Anderson Video - Faraday's Law Example

Professor Anderson
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And it's pointing down so we have our surface normal n hat pointing down. And let's give you the following information: When we start this experiment at T= 0 this angle is of course 90 degrees. But this thing is going to rotate into a new position and at some later time it's sitting right here. N hat is now in this direction and this angle between those two is theta 1= 45 degrees. And it does this at a time t1 of a hundredth of a second. Okay? So this thing is rotating pretty quickly. It rotates into this position in a hundredth of a second. And let's give you a little bit more information: The number of turns on this coil is 950. The coil has a radius, so this is information about the coil, the radius of the coil R is 6 centimeters. So 0.06 meters. And let's also tell you that the EMF generated is 0.065 volts. And this is all the information that we're given. Let's say that we are looking for B. What is the strength of the B field that's going to generate these results? Okay, so we go back to our definition for EMF. Let's make a little room. So the EMF we said was negative N delta phi over delta T. Alright and we're only really worried about the magnitude here so we'll just put some magnitude bars on it. Not worry about the negative sign. It's delta phi over delta T but what is delta phi? Well phi we said was B times A times the cosine of theta. So a delta phi is just going to be phi in the first case which is B times A times cosine of some theta one. And then we're gonna subtract phi naught. Okay? So this is B times A times cosine of, we just called it, theta that's what we get on the top. On the bottom we have delta T. And now we have some common factors, so we can factor out some stuff. We've got N times B times A. We have the cosine of theta one, the cosine of theta, and we're all dividing that by delta T. And so now we can solve this thing for B. What do we get? Well I've got to multiply across by delta T. So I get EMF epsilon times delta T. And then I have to divide by N, divide by A and divide by cosine theta one minus cosine of theta. And now we should have just about all those numbers. Except it's a circular loop right so we need to put in for the area A. So this is pi R squared for our area A, cosine theta one minus cosine of theta. Okay, and now we can solve this. Alright so we've got B equals epsilon we said was 0.065 volts. Delta T was 0.01 seconds. Down on the bottom we've got N which we said was 950. We've got pi which is pi. We've got R which is 0.06 squared. And then we have cosine of theta 1 which we said was 45 degrees, theta was zero degrees. Okay. And if somebody punches in this stuff into your calculator tell me what you get. So we've got 6.5 times 10 to the minus two. We've got another ten to the minus two right there We've got 950, we've got a pi, we've got six times ten to the minus two squared. Cosine of 45 is one over root two, cosine of zero is what? >>Can I ask a question? >>Yeah. >>Wasn't it cosine of 90? >>Oh. Let's sneak in a nine right there. Cosine of 90. Okay good. Yeah that's right because the surface normal is pointing down yeah thank you. So cosine of 90 we know that cosine of 90 is zero. Okay? So this is what we end up with and let's approximate it here while you guys punch it into your calculator. So we've got 6.5 times 10 to the minus four. Up top. We've got 9.5 times ten to the two, we've got a pi, we have 36 times ten to the minus four. And then we have a 1 over root two and let's see what this becomes. 10 to minus 4 drops out with that. We've got 6.5 over this stuff. Which is 9.5 times pi times 36. This is a lot of numbers here but let's keep going. We're going to say this is ten so this becomes the 10 to the 3. And then we've got pi times 36 that's got to be really close to a hundred. Right? So that's 10 to the 3 times the 10 to the 2 so that's the 10 to the 5. And then we got a 1 over root 2 and that's the same as multiplying by the square of 2 at the top. Which has to be really close to 10 up in the top. And so I'm gonna say my guess is 10 to the minus 4. That's what I'm guessing. Anybody punch into your calculator and get an answer? >>8.56 times 10 to the negative 5. >>8.56 times 10 to the negative 5. Good so that was a decent guess right? Because 10 times 10 to the negative 5 would be 10 to the minus 4. Okay. So that is the strength of the B field. What are the units on B field? Tesla right? So we put a capital T right there. Okay. Questions about that one?
