A toroid with cross section of radius 2 cm has a silicon steel ... | Filo
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SOLVED: A toroid is formed by winding a wire which carries current with N turns. The toroid has a mean radius R, and a cross-section radius Tc which is smaller than R. Using Ampere's law, derive an expression for B inside the toroid as the function of radial distance r which is the distance from the center of the toroid. Again, draw a figure with an amperian loop, direction of the path, and the direction of the magnetic field.Toroid:R = mean radiusTc = cross-section radiusN = 2000 turnsI = 2AR =
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A toroid of n turns, mean radius R and cross-sectional radius a carries current I. It is placed on a horizontal table taken as x-y plane. Its magnetic moment m ______. - Physics | Shaalaa.com
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A toroid has a major radius R and a minor radius r, and is tightly wound with N turns of wire, as shown in Figure P32.12. If R ≫r, the magnetic field in the region enclosed by the wire of the torus, of cross-sectional area A=πr^2, is essentially the same as the magnetic field of a solenoid that has been bent into a large circle of radius R Modeling the field as the uniform field of a long solenoid, show that the self-inductance of such a toroid is approximately L ≈(μ0 N^2 A)/(2 πR) (An exact exp
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