the-inductance-l-in-mu-mathrmh-of-a-coaxial-cable-is-given-by-l0032015-log-a-x-where-a-and-x-are-the

Name: the inductance l in mu mathrmh of a coaxial cable is given by l0032015 log a x where a and x are the
Uploaded: 2020-02-15
Duration: 1 min 19 s
Channel: Numerade
Description: The inductance $L$ (in $\mu \mathrm{H}$ ) of a coaxial cable is given by $L=0.032+0.15 \log (a / x),$ where $a$ and $x$ are the radii of the outer and inner conductors, respectively. For constant $a,$ find $d L / d x$.

Question

The inductance $L$ (in $\mu \mathrm{H}$ ) of a coaxial cable is given by $L=0.032+0.15 \log (a / x),$ where $a$ and $x$ are the radii of the outer and inner conductors, respectively. For constant $a,$ find $d L / d x$.

Salamat Ali · Accepted Answer

the capacitance off a cylindrical capacitor is given by C is equal to copper times. See Nord, where copper is the dialect of constant. Seen on is the capacitance without the dye electric This weekend, Right to buy downs Carbide&#x27;s Long Times L, divided by Ellen off be divided by a where l is the length off the cable ace. The Radius and B is the outer radius with the cap of value, which is 2.6 fool polystyrene. The capacitance party unique lent off. The cable is given by C D. Bye bye Help that is, to buy K. It&#x27;s long divided by Ellen. Off be divided by a so l l will cancel out. We substitute the values to pi Times copper, which is 2.6 times you 8.85 times 10 to the power minus 12. Divided by Ellen off 0.6 year old millimeter divide by 0.10 millimeter. This gives us, um 8.1 times 10 to the power minus 11 for rod for a meter

Farhanul Hasan · Answer

first we use NPR&#x27;s law to calculate a magnetic field strength. So, um, be integrated pinnacle of be over our is a gold you not times current enclosed. Um, just gonna write that as I on. So be time. So it&#x27;s going along loops a two part times two pi r is equal immuno I and therefore we expect vehicles you not I over two pi r Next step is to compute the flux So flux will be inequality over d a And so you have you not I over two pi r times l d r um And so this is going from A to B articles in a radius anto articles out of its B on. So we pull out the terms that are not involved in the new goals. You&#x27;re not ill over two pi you&#x27;re left with dinner. Go from a to B of one of her RDR one of her are is integrated to be log of are So this is me not I l number two pi times log looks, uh, natural log of, um are valued it from a to B. Yeah, and so this is me not I l over two pi Clinton&#x27;s log of be over And so that&#x27;s your flux. And then we know that self inducted physical too flux over current. So this is this stuff of a current And so the current cancels out you have you not? No By all over two pi times eyes the eyes cancel times log Be over. Ah, hence self inductions per unit length will just divide this by l ah is me not, uh over two pi times log of be over.