And it's pointing down so we have our surface normal n hat pointing down. And let's give you the following information: When we start this experiment at T= 0 this angle is of course 90 degrees. But this thing is going to rotate into a new position and at some later time it's sitting right here. N hat is now in this direction and this angle between those two is theta 1= 45 degrees. And it does this at a time t1 of a hundredth of a second. Okay? So this thing is rotating pretty quickly. It rotates into this position in a hundredth of a second. And let's give you a little bit more information: The number of turns on this coil is 950. The coil has a radius, so this is information about the coil, the radius of the coil R is 6 centimeters. So 0.06 meters. And let's also tell you that the EMF generated is 0.065 volts. And this is all the information that we're given. Let's say that we are looking for B. What is the strength of the B field that's going to generate these results? Okay, so we go back to our definition for EMF. Let's make a little room. So the EMF we said was negative N delta phi over delta T. Alright and we're only really worried about the magnitude here so we'll just put some magnitude bars on it. Not worry about the negative sign. It's delta phi over delta T but what is delta phi? Well phi we said was B times A times the cosine of theta. So a delta phi is just going to be phi in the first case which is B times A times cosine of some theta one. And then we're gonna subtract phi naught. Okay? So this is B times A times cosine of, we just called it, theta that's what we get on the top. On the bottom we have delta T. And now we have some common factors, so we can factor out some stuff. We've got N times B times A. We have the cosine of theta one, the cosine of theta, and we're all dividing that by delta T. And so now we can solve this thing for B. What do we get? Well I've got to multiply across by delta T. So I get EMF epsilon times delta T. And then I have to divide by N, divide by A and divide by cosine theta one minus cosine of theta. And now we should have just about all those numbers. Except it's a circular loop right so we need to put in for the area A. So this is pi R squared for our area A, cosine theta one minus cosine of theta. Okay, and now we can solve this. Alright so we've got B equals epsilon we said was 0.065 volts. Delta T was 0.01 seconds. Down on the bottom we've got N which we said was 950. We've got pi which is pi. We've got R which is 0.06 squared. And then we have cosine of theta 1 which we said was 45 degrees, theta was zero degrees. Okay. And if somebody punches in this stuff into your calculator tell me what you get. So we've got 6.5 times 10 to the minus two. We've got another ten to the minus two right there We've got 950, we've got a pi, we've got six times ten to the minus two squared. Cosine of 45 is one over root two, cosine of zero is what? >>Can I ask a question? >>Yeah. >>Wasn't it cosine of 90? >>Oh. Let's sneak in a nine right there. Cosine of 90. Okay good. Yeah that's right because the surface normal is pointing down yeah thank you. So cosine of 90 we know that cosine of 90 is zero. Okay? So this is what we end up with and let's approximate it here while you guys punch it into your calculator. So we've got 6.5 times 10 to the minus four. Up top. We've got 9.5 times ten to the two, we've got a pi, we have 36 times ten to the minus four. And then we have a 1 over root two and let's see what this becomes. 10 to minus 4 drops out with that. We've got 6.5 over this stuff. Which is 9.5 times pi times 36. This is a lot of numbers here but let's keep going. We're going to say this is ten so this becomes the 10 to the 3. And then we've got pi times 36 that's got to be really close to a hundred. Right? So that's 10 to the 3 times the 10 to the 2 so that's the 10 to the 5. And then we got a 1 over root 2 and that's the same as multiplying by the square of 2 at the top. Which has to be really close to 10 up in the top. And so I'm gonna say my guess is 10 to the minus 4. That's what I'm guessing. Anybody punch into your calculator and get an answer? >>8.56 times 10 to the negative 5. >>8.56 times 10 to the negative 5. Good so that was a decent guess right? Because 10 times 10 to the negative 5 would be 10 to the minus 4. Okay. So that is the strength of the B field. What are the units on B field? Tesla right? So we put a capital T right there. Okay. Questions about that one?