00000_S03252_W000218.mp3"
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EN_B00000_S03252_W000219 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"duration": 8.438,
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"language": "en",
"speaker": "EN_B00000_S03252",
"text": " And if that electron has an energy of one electron volt, its voltage is one volt, which means its energy is equal to this number.",
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EN_B00000_S03252_W000220 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.3047,
"duration": 6.384,
"id": "EN_B00000_S03252_W000220",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " And that's why one electron volt is 1.6 times 10 to the negative 19 joules. Just in case you were wondering.",
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EN_B00000_S03252_W000221 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.3086,
"duration": 5.569,
"id": "EN_B00000_S03252_W000221",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " Now let's say if we have two wires parallel to each other.",
"wav": "EN_B00000/EN_B00000_S03252/mp3/EN_B00000_S03252_W000221.mp3"
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EN_B00000_S03252_W000222 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"language": "en",
"speaker": "EN_B00000_S03252",
"text": " Let's say that there's a current in wire 1 and in wire 2. And these two currents are in the same direction.",
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EN_B00000_S03252_W000223 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"speaker": "EN_B00000_S03252",
"text": " Will these two wires attract each other or will they repel? It turns out that these two wires will attract each other if they have a current in the same direction. Now, if the... | EN_B00000_S03252_W000223.mp3 | [
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EN_B00000_S03252_W000224 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"language": "en",
"speaker": "EN_B00000_S03252",
"text": " With the opposite current, then the situation is different.",
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EN_B00000_S03252_W000225 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S03252_W000227 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S03252_W000228 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S03252_W000229 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S03252_W000230 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S03252_W000231 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S03252_W000232 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S03252_W000233 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " Is you want your fingers to point into the page.",
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EN_B00000_S03252_W000234 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S03252_W000235 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " Which is b1 that's created by i1. You want that to be in the page.",
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} | EN_B00000_S03252_W000235.mp3 | [
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EN_B00000_S03252_W000236 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"duration": 5.466,
"id": "EN_B00000_S03252_W000236",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " Now what's gonna happen is, the force is gonna come out of the palm of your hand. And it turns out...",
"wav": "EN_B00000/EN_B00000_S03252/mp3/EN_B00000_S03252_W000236.mp3"
} | EN_B00000_S03252_W000236.mp3 | [
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EN_B00000_S03252_W000237 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.3674,
"duration": 5.654,
"id": "EN_B00000_S03252_W000237",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " That force is directed towards wire 1. So it's a force of attraction.",
"wav": "EN_B00000/EN_B00000_S03252/mp3/EN_B00000_S03252_W000237.mp3"
} | EN_B00000_S03252_W000237.mp3 | [
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EN_B00000_S03252_W000238 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.3108,
"duration": 5.229,
"id": "EN_B00000_S03252_W000238",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " And so anytime you have two wires with the current going in the same direction,",
"wav": "EN_B00000/EN_B00000_S03252/mp3/EN_B00000_S03252_W000238.mp3"
} | EN_B00000_S03252_W000238.mp3 | [
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EN_B00000_S03252_W000239 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.3425,
"duration": 29.915,
"id": "EN_B00000_S03252_W000239",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " It's going to create an attractive force. The two wires will be attracted to each other. If the current is in the opposite direction, then the two wires will repel each other.... | EN_B00000_S03252_W000239.mp3 | [
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EN_B00000_S03252_W000240 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.3276,
"duration": 3.395,
"id": "EN_B00000_S03252_W000240",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " And also, the second wire creates a magnetic field.",
"wav": "EN_B00000/EN_B00000_S03252/mp3/EN_B00000_S03252_W000240.mp3"
} | EN_B00000_S03252_W000240.mp3 | [
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EN_B00000_S03252_W000241 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.2121,
"duration": 6.537,
"id": "EN_B00000_S03252_W000241",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " That exerts a force on the first wire causing the first wire to move towards the second if the currents are in the same direction.",
"wav": "EN_B00000/EN_B00000_S03252/mp3/EN... | EN_B00000_S03252_W000241.mp3 | [
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EN_B00000_S03252_W000242 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.4331,
"duration": 9.27,
"id": "EN_B00000_S03252_W000242",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " Let's start with this equation. Wire 1 creates a monadic field b1, which is u0 times i1 over 2 pi r.",
"wav": "EN_B00000/EN_B00000_S03252/mp3/EN_B00000_S03252_W000242.mp3"
} | EN_B00000_S03252_W000242.mp3 | [
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EN_B00000_S03252_W000243 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.3796,
"duration": 4.227,
"id": "EN_B00000_S03252_W000243",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " Where r is the distance between the two wires.",
"wav": "EN_B00000/EN_B00000_S03252/mp3/EN_B00000_S03252_W000243.mp3"
} | EN_B00000_S03252_W000243.mp3 | [
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EN_B00000_S03252_W000244 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.4369,
"duration": 5.331,
"id": "EN_B00000_S03252_W000244",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " So wire1, which produces a current one, will generate a monadic field",
"wav": "EN_B00000/EN_B00000_S03252/mp3/EN_B00000_S03252_W000244.mp3"
