FIGURE EX30.19 shows the current as a function of time through...

Question

FIGURE EX30.19 shows the current as a function of time through a 20-cm-long, 4.0-cm-diameter solenoid with 400 turns. Draw a graph of the induced electric field strength as a function of time at a point 1.0 cm from the axis of the solenoid.

Accepted Answer

Welcome back. Everyone in this problem. A cylindrical solenoid measures 16 centimeters in length and five centimeters in diameter and is wound with 200 turns. The figure below shows the current as it changes over time for the solenoid. And here it shows a graph of current versus time. Using this information determine and graphically represent the magnitude of the induced electric field at a 0.2 centimeters from the solenoid central axis as a function of time. Now let's try and make sense of what's really going on here. OK. So we're talking about a cylindrical solenoid that probably looks something like this. OK? And in the cylindrical cleo, we know that it measures 16 centimeters in length. All right. So its length is 16 centimeters. We also know that it's wound from 200 wound with 200 turns. So N equals 200 it's five centimeters in diameter. So its diameter is five centimeters which would probably be somewhere right here. What else do we know? Well, let's put our lines of magnetic flux in here. OK. And with our lines, we know that an electric field is induced at a 0.2 centimeters from the solenoid central axis. So we can imagine the electric field looks something like that where I can draw my arrows here for the field and let me make it a dotted line just to signify what's going on here. Ok. So we know that that electric field is induced. And if we were to look at our solenoid from, from the top, ok. If we were to look overhead, what's really going on here is that we have our solenoid, OK. We have our electric field. So let's draw our electric field here. OK. Of course, we're only interested in the magnitude of our electric field. But I just put the vectors on this diagram and from the solenoid central axis outward, we're in interested in the electric field that's induced two centimeters away. OK. So I represent my magnetic field here. We want that electric field that's at a distance of two centimeters from the solenoids central axis. OK. So this is like an overhead view. No, for that electric field, we want to determine graphically the magnitude of the induced electric field as a function of time. What does that mean? Well, if you look at our graph here, we have a graph of the induced current, right, for, for the solenoid over time. So essentially we want a graph that will look the same as this one. But it's why axis will be the electric field as a function of time. So if we were to put that graph here essentially what we want is a graph that looks something like this. OK. So we have our time and we're talking about the same time. So from zero to 0.6 seconds, so that's 0.6. Say this is 0.10 0.20 0.3 0.4 and 0.5. And we want the electric fuel strength. No, what do we know here? We're talking about electric field strength. So we have to think how is, how are we going to find the electric field strength at these different intervals? How is electric field strength related to a change in current? We recall that a changing magnetic field induces a current? Ok. And that means wherever we have a changing magnetic field, it creates an electric field around it. So do we know any formulas that relates a change in magnetic field to electric field strength? Well, recall that the magnitude of an electric field equals half of R multiplied by the change in the magnetic field divided by sorry, the change in the magnetic field over time. OK. And R is our distance from the center of the solenoid to the electric field. Now, we already know that we already know that R is two centimeters. OK. But we don't know what the changing magnetic field is. So let's think, is there any way that we can figure that out? What do we know about the magnetic field? Well, also recall that the formula for the magnetic field equals the permeability of free space multiplied by the number of turns multiplied by the current divided by the length of the sole for a cylindrical solenoid. OK. So if that's our magnetic field, that means the change in magnetic field, OK, would be equal to Muno multiplied by the number of turns multiplied by the change in current. Because remember that it is a change in magnetic field that induces the current and all of that divided by L. So since that's the case right, we can substitute our value here or change in magnetic field into our formula for the electric field strength. So by doing that, our formula now becomes our electric field strength, right magnitude of our electric field equals R divided by two multiplied by our change in magnetic field strength, which is Muno multiplied by N multiplied by our change in current, right, divided by L divided by our change in time camp. And let me put that right, that a little better OK, change in current over the change in time. And that's very important because no, if we know the change in current and the change in time, we'll be able to find the value of our electric field strength. Now notice that we're not interested in any changes from muat and, and the length of the senno because