Question

A toroidal solenoid has an inner radius of $12\text{cm}$ and an outer radius of $15\text{cm}$. It carries a current of $1.50\text{A}$. It must have turns (equally spaced), so that it will produce a magnetic field of $3.75\text{mT}$ at radial distance $14\text{cm}$ from its centre.

Answer 1

Given:
Inner radius, $r_1=12.0\text{cm}$
Outer radius, $r_2=15.0\text{cm}$
Current, $I=1.50\text{A}$
Magnetic field, $B=3.75\text{mT}$
$r=14.0\text{cm}=0.14\text{m}$
Total number of turns, $N=?$

As $r_1<r<r_2$, so the magnetic field inside a toroid is given by,

$B=\frac{{\mu }_{0}NI}{2\pi r}$

Putting all given data in the equation,

$3.75\times {10}^{-3}=\frac{4\pi \times {10}^{-7}\times N\times 1.5}{2\pi \times 0.14}$

$\Rightarrow N=\frac{3.75\times {10}^{-3}\times 0.14}{2\times 1.5\times {10}^{-7}}$

$\therefore N=1750\text{turns}$