} | EN_B00000_S03252_W000244.mp3 | [
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EN_B00000_S03252_W000245 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.3577,
"duration": 5.094,
"id": "EN_B00000_S03252_W000245",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " That is at wire 2. And so R is the distance between the two wires.",
"wav": "EN_B00000/EN_B00000_S03252/mp3/EN_B00000_S03252_W000245.mp3"
} | EN_B00000_S03252_W000245.mp3 | [
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EN_B00000_S03252_W000246 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " Now we need to use the other equation, that is the force that",
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EN_B00000_S03252_W000247 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " Axon, a current carrying wire. That's F is equal to ILB sine theta. But we're not going to be worried about the angle in this problem.",
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EN_B00000_S03252_W000248 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " So the force on wire 2, which we'll call F2, is equal to the current on that wire times the length times the magnetic field created by wire 1. So what we're going to do now is ... | EN_B00000_S03252_W000248.mp3 | [
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EN_B00000_S03252_W000249 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"speaker": "EN_B00000_S03252",
"text": " So it's going to be I2 times L times mu0 I1 over 2 pi r. So that's how you can calculate the force between the two wires. F2 and F1, they have the same magnitude.",
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EN_B00000_S03252_W000250 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " What is the magnitude and direction of the force between two parallel wires that are 30 meters long and 2 centimeters apart, each carrying a current of 50 amps in the same dir... | EN_B00000_S03252_W000250.mp3 | [
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EN_B00000_S03252_W000251 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"language": "en",
"speaker": "EN_B00000_S03252",
"text": " So I1 and I2 both equal 50 amps. Now we have the length of the wire, which is L. And the length of both wires is 30 meters. And the distance between them, which is R, that's 2... | EN_B00000_S03252_W000251.mp3 | [
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EN_B00000_S03252_W000252 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " Now, for these type of problems, there's only two answers for the direction. Either the force is attractive, or they will repel.",
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EN_B00000_S03252_W000253 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " Because the currents are in the same direction, we have a force of attraction. And so that answers the direction of the force. It's simply attraction. Now all we need to do is... | EN_B00000_S03252_W000253.mp3 | [
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EN_B00000_S03252_W000254 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"speaker": "EN_B00000_S03252",
"text": " Times 50 times 50, which we can write as 50 squared, times L, which is 30 meters, divided by 2 pi, times the radius of .02 meters. Now 4 pi divided by 2 pi is gonna be 2.",
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EN_B00000_S03252_W000255 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " So the force between these two wires is .75 newtons.",
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EN_B00000_S03252_W000256 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.1438,
"duration": 3.497,
"id": "EN_B00000_S03252_W000256",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " And a direction is a force of attraction.",
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EN_B00000_S03252_W000257 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.2213,
"duration": 3.481,
"id": "EN_B00000_S03252_W000257",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " These two forces will be pointing towards each other.",
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EN_B00000_S03252_W000258 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"duration": 4.21,
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"language": "en",
"speaker": "EN_B00000_S03252",
"text": " Now the next thing that we need to talk about is Ampere's Law.",
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EN_B00000_S03252_W000259 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"language": "en",
"speaker": "EN_B00000_S03252",
"text": " Ampere's law describes the relationship between the current and the magnetic field produced by that current. Perhaps you've seen this equation.",
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EN_B00000_S03252_W000260 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"duration": 7.42,
"id": "EN_B00000_S03252_W000260",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " The sum of all the magnetic fields that is parallel to any, uh, segments that the magnetic field passes through.",
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EN_B00000_S03252_W000261 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.4712,
"duration": 9.389,
"id": "EN_B00000_S03252_W000261",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " That's gonna be equal to mu0 times the current enclosed by the path that the magnetic field makes. It has to be a closed path.",
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EN_B00000_S03252_W000262 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"duration": 6.978,
"id": "EN_B00000_S03252_W000262",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " A good example is using a wire. So let's say if we have a current that passes through this wire.",
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EN_B00000_S03252_W000263 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"duration": 6.299,
"id": "EN_B00000_S03252_W000263",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " Now this current will create a magnetic field that travels around the wire in a circular path.",
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EN_B00000_S03252_W000264 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"language": "en",
"speaker": "EN_B00000_S03252",
"text": " So if we take the monadic field,",
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EN_B00000_S03252_W000265 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"speaker": "EN_B00000_S03252",
"text": " And multiply it by the path that is parallel to it. Let's call it, uh, delta out.",
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EN_B00000_S03252_W000266 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " Get an equation that will give us the magnetic field created by this wire.",
