Question

A toroid having a square cross-section 5 cm on each side and inner radius of 15 cm has 500 turns and carries 0.8 A. what is the magnetic field inside the toroid at the inner radius.

Answer 1

(a) The magnetic field at distance r from the center of a toroid is given by:

$B=\frac{{\mu }_{0}IN}{2\pi r}$ . . . . (1)

where N is the number of the turns, the magnetic field at the innefr radius

of the toroid $r_{in}$ = 15.0 cm is:

$B=\frac{{\mu }_{0}IN}{2\pi r_{in}}$

$=\frac{(4\pi \times {10}^{-7}T.\frac{m}{A})\left(0.800A\right)\left(500\right)}{2\pi (0.150m}$

$=5.33\times {10}^{-4}T$

$B=5.33\times {10}^{-4}T$