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

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.

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30. Induction and Inductance

Inductors

Physics

30. Induction and Inductance

Inductors

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3:09 minutes

Problem 13a

Textbook Question

A toroidal solenoid has mean radius 12.0 cm and crosssectional area 0.600 cm². How many turns does the solenoid have if its inductance is 0.100 mH?

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Start by recalling the formula for the inductance of a toroidal solenoid: \( L = \frac{\mu_0 N^2 A}{2\pi r} \), where \( L \) is the inductance, \( \mu_0 \) is the permeability of free space \( (4\pi \times 10^{-7} \text{ Tm/A}) \), \( N \) is the number of turns, \( A \) is the cross-sectional area, and \( r \) is the mean radius.

Convert the given measurements to meters: the mean radius \( r = 12.0 \text{ cm} = 0.12 \text{ m} \) and the cross-sectional area \( A = 0.600 \text{ cm}^2 = 0.0000600 \text{ m}^2 \).

Rearrange the formula to solve for the number of turns \( N \): \( N = \sqrt{\frac{2\pi r L}{\mu_0 A}} \).

Substitute the known values into the rearranged formula: \( L = 0.100 \text{ mH} = 0.0001 \text{ H} \), \( r = 0.12 \text{ m} \), \( A = 0.0000600 \text{ m}^2 \), and \( \mu_0 = 4\pi \times 10^{-7} \text{ Tm/A} \).

Calculate the number of turns \( N \) using the substituted values in the formula: \( N = \sqrt{\frac{2\pi \times 0.12 \times 0.0001}{4\pi \times 10^{-7} \times 0.0000600}} \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Inductance

Inductance is a property of an electrical conductor that quantifies its ability to induce an electromotive force (EMF) when the current flowing through it changes. It is measured in henries (H) and is a crucial factor in the design of coils and solenoids, as it determines how effectively they can store magnetic energy.

Toroidal Solenoid

A toroidal solenoid is a coil of wire shaped like a doughnut, with the wire wound around a circular core. This configuration confines the magnetic field within the core, minimizing external magnetic interference. The inductance of a toroidal solenoid depends on its number of turns, cross-sectional area, and the core's permeability.

Magnetic Permeability

Magnetic permeability is a measure of how easily a material can support the formation of a magnetic field within itself. It is a key factor in determining the inductance of a solenoid, as materials with higher permeability allow for stronger magnetic fields, thus increasing the solenoid's inductance. The permeability of free space (vacuum) is a constant used in calculations involving inductance.

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A toroidal solenoid has mean radius 12.0 cm and crosssectional area 0.600 cm2. How many turns does the solenoid have if its inductance is 0.100 mH?

Key Concepts

Inductance

Toroidal Solenoid

Magnetic Permeability

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A toroidal solenoid has mean radius 12.0 cm and crosssectional area 0.600 cm². How many turns does the solenoid have if its inductance is 0.100 mH?