Consider a toroid having a barrel diameter which is much less than the inner rad
ID: 2123841 • Letter: C
Question
Consider a toroid having a barrel diameter which is much less than the inner radius of the toroid; in such as case, it is reasonable to consider the magnetic field strength in the barrel of the toroid to be constant.
(a) If such a toroid has a mean radius (the average of the inner and outer radii) of 7.50 cm, carries 640 turns, and if the cross-sectional area of the barrel of the toroid is 0.0310 square cm, then find the inductance of the toroid. HINT: Invent a current, calculate the field strength at the mean radius, and assume this field strength is constant for your flux calculation.
H
(b) If such a toroid carries a current of 2.90 A, what is the energy stored in the magnetic field of the toroid?
J
(c) What is the volume of the barrel of this toroid?
m3
(d) If such a toroid carries a current of 2.90 A, what is the magnetic energy density in the barrel of the toroid?
J/m3
Explanation / Answer
a)the self inductance of the toroid is
L = (u_o x n1 x A/L)
where u_o = 4pi x 10^-7 Tm/A,n1 = 640 turns,A = 0.0310 cm^2 = 0.0310 x 10^-4 m^2 and L = 7.50 cm = 7.50 x 10^-2 m
b)the energy stored is
U = (1/2)L x I^2
where I = 2.90 A
c)the volume of the barrel is
V = A x L
d)the magnetic energy density is
J = (U/V)