Zachary Warner · Answer

for this situation. We&#x27;re asked to consider a co axial cable with a solid inter radius. Um, conductor of a radius are one, and it&#x27;s surrounded by a concentric cylindrical tube or in your radius are too. And then an outer radius are three. You could look at figure 42 to get a visual representation of this. So it carries a current I not distributed uniformly across the cross section. So I not carried uniformly across and it descends. Determine the magnetic field at different distances are from the access for part A. It&#x27;s when are is less than are one. Okay, so indicate this is part A when our is less than our one. So first thing we&#x27;re gonna do is we&#x27;re gonna use amperes law because NPR&#x27;s law will give us the magnetic field from a current. So amperes law says that the integral over closed loops that&#x27;s what this little circle represents on the integral to redraw that, to make it look nicer looks better. The integral of be dot d l. Where d l is. The unit length here is equal to mu, not magnetic permeability of free space times I enclosed, so we&#x27;ll call this. I e n c. Okay, so we carry out the integral in the left side. This simply ends up becoming magnetic field times. Um, the area of the integration, which is two pi R. Since we&#x27;re just integrating over the surface here, uh, the area is actually doesn&#x27;t give up, Doesn&#x27;t get area two pi r gives you the circumference of the circle that were integrating over eso This be times two pi r. Since the two pirates, what comes from the integration over the close circle is still equal to mu. Not times I enclosed so we can rearrange this to solve for be in general when we find that be is equal to you Not times I enclosed divided by two times pi times the radius are, um where r is the distance we are from the from the access that the distance we&#x27;re trying to determine the magnetic field strength at. So the next question becomes Then what is, uh what is I enclosed? Okay, So the current enclosed in the galaxy and surface is the product of the area of the calcium surface and the current density. So we can say that I enclosed here is equal to the area A times the current density J But what&#x27;s the area here? The area since we&#x27;re considering it to be circular is just gonna be pie are square, right? And, um, the current density is going to be the current that&#x27;s running throughout my not divided by the area in which it&#x27;s running. We&#x27;ll call this area a prime. But since we&#x27;re considering it, when our is let&#x27;s see what our is less than our one this a prime is just gonna be the area of the R one section, which is just pie Times are one square. So now we have I enclosed. We can plug that back in above, into our magnetic field and what we end up with is mu knock I enclosed would be pi r squared times I not divided by pi r one squared so we can plug that in. So this is mu not over two pi r Then this is going to get multiplied by applied r squared. Oh, um and then times I not divided by high are one squared so we can cancel out the light terms like the pies and the ours. Um, noticed that our one is only in one place in the denominator there. But there&#x27;s an R squared on top as well as in our own bottom. So if you cancel out the light terms, but we end up with is mu not times are times I not divided by two times pi times are are one square. We can go ahead and box set in as our solution for part A Barbie wants us to find the magnetic field. But this time for the regions, where are one is less than R and R is less than our two so essentially between our one and are too. Okay, well, indicate this is part B. It&#x27;s the exact same equation that we&#x27;re starting out with, You know, for the magnetic field where B is equal toe you not I am closed over two pi r. So all point to it here, Or maybe all just underlying it with the red line here. So this equation that I&#x27;m underlining here, this b equals, um you know, I am closer to pi R. This is the one we&#x27;re gonna keep referring back to. But in each case, we just have to figure out what the I enclosed is well, in this case, the I enclose is pretty simple to find. It&#x27;s just equal to Ay, not since In that region, the current that&#x27;s running is just the guy, not current. So all we have to do in our equation for the magnetic field is replaced. I enclosed with I not and we&#x27;ve solved for part B. This is mu not times I not divided by two pi r again, that&#x27;s for the region between our one and are too part see would like us to calculate the magnetic field in the, uh, in the region where are too ours between our two and our three. So when ours between our two and our three, the uh enclosed to current is equal to I know it multiplied by, uh, the ratio of the different distant. Excuse me. The ratio of the differences between the area made by the distance the magnetic field is trying to be calculated at that our value not are one or two or three, but the R value were trying to calculate that I feel that that area minus the area a two, um, the area that&#x27;s made by, uh, by the R to value. And the reason this is true is because here are is greater than or two right. And then this is gonna be divided by the difference between, uh, the area A three minus a two. And that&#x27;s because a three is also larger than a two so we can plug in the values for these areas. Where a is pi r squared a two is pi r two squared and a three is pi r three squared so you can see that each one has a pie term for all these areas. So that pie term is just gonna cancel when we can just simply write the are Turks. This is r squared, minus are two squared divided by, um are three squared minus R two squared again. I canceled all the pies that would normally be written in front because each one would have a pie term. We could just cancel those out. Okay, so now we just need to replace our I enclosed in our equation for the magnetic field with this value that we just found and we&#x27;ve sold for the magnetic field apart. See? So here b is equal to you, not over two pi r But me, not times I not right because I enclosed still has that I not value. But then we also have this r squared minus r two squared, divided by our three squared minus r two squared. All right, then we can go ahead and box set in is our solution for parts. Uh, see? Okay. Part D would like us to find the magnetic field in the region where are is greater than our three. Well, when our is greater than our three, the I enclosed value is equal to zero. Therefore, if I enclosed is equal zero me plug zero in to our expression for a B, then B is also gonna be equal to zero. So party is simple. Box it in. And then part E would like us to graft the magnetic field as a function of our So we can go ahead and do that where the y axis is our magnetic field be next. Access is our value. So we have different. We have our one right. We have our two and we have our three. Well, if you go back to look at the magnetic field in the region where are is less than our one. We found that it was Muna Are times I not divided by two pi r one square. But he did, um here we&#x27;re considering where are is less than our one. So as our creases increases, if you look at that expression are is linear. Excuse me. B is linear with our if you look at that expression, right, So we&#x27;re gonna increase at a linear rate, something like this. So you hit a maximum value at our one Now are in the region where are is greater than our one, but less than our two. That&#x27;s what we found in part be so when our is greater than our one, but less than our to its muna I over two pi r. So it decreases as our increases. So now if you go back to autograph, this is going to start to drunk. Okay, then we get to the region were between two and three, right, which is, uh, what we found in part. See? So if you go back to our solution for park Si, let&#x27;s see, here it is at the top. We have are squaring minds are two squared divided by our three squared minus R two squared. So this is more a little bit more difficult. So I went ahead and just kind of plug this into a, uh into a graphic calculator, and it&#x27;s still decreased. But it looked something more like We go back to the graph page. Look, something more like this faster rate went down. And then once you hit our equals three, the magnetic field decreases to zero. So goes to zero, and that is the solution for party.

