A toroid having a square cross section, 5.27 cm on a side, and an inner radius o

Question

A toroid having a square cross section, 5.27 cm on a side, and an inner radius of 17.9 cm has 730 turns and carries a current of 1.02 A. (It is made up of a square solenoid instead of round one bent into a doughnut shape.) What is the magnitude of the magnetic field inside the toroid at (a) the inner radius and (b) the outer radius?

Chapter 29, Problem 049 A toroid having a square cross section, 5.27 cm on a side, and an inner radius of 17.9 cm has 730 turns and carries a current of 1.02 A. (It is made up of a square solenoid instead of round one bent the outer radius? into a doughnut shape.) What is the magnitude of the magnetic field inside the toroid at (a) the inner radius and (b) the outer radius? (a) Number (b) Number Units Units

Explanation / Answer

Applying Ampere's law for this toroid

B = uo*N*I/2*pi*r

inner radius = 0.179m

outer radius = 0.179 + 0.527 = 0.706 m

a)

Bi = (4pi x 10^-7 x 730 x 1.02) / (2pi x 0.179 )

Bi = 0.00083 T

b)

Bo = (4pi x 10^-7 x 730 x 1.02) / (2pi x  0.706)

B0 = 0.00021 T

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