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 2 years ago
A 3.00 toroidal mH solenoid has an average radius of 5.60 cm and a crosssectional area of 2.40 cm^2.
How many coils does it have? In calculating the flux, assume that B is uniform across a cross section, neglect the variation of B with distance from the toroidal axis.
At what rate must the current through it change so that a potential difference of 2.60 V is developed across its ends?
 2 years ago
A 3.00 toroidal mH solenoid has an average radius of 5.60 cm and a crosssectional area of 2.40 cm^2. How many coils does it have? In calculating the flux, assume that B is uniform across a cross section, neglect the variation of B with distance from the toroidal axis. At what rate must the current through it change so that a potential difference of 2.60 V is developed across its ends?

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85295james
 2 years ago
Best ResponseYou've already chosen the best response.0WOW you got me on this one

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1\[\epsilon = L \frac{di}{dt}\]\[\epsilon=N\frac{d\Phi_B}{dt}=N\pi r^2 B\] You're given the emf, L, and r. You can already find di/dt. All you need to find N is to use Ampere's Law to find B, with C being the circle with the toroid's average radius. \[\int_C \vec B.d\vec l=\mu_0 i_{enc}\]Hint: Visualize the torus and what part of it is going through the circle in order to get ienc.

tvvv1012
 2 years ago
Best ResponseYou've already chosen the best response.1i have 8 mins and i dont think i can do that in time lol

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1I did basically all of it!

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1First equation: solve for di/dt. Second equation: Find B using Ampere's Law with C being the path along the center of the circle cross sections of the torus, then solve for N.

tvvv1012
 2 years ago
Best ResponseYou've already chosen the best response.1i tried pluging in and solving and i got the wrong answer 5mins!
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