none of those change. So we can even find a value for those when solving for the electric field strength that will make it easier to work with. So let's go ahead and do that. So using that formula substituting the values we know then RR which is the distance from the center is two centimeters that's divided by two multiplied by Muno the permeability of free space which is four pi times 10 to the negative seven tesla meter tesla meters per amper multiplied by our chain or, or the number of turns, which is 200 and all of that is divided by the length of our cleo which is 16 centimeters, 16 times 10 to the negative 2 m all multiplied by a change in current over time. Now we'll be able to work all what we have here or constants. And when we do that, if you put that into your calculator, you'll find that your electric field strength equals 1.571 times 10 to the negative five volt seconds per per meter multiplied by a change in current over time. So no, we'll be able to find the electric fuse trends once we know the change in current and the change in time. So let's go back to our graph. We can figure those out from the graph that's given. Now, when we look at our intervals here, notice that there are four distinct intervals we have from zero seconds to 0.2 where our current is constant. Then from 0.2 to 0.3. Let's put that in red where our current changes again, from 0.3 to 0.4 our current is constant and then from 0.4 to 0.6 again, our current changes. Now there's something very important to note here or the points at which our current is constant, there is no induced current. Therefore, there was no change in the magnetic field. And if there's no change in magnetic field, that means an electric field wasn't created. So at the points where our current is constant, our electric field strength equals zero. So now we know what our electric field strength is for the intervals from zero to 0.2 we can put it on our graph and from 0.3 to 0.4. So that information we already have. No, we just need to figure out our electric field strength for the intervals, 0.2 to 0.3 and 0.4 to 0.6. So let's start with our interval from 0.2 to 0.3. For this interval, we know that the change in current it goes from 5 to 8 A MS. So the change in current is 3 a.m. here and it goes from 0.2 to 0.3 seconds. So the change in time is 0.1 seconds. So now since we have that information, we can find its electric field strength so far, 0.2 to 0.3 seconds. The electric field strength will be equal to 1.571 times 10 to the negative five volt seconds per per meter multiplied by a change in current which is three umpires divided by a change in time which is 0.1 seconds. Now, when we put this into our calculator, you will find that the electric field strength is five times 10 to the negative four volts per meter. So let's put that on our graph. So if we come back to our graph and let's draw some ah vertical lines here. OK? And we know that we let's let's say that we have five volts per meter here are five times 10 than the negative four. OK? Oh You know what, we can put that in our, in our axis. OK? Because we want to make sure we we represent that it's times 10 to the negative four. So let me adjust my axis here. So this is actually the electric sphere strength and times 10 to the negative four volts per meter, right? So we have it at five. OK. So five, we can represent it with this straight line here and I'm going to rob out what's above and use a dotted line to represent what's below just to show that it's coming from those parts. So this would represent the electric fuel change from 0.2 to 0.3 seconds. No, we just need it for 0.4 to 0.6 seconds. What do we know? Well, from 0.4 to 0.6 seconds, notice that our change in current goes from eight pires to zero amps. So our change is eight S and our change in time goes from 0.4 to 0.6. So that's 0.2 seconds. So now that we have that information, we can substitute it into our formula because now that tells us for 0.4 to 0.6 seconds. Ok. Are electric fuel strength will be equal to 1.5 of one times 10 to the negative five W seconds per per meter multiplied by our change in current, which is eight s divided by our change in time, which is 0.2 seconds. And again, if you put that into your calculator, you'll find that the electric field strength is six times 10 to the negative four volts per meter so that we can also put on our graph. So in doing that, well, if five is right there, that means six is a bit above it. So I could put six right here and no, not six would be safe from this point to this point. Ok? And just let me just draw those vertical lines just to make sure I get it right. Ok. Oh, ok. And if I rub that out just to show that they aren't really there, then this is what our graph would look like for the the induced electric field two centimeters from the solenoid central axis as a function of time. Thanks a lot for watching everyone. I hope this video helped.

FIGURE EX30.19 shows the current as a function of time through a 20-cm-long, 4.0-cm-diameter solenoid with 400 turns. Draw a graph of the induced electric field strength as a function of time at a point 1.0 cm from the axis of the solenoid.

Key Concepts

Electromagnetic Induction

Induced Electric Field

Solenoid Characteristics

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