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A 2.50 $\mathrm{mH}$ toroidal solenoid has an average radius of 6.00 $\mathrm{cm}$ and a cross-sectional area of 2.00 $\mathrm{cm}^{2} .$ (a) How many coils does it have? (Make the same assumption as in Example $21.10 .$ ) (b) At what rate must the current through it change so that a potential difference of 2.00 $\mathrm{V}$ is developed across its ends?

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a. 1937 coilsb. 800 $\mathrm{A} / \mathrm{s}$

Physics 102 Electricity and Magnetism

Chapter 21

Electromagnetic Induction

Current, Resistance, and Electromotive Force

Direct-Current Circuits

Magnetic Field and Magnetic Forces

Sources of Magnetic field

Inductance

Alternating Current

Cornell University

Rutgers, The State University of New Jersey

University of Washington

Lectures

03:27

Electromagnetic induction is the production of an electromotive force (emf) across a conductor due to its dynamic interaction with a magnetic field. Michael Faraday is generally credited with the discovery of electromagnetic induction in 1831.

08:42

In physics, a magnetic field is a vector field that describes the magnetic influence of electric currents and magnetic materials. The magnetic field at any given point is specified by both a direction and a magnitude (or strength); as such it is a vector field. The term is used for two distinct but closely related fields denoted by the symbols B and H, where H is measured in units of amperes per meter (usually in the cgs system of units) and B is measured in teslas (SI units).

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this problem. We have a total still annoyed, and we're told that it has a self inducting of 2.5 million Henry's on average radius of six centimeters and a cross sectional area, two centimeters squared. The first part. We want to find the number of coils and that's associated with this. So annoyed so since we are told. But the soap induct in't is Let's recall that self induct Ince's given by the Net fucks so in number of turns multiplied by the folks going through one of those turns divided by the current. Our flux is given by the cross sectional area multiplied by air. Be field about it by our current and R B field for a tour. Oh, it'll set up. We used the approximation. That's mean radius approximation. But it's equal to new, not times the number of turns multiplied by our current on divided by two pi R. And that's forever using that approximation. Really, that's just valid at the center of the Toronto, So annoyed. But we're saying let's just have it be in that value everywhere, so we see that the currents cancel and we have you not cross sectional area divided by two pi r and we have not multiplied by n squared. And this is nice because we have all of the parameters in this equation besides end. So if we rearrange this, we have n is equal to the square root of two pi R times at all, all devoted by, you know, times the cross sectional area. We move this over this entering it in might be a little bit long. Now let's let's plug everything in and see what this gets us so to Pie won't apply by mean radius six. Put that in the meters and we have 2.5 million Henry's put that into Henry's and then we have immunity is four pi times 10 to the negative seven and then our cross sectional area is two centimeters squared with that into meter squared on every plug. Off this in, we get 1937 turns to the second part. We want to find what the rate in change of current is in order to have an EMF of two votes. So the relevant equation here that we want to use is are Emma being equal to herself, induct INTs well supplied by the great and change in the current. You know, just so for what this is we know we know both. What we want are the IMF to be We know what are, uh, ourself induct Ince's so just her The IMF, divided by herself, induct in CE We want two bolts and that needs to come from a two went by. Really? Henry Dr. And if we plug this in, we get 800 amps per second.

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