A 2.50-mH toroidal solenoid has an average radius of 6.00 cm and ...

Question

A 2.50-mH toroidal solenoid has an average radius of 6.00 cm and a cross-sectional area of 2.00 cm^2. (a) How many coils does it have? (Make the same assumption as in Example 30.3.)

Accepted Answer

Welcome back, everybody. We are making observations about a Tor Oid and we are told a couple different things. We're told that it has an induct inside of 48.5 mil, it hurts. We're told that it has a diameter of 15 centimeters. Also told that it has a cross sectional area of 1.8 centimeters squared. And we are tasked with finding the total number of terms in the toy Roid. Well, we're gonna have to use a couple different formulas here. We know that conductance is related to the number of terms times our magnetic flux all divided by the current. Now the magnetic flux, we know to be equal to mu not times the number of terms, times the current times the cross sectional area all divided by two PI R. But here's what I'm gonna do here. I'm gonna sub this formula in right here. So we have that eh is equal to N over I times mu not N I A divided by two pi R. You can see that the current cancels out on top and bottom and we are left with that inducted since is equal to, let's see here. We have and squared mu not a all divided by two pi R. Here's what I'm gonna do. I'm gonna circle this guy and just move over. That's the tiniest bit. And here's why, what we wanna do is we want to isolate this term right here because we're trying to find the number of terms. So I'm going to multiply both sides by two pi R times mu not over the cross or time over time, the cross sectional areas, let me go ahead and write this term on this side as well. And you'll see that on the left hand side, all of this cancels out. The last thing to do is to take the square root of both sides to get rid of the power on the number of turns and this disappears. So now what we are left with is that the number of terms is equal to the square root of two pi times R radius times are conductance all divided by new, not times the cross sectional area. Now, before we can start plugging in values, we need to make sure that we have all of these values and in the right term, right units, sorry. So let's start with our area here. We're given our area as 1.8 centimeters squared, but we need this to be in meters squared when we know that there's 100 centimeters in one m. And what we'll do is we'll just square the top and bottom so that these units will cancel out multiplying straight across. This gives us an area of 1.8 times 10 to the negative fourth meters squared. Now, we also need to find our radius while our radius is just going to be half of our diameter. This is gonna be 15 divided by two. But if you'll notice our diameter is given to us in centimeters, but we need this in meters. So we're gonna multiply this by 10 to the negative second, giving us 7.5 times 10 to the negative second eaters. Great. So now we have all of our terms. We have our formula. Let's just go ahead and plug it in here. We have that. The number of turns is equal to and I'm gonna draw this square root sign pretty big equal to two pi times R radius of 7.5 times 10 to the negative. Second times are inducted of 48.5 mil. It hurts but we need this and hurts. So I'm gonna multiply this by 10 to the negative three. This of course is divided by our constant of you not which is four pi times 10 to the negative seventh times are cross sectional area of 1.8 times 10 to the negative fourth. When you plug all of this into your calculator, we get that the number of turns in the Tor Oid is 1.1 times 10 to the fourth turns corresponding to our final answer choice of c Thank you all so much for watching. Hope this video helped. We'll see you all in the next one.

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A 2.50-mH toroidal solenoid has an average radius of 6.00 cm and a cross-sectional area of 2.00 cm^2. (a) How many coils does it have? (Make the same assumption as in Example 30.3.)

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