Magnetic Field at the Toroids Problems and Solutions

 Problem#1

A toroid having a square cross section, 5.00 cm on a side, and an inner radius of 15.0 cm has 500 turns and carries a current of 0.800 A. (It is made up of a square solenoid) What is the magnetic field inside the toroid at (a) the inner radius and (b) the outer radius?

Answer:
Known:
inner radius, r = 15.0 cm
carries a current, i = 0.800 A.

(a) the magnetic field inside the toroid at the inner radius is

B = μ₀iN/2πr = (4π x 10^─7 T.m/A)(0.800 A)(500)/(2π x 0.15 m) = 5.33 x 10^─4 T

(b) the magnetic field inside the toroid at the outer radius. The outer radius is r = 20.0 cm. The field there is

B = μ₀iN/2πr = (4π x 10^─7 T.m/A)(0.800 A)(500)/(2π x 0.20m) = 4.00 x 10^─4 T

Problem#2
A toroidal solenoid has an inner radius of 12.0 cm and an outer radius of 15.0 cm. It carries a current of 1.50 A. How many equally spaced turns must it have so that it will produce a magnetic field of 3.75 mT at points within the coils 14.0 cm from its center?

Answer:
know
Known:
inner radius, r₁ = 12.0 cm
outer radius, r₂ = 15.0 cm
carries a current, I = 1.50 A

magnetic field, B = 3.75 mT = 3.75 x 10^─3 T at points within the coils r = 14.0 cm = 0.14 m from its center

Field can be calculated between inner and outer radius where it can be given by

B = μ₀NI/2πr
3.75 x 10^─3 T = (4π x 10^─7 T.m/A)N x 1.50 A/(2π x 0.14 m)
N = 1750 turns

Problem#3
A magnetic field of 37.2 T has been achieved at the MIT Francis Bitter National Magnetic Laboratory. Find the current needed to achieve such a field (a) 2.00 cm from a long, straight wire; (b) at the center of a circular coil of radius 42.0 cm that has 100 turns; (c) near the center of a solenoid with radius 2.40 cm, length 32.0 cm, and 40,000 turns.

Answer:
Known:
magnetic field, B = 37.2 T

(a) we use B = μ₀I/2πr
So,
I = 2πrB/μ₀ = 2π x  2.00 x 10^─2 m x 37.2 T/(4π x 10^─7 T.m/A) = 3.72 x 10⁶ A

(b) we use B = μ₀NI/2R
So,
I = 2RB/μ₀N = 2 x  42.00 x 10^─2 m x 37.2 T/(4π x 10^─7 T.m/A x 100) = 1.24 x 10⁵ A

(c) we use B = μ₀NI/L
So,
I = LB/μ₀N = 32.0 x 10^─2 m x 37.2 T/(4π x 10^─7 T.m/A x 40,000) = 237 A

Problem#4
A toroidal solenoid has inner radius r₁ = 15.0 cm and outer radius r₂ = 18.0 cm. The solenoid has 250 turns and carries a current of 8.50 A. What is the magnitude of the magnetic field at the following distances from the center of the torus: (a) 12.0 cm; (b) 16.0 cm; (c) 20.0 cm?

Fig.1

Answer:
Known:
inner radius r₁ = 15.0 cm
outer radius r₂ = 18.0 cm
number of turns, N = 250
carries a current, I = 8.50 A

Outside and inside torsional coil there is no currnt so according to amperes rule there is no magnetic field outside or inside torsional coil.

The torus extends from r₁ = 15.0 cm to r₂ = 18.0 cm.

(a) for r = 12.0 cm (Path 1), which is outside the torus, so B = 0

(b) for r = 16.0 cm = 0.16 m (Path 2), we use

B = μ₀IN/2πr = (4π x 10^─7 T.m/A)(8.50 A)(250)/(2π x 0.16 m)
B = 2.66 x 10^─3 T

(c) for r = 20.0 cm (Path 3),  which is outside the torus, so B = 0