The toroid in the figure below consists of 474 turns and has a rectangular cross

Question

The toroid in the figure below consists of 474 turns and has a rectangular cross section. Its inner and outer radii are a and b, respectively, where a = 7.0 cm and b = 11.0 cm. The figure shows half of the toroid to allow us to see its cross-section. The height of the toroid is h = 3.00 cm.

(a) Suppose that a current of 4.00 A were flowing through the toriod. What would be the magnetic flux through one single turn of the toroid? Wb

(b) For the same current as in part (a), what would be the magnetic flux through the entire toroid? Wb

(c) Compute the inductance of the toroid. Give your answer in microhenries. uH

(d) Suppose that, instead of the current given in part (a), a current of 12.00 A were flowing through the toriod. In this case, what would be the inductance of the toroid? Give your answer in microhenries. uH

(e) For the amount of current given in part (d), what is the magnetic flux through the entire toriod? Wb

(f) For each of the assumed currents in this question, what would be the value of the magnetic energy stored in the toroid?

for the 4.00-A current: J

for the 12.00-A current: J

Explanation / Answer

a.) Using Ampere's law,

B.ds = u0*Ienc

B(2pir) = u0*N*I

So B = u0*N*I/(2pir)

The magnetic flux through one turn of the toroid may be obtained by integrating over the rectangular cross section with dA = hdr as the differential area element:

Flux through one turn = integral B.dA = integral from a to b (u0NI/2pir)*hdr = (u0NIh/2pi)ln(b/a)

So Flux through one turn = 2 x 10^-7*474*4*0.03ln(11/7) = 5.14 x 10^-6 Wb

b.)Flux through entire toroid = N*Flux through one turn = 474*5.14 x 10^-6 = 2.44 x 10^-3 Wb

c.)Inductance of toroid = Total flux/I = 2.44 x 10^-3/4 = 6.093 x 10^-4 H = 609.3 x 10^-6 H = 609.3 uH

d.)Inductance L = (u0*N^2*h/2pi)ln(b/a) which is independent of current(I), so L will be same

L = 609.3 uH

e.) Flux though entire toroid = (u0*N^2*I*h/2pi)ln(b/a) which is proportional to current(I)

So Flux though entire toroid when I = 12 A is = 3*2.44 x 10^-3 Wb = 7.32 x 10^-3 Wb

f.)ub = B^2/(2u0) = u0*N^2*I2/(8pi^2*r^2)

The total energy stored can be obtained by:

dV = (2pirhdr)

Ub = integral ub*dV = integral a to b [u0*N^2*I2/(8pi^2*r^2)]*(2pirhdr) = (u0*N^2*I^2*h/4pi)*ln(b/a)

For I = 4A, Ub = 10^-7*474^2*4^2*0.03ln(11/7) = 4.874 x 10^-3 J

For I = 12 A, Ub = 10^-7*474^2*12^2*0.03ln(11/7) = 0.04387 J

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