Magnetic Field at the Toroids Problems and Solutions

 Problem#1

A wooden ring whose mean diameter is 14.0 cm is wound with a closely spaced toroidal winding of 600 turns. Compute the magnitude of the magnetic field at the center of the cross section of the windings when the current in the windings is 0.650 A.

Answer:
Known:
Diameter, d = 14.0 cm, radius r = 7.0 cm = 0.07 m
number of turns, N = 600
carries a current, I = 0.650 A

inside a toroidal solenoid

B = μ₀IN/2πr

If the radial thickness of the torus is small compared to its mean diameter, B is approximately
uniform inside its windings.

B = (4π x 10^─7 T.m/A)(0.650 A)(650)/(2π x 0.070 m) = 1.11 x 10^─3 T

Problem#2
A toroidal solenoid with 400 turns of wire and a mean radius of 6.0 cm carries a current of 0.25 A. The relative permeability of the core is 80. (a) What is the magnetic field in the core? (b) What part of the magnetic field is due to atomic currents?

Answer:
Known:
mean radius, r = 6.0 cm = 0.060 m
number of turns, N = 400
carries a current, I = 0.25 A
relative permeability of the core, K_m = 80

(a) the magnetic field in the core is

B = K_mμ₀NI/2πr
B = 80 x (4π x 10^─7 T.m/A)(0.25 A)(400)/(2π x 0.060 m) = 2.67 x 10^─2 T

(b) The magnetic susceptibility is χ_m= K_m – 1 = 79, so the magnetic field due to atomic currents is

B’ = (79/80)B = (79/80)(2.67 x 10^─2 T) = 2.63 x 10^─2 T

Problem#3
A toroidal solenoid with 500 turns is wound on a ring with a mean radius of 2.90 cm. Find the current in the winding that is required to set up a magnetic field of 0.350 T in the ring (a) if the ring is made of annealed iron (K_m = 1400) and (b) if the ring is made of silicon steel (K_m = 1500).

Answer:
Known:
mean radius, r = 2.90 cm = 0.029 m
number of turns, N = 500
magnetic field, B = 0.350 T

(a) if the ring is made of annealed iron (K_m = 1400)

the magnetic field in the iron is

B = K_mμ₀NI/2πr
So,
0.350 T = 1400 x (4π x 10^─7 T.m/A)(I)(500)/(2π x 0.029 m)

I = 0.0725 A = 72.5 mA

(b) if the ring is made of silicon steel (K_m = 1500)
0.350 T = 1500 x (4π x 10^─7 T.m/A)(I)(500)/(2π x 0.029 m)
 I = 0.0677 A = 6.77 mA

Problem#4
The current in the windings of a toroidal solenoid is 2.400 A. There are 500 turns, and the mean radius is 25.00 cm. The toroidal solenoid is filled with a magnetic material. The magnetic field inside the windings is found to be 1.940 T. Calculate (a) the relative permeability and (b) the magnetic susceptibility of the material that fills the toroid.

(a) The magnetic field inside a tightly wound toroidal solenoid is

B = K_m μ₀nI = K_mμ₀NI/2πr

Where n is the number of turns per unit length and N is the total number of turns.
Solving the last equation for Km, we get

K_m = 2πrB/μ₀NI
= 2π x 0.25 x 1.94/(4π ×10^-7) 500 x 2.4 = 2021

(b) The magnetic susceptibility is χ_m= K_m – 1 = 2020 

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