A toroid of mean radius 28 cm andcircular cross section of radius 1.5cm is wound

Question

A toroid of mean radius 28 cm andcircular cross section of radius 1.5cm is wound with a superconducting wire of length 1000 m thatcarries a current of 390 A. (a) What is the number of turns on thecoil?
1 turns

(b) What is the magnitude of the magnetic field at the meanradius?
2 T

(c) Assuming that B is constant over the area of the coil,calculate the magnetic energy density and the total energy storedin the toroid.
3 J/m2
4 kJ (a) What is the number of turns on thecoil?
1 turns

(b) What is the magnitude of the magnetic field at the meanradius?
2 T

(c) Assuming that B is constant over the area of the coil,calculate the magnetic energy density and the total energy storedin the toroid.
3 J/m2
4 kJ

Explanation / Answer

(a)the magnetic field created by a toroid is B = (oNI/2r) the mean radius r is very large compared with thecrossectional radius a of the torus,therefore the field isapproximately uniform inside the torus. or N = (B * 2r/oI) B = (oI/2r) or N = ((oI/2r) *2r/oI) = 1 the number of turns per unit length is n = (N/L) L = 1000 m or N = n * L = n * 1000 = 1 * 1000 = 1000 turns (b)the magnitude of the magnetic field at the meanradius B = (oNI/2r) o= 4 * 10-7N/A2,N = 1000 turns,I = 390 A and r = 28 cm = 28 *10-2 m (c)the self inductance of the toroid is L = o* (N2/l) * A N = 1000 turns,l = 1000 m and A = * r2 wherer = 28 cm = 28 * 10-2 m the total energy stored in the toroid is E = (1/2)L * I2 or E = (1/2) * o* (N2/l) * A *I2 the magnetic energy density is U = (E/A) = ((1/2) * o* (N2/l) * A* I2/A) = (1/2) * o* (N2/l)* I2 the magnetic energy density is U = (E/A) = ((1/2) * o* (N2/l) * A* I2/A) = (1/2) * o* (N2/l)* I2

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