Question

The inner and outer radius of a toroid core are 28 cm and 29 cm respectively and around the core 3700 turns of a wire are wounded. If the current in the wire is 10 A, then the magnetic field inside the core of the toroid is :

• A. 2.60 x ${10}^{-2}$ T
• B. 2.60 x ${10}^{-3}$ T
• C. 4.52 x ${10}^{-2}$ T
• D. 4.52 x ${10}^{-3}$T

Answer 1

Answer: (A)

2.60 x ${10}^{-2}$ T

The number of turns per unit length for the given toroid $n=\frac{N}{2\pi r_{av}}$
The average radius of toroid
$r_{av}=\frac{28+29}{2}$ = 28.5 cm = 28.5 x ${10}^{-2}$m
$\therefore$ $n=\frac{3700}{2\times 3.14\times 28.5\times {10}^{-2}}$ = 2067.27 $\simeq 2067$
Now, $B={\mu }_{0}nI=4\pi$ x ${10}^{-7}$ x 2067 x 10
=259615.2 x ${10}^{-7}$T = 2.60 x ${10}^{-2}$ T