Question

A toroid made of CRGO has an inner diameter of 10 cm and an outer diameter of 14 cm. The thickness of the toroid is 2 cm. 200 turns of copper wire is wound on the core. \( \mu_0 = 4\pi \times 10^{-7} \, \text{H/m} \) and \( \mu_r \) of CRGO is 3000. When a current of 5 mA flows through the winding, the flux density in the core in millitesla is \(\underline{\hspace{2cm}}\).

Hint:

To calculate the flux density in a toroid, use the formula \( B = \mu_0 \mu_r \frac{N I}{l} \), where \( N \) is the number of turns, \( I \) is the current, and \( l \) is the length of the magnetic path.

Answer 1

Correct Answer: 10

The magnetic field strength \( H \) is given by: \[ H = \frac{N I}{l} \] where \( N = 200 \), \( I = 5 \, \text{mA} = 5 \times 10^{-3} \, \text{A} \), and \( l \) is the length of the magnetic path, which is the average circumference of the toroid. The average radius \( r = \frac{10 + 14}{2} = 12 \, \text{cm} = 0.12 \, \text{m} \), so: \[ l = 2\pi r = 2\pi \times 0.12 = 0.754 \, \text{m}. \] Thus, \[ H = \frac{200 \times 5 \times 10^{-3}}{0.754} \approx 1.32 \, \text{A/m}. \] The flux density \( B \) is given by: \[ B = \mu_0 \mu_r H = 4\pi \times 10^{-7} \times 3000 \times 1.32 \approx 1.66 \times 10^{-3} \, \text{T} = 1.66 \, \text{mT}. \] Thus, the flux density is approximately \( 1.66 \, \text{mT} \).