91Ó°ÊÓ

Inner radius is ${\mathrm{R}}_{\mathrm{i}}=5.0\text{ }\mathrm{cm}$

Outer radius is${\mathrm{R}}_{\mathrm{o}}=6.0\text{ }\mathrm{cm}$

Primary windings are${\mathrm{N}}_1=400$

Toroidal field, ${\mathrm{B}}_0=0.20\mathrm{mT}$

Secondary windings are ${\mathrm{N}}_2=50$

Resistance is $\mathrm{R}=8.0\text{ }\mathrm{Î\copyright }$

We use the concept of magnetic field of toroid. We findthenumber of turns per unit length, and then we find the current. For charge, we usetherelation between charge, number of turns, magnetic flux, and resistance.

Toroidal field is given as-

${\mathrm{B}}_0={\mathrm{μ}}_0{\mathrm{ni}}_{\mathrm{p}}$

Current is given as-

$\mathrm{i}=\frac{\mathrm{dq}}{\mathrm{dt}}$

Number of turns per unit length-

$\mathrm{n}=\frac{\mathrm{N}}{\mathrm{l}}$

Using the equation,

${\mathrm{B}}_0={\mathrm{μ}}_0{\mathrm{ni}}_{\mathrm{p}}$

We can calculate${\mathrm{i}}_{\mathrm{p}}$, as-

${\mathrm{i}}_{\mathrm{p}}=\frac{{\mathrm{B}}_0}{{\mathrm{μ}}_0\mathrm{n}}\text{ â¶Ä‰}â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(1\right)$

Here, $\mathrm{n}$ is given as-

$\mathrm{n}=\frac{{\mathrm{N}}_1}{\mathrm{l}}$

Length of the circular path is $=2{\mathrm{Ï€°ù}}_{\mathrm{avg}}$ .

Here we have to take the average radius.

$\begin{array}{c}{\mathrm{r}}_{\mathrm{avg}}=\frac{({\mathrm{R}}_{\mathrm{i}}+{\mathrm{R}}_0)}{2}\\ =\frac{(5.0\text{ c³¾}+6.0\text{ c³¾})}{2}\\ =\frac{11}{2}\text{ c³¾}\\ =5.5\mathrm{cm}\end{array}$

We can convert it in m:

$\begin{array}{c}{\mathrm{r}}_{\mathrm{avg}}=5.5\mathrm{cm}×\left(\frac{1\mathrm{m}}{100\text{ }\mathrm{cm}}\right)\\ =5.5×{10}^{-2}\mathrm{m}\end{array}$

We get the length as

$\begin{array}{c}\mathrm{l}=2\mathrm{Ï€}(5.5×{10}^{−2}\text{″¾})\text{ }\\ =0.34\text{ }\mathrm{m}\end{array}$

Substituting it in equation of n, we get

$\begin{array}{c}\mathrm{n}=\frac{400}{0.34\text{″¾}}\\ =1.16×{10}^3\mathrm{turns}/\mathrm{m}\end{array}$

Using the given values in equation (1),

$\begin{array}{c}{\mathrm{i}}_{\mathrm{p}}=\frac{0.20×{10}^{−3}\text{â€Í¿}}{(4\mathrm{Ï€}×{10}^{−7}\text{ H}/\text{m})(1.16×{10}^3\text{â€Í¿urns}/\text{m})}\\ =\frac{0.20×{10}^{−3}}{1.46×{10}^{−3}}\text{‼î}\\ =0.14\mathrm{A}\end{array}$

Current in the windings to attain a toroidal field of magnitude ${\mathrm{B}}_0=0.20\text{ }\mathrm{mT}$ is ${\mathrm{i}}_{\mathrm{p}}=0.14\mathrm{A}$

Emf induced in secondary coil is

$\mathrm{Î\mu }={\mathrm{N}}_2\frac{\mathrm{»åÏ•}}{\mathrm{dt}}$

Using Ohm’s law we can write the above equation as-

${\mathrm{i}}_{\mathrm{s}}\mathrm{R}={\mathrm{N}}_2\frac{\mathrm{»åÏ•}}{\mathrm{dt}}$

${\mathrm{i}}