A toroidal solenoid has mean radius 12.0 cm and crosssectional ar...

Question

A toroidal solenoid has mean radius 12.0 cm and crosssectional area 0.600 cm^2. (a) How many turns does the solenoid have if its inductance is 0.100 mH?

Accepted Answer

Welcome back, everybody. We are making observations about a Tor Oid here and we are told that the diameter of our Tor Oid is centimeters that has a cross sectional area of 220.5 centimeters squared. And we are tasked with finding the number of turns in the Tor Oid needed in order to have an inductive of 0. Mil Hertz. So we actually have a formula that relates all of these terms. And the formula is given as follows. It is the inductive is equal to a constant mu not times our number of turns squared times the cross sectional area all divided by two pi times our radius. And here's what I'm gonna do. We need to isolate this term right here. So I'm gonna multiply both sides by two pi R over mu not times the area, what we do to one side we must do to the other. So let me copy this over here as well. You'll see that on the right hand side here, these terms are all going to cancel out. Now, we just have to get rid of this power on the number of turns here. I'm gonna take the square root of both sides. And you'll see on the right hand side here that just gets rid of the power. So what we are left with is the final formula that we're gonna get to use of. The number of turns is equal to the square root of two pi times radius times are inducted all divided by our constant new, not times our area, we know all these values. So let's go ahead and plug them in here. So we have on top two pi times our radius. Well, what is our radius? Radius is just going to be half of the diameter. So it'll be 22 divided by two. But we need this in meters not centimeters. So I'm gonna multiply this by 10 to the negative second. Now this is going to be multiplied by our conductance which is given to us in middle hurts but we need, it hurts. So it'll be 0.2 times 10 to the negative third divided by our constant of mu naught, which is four pi times 10 to the negative seventh times our cross sectional area. But we need our cross sectional area in meters squared. This one's a little bit tougher. So I'm gonna go ahead and pull this aside so that we can convert it here. We know that there are 100 centimeters per one m, but we need to cancel out the units of centimeters squared. So let's square both the top and bottom, you'll see that these units cancel out and we are left with a cross sectional area of 0. times 10 to the negative fourth meters which we will plug in right here right here, 100.5 times 10 to the negative four. And when you plug this entire thing into our calculator, you get that the number of turns in the Tor oid is 1. times 10 to the third corresponding to our final answer. Choice of a. Thank you all so much for watching. Hope this video helped. We will see you all in the next one.

A toroidal solenoid has mean radius 12.0 cm and crosssectional area 0.600 cm^2. (a) How many turns does the solenoid have if its inductance is 0.100 mH?

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