Mayukh Banik · Answer

we have TV. It was e does D. Now we have e equals to kill them over a and Delta V equals to lambda. Long be over a So from there we can find that e equals Delta V over a Don be over a So from these we find that Delta V equals E a. Don be over a putting the numbers in. He&#x27;s 18 times. Stand to the six Eddie Iea&#x27;s 0.8 millimeter or 10 to the minus three middle long Be over a is 3/0 0.8 putting the numbers in we get Delta V. It was 1.9 times 10 to the four World are 19.

Nolan Smyth · Answer

given that the outer conductor of the co axial cable has the greatest radius of three millimeters, which should be the radius of the inner conductor so that the inductions pregnant Lee does not exceed 55 Nano Henry&#x27;s per meter. Okay, the first thing we need to know is that the induction screening at length with co ax of people going by the following the expression inducted immediately. Goodbye. You know, over to pie the natural log of the ratio of the outer inner radio. That&#x27;s gonna be our two over our one. We want to find the value of our one such like this. You do not exceed 5500. So we want to solve our one the radius of the inner and Dr Start Well, first, ice looting. And that&#x27;s a log. That&#x27;s a lot of part two. One visit to is just algebra here to five times duck pins for you&#x27;re not taking that log on both sides. Art scene off. One is equal to e explode. Exponential. How well you&#x27;re not. Now we can see that our one is something people Thio are too. This terms here exponential too high. I was in darkness but you&#x27;re not. So even though how big Al the inductive over little Elian at length. You&#x27;re not hungry. It&#x27;s given that we know the rest of cranberries, we can simply plug in values and stall. If you value our one. It turns out this critical value. 2.2 8,000,000 years. So the inter radius of the cable should not be less than 2.28 millimeters, nor it&#x27;s for induction unit length of 55 Now Henry&#x27;s or less. That&#x27;s my final answer.

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If the loudness $b$ (in decibels) of a sound of i…

Problem 50 Easy Difficulty

The inductance $L$ (in $\mu \mathrm{H}$ ) of a coaxial cable is given by $L=0.032+0.15 \log (a / x),$ where $a$ and $x$ are the radii of the outer and inner conductors, respectively. For constant $a,$ find $d L / d x$.

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Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

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Basic Technical Mathematics with Calculus