Consider a toroid having a barrel diameter which is much less than the inner rad
ID: 2123900 • 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 7.10 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.
1 H
(b) If such a toroid carries a current of 3.30 A, what is the energy stored in the magnetic field of the toroid?
2 J
(c) What is the volume of the barrel of this toroid?
3 m3
(d) If such a toroid carries a current of 3.30 A, what is the magnetic energy density in the barrel of the toroid?
4 J/m3
Explanation / Answer
a)Inductance = mew*N^2 *A/(2*pi*r) = 4*pi*10^-7 *640^2 *0.031*10^-4)/(2*pi*0.071) = 3.576*10^-6 H
b)Energy = 0.5*L*I^2 = 0.5* 3.576*10^-6 *3.3^2 = 1.947*10^-5 J
c)
pi*r^2 = 0.031 *10^-4
r = 9.936*10^-4
R = 0.071
volume = 2*pi^2 *R*r^2 = 2*pi^2*0.071 *(9.936*10^-4 )^2 = 1.3822*10^-6 m^3
d)magnetic energy density = energy/volume = (1.947*10^-5 J)/(1.3822*10^-6 m^3) = 14.085 J/m^3