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EN_B00000_S03252_W000267 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S03252_W000268 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S03252_W000269 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " times 2 pi r, which equals mu is not over i. I mean times i. So the monadic field",
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EN_B00000_S03252_W000270 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " If you divide both sides by 2 pi r, we get this familiar equation, nu zero times i over 2 pi r.",
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EN_B00000_S03252_W000271 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"speaker": "EN_B00000_S03252",
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EN_B00000_S03252_W000272 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"speaker": "EN_B00000_S03252",
"text": " By this loop. So that loop has an area, and the current that passes through that area, that's this current, is the current enclosed by that circular loop. And so that's how yo... | EN_B00000_S03252_W000272.mp3 | [
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EN_B00000_S03252_W000273 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S03252_W000274 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"speaker": "EN_B00000_S03252",
"text": " And the reason why it's advantageous to create a solenoid, anytime you create a loop of wire,",
"wav": "EN_B00000/EN_B00000_S03252/mp3/EN_B00000_S03252_W000274.mp3"
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EN_B00000_S03252_W000275 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S03252_W000276 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.4155,
"duration": 4.839,
"id": "EN_B00000_S03252_W000276",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " And so solenoids are very useful for creating powerful magnetic fields.",
"wav": "EN_B00000/EN_B00000_S03252/mp3/EN_B00000_S03252_W000276.mp3"
} | EN_B00000_S03252_W000276.mp3 | [
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EN_B00000_S03252_W000277 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.3662,
"duration": 8.863,
"id": "EN_B00000_S03252_W000277",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " So the magnetic field will travel in a circular pattern. So to calculate the, or to divide the formula for a solenoid,",
"wav": "EN_B00000/EN_B00000_S03252/mp3/EN_B00000_S032... | EN_B00000_S03252_W000277.mp3 | [
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EN_B00000_S03252_W000278 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.1855,
"duration": 6.52,
"id": "EN_B00000_S03252_W000278",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " We need to create a path of the magnetic field. So let's draw a rectangular path.",
"wav": "EN_B00000/EN_B00000_S03252/mp3/EN_B00000_S03252_W000278.mp3"
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EN_B00000_S03252_W000279 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"duration": 6.706,
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"language": "en",
"speaker": "EN_B00000_S03252",
"text": " So what we need to do is add up the monadic field that is parallel to each segment.",
"wav": "EN_B00000/EN_B00000_S03252/mp3/EN_B00000_S03252_W000279.mp3"
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EN_B00000_S03252_W000280 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"language": "en",
"speaker": "EN_B00000_S03252",
"text": " So that's going to be the monadic field times the length of segment AC.",
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EN_B00000_S03252_W000281 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"duration": 8.285,
"id": "EN_B00000_S03252_W000281",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " plus the magnetic field times the length of segment CD plus the magnetic field times the length of segment",
"wav": "EN_B00000/EN_B00000_S03252/mp3/EN_B00000_S03252_W000281.mp... | EN_B00000_S03252_W000281.mp3 | [
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EN_B00000_S03252_W000282 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.3076,
"duration": 5.857,
"id": "EN_B00000_S03252_W000282",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " db plus the magnetic field times the length of segment ba.",
"wav": "EN_B00000/EN_B00000_S03252/mp3/EN_B00000_S03252_W000282.mp3"
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EN_B00000_S03252_W000283 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"duration": 6.791,
"id": "EN_B00000_S03252_W000283",
"language": "en",
"speaker": "EN_B00000_S03252",
"text": " Now, let's focus on segment BD. Segment BD is outside of the solenoid.",
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EN_B00000_S03252_W000284 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"duration": 3.43,
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"language": "en",
"speaker": "EN_B00000_S03252",
"text": " And the monadic field is very weak outside of the solenoid, so...",
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EN_B00000_S03252_W000285 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " We can say that the contribution for, uh, BL in segment BD is very small. So it's negligible.",
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EN_B00000_S03252_W000287 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S03252_W000288 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " So it's because it's perpendicular to the magnetic field that's inside the solenoid.",
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EN_B00000_S03252_W000289 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"language": "en",
"speaker": "EN_B00000_S03252",
"text": " It's contribution is going to be nudgeable. So we can eliminate BA and DC.",
"wav": "EN_B00000/EN_B00000_S03252/mp3/EN_B00000_S03252_W000289.mp3"
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EN_B00000_S03252_W000290 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S03252_W000291 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S03252_W000292 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " In segment AC, we'll have the greatest contribution towards the sum of all the monadic fields that is parallel to each segment.",
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EN_B00000_S03252_W000293 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"id": "EN_B00000_S03252_W000293",
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"text": " Which is basically this equation. We can now replace this term with b times L. Where L is simply the length of the solenoid.",
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EN_B00000_S03252_W000294 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"language": "en",
"speaker": "EN_B00000_S03252",
"text": " The length of the solenoid being segment AC. But we'll just call it L. So BL is equal to mu0 times i. But this is 4 if we only have one loop. Let's say if you have a wire with... | EN_B00000_S03252_W000294.mp3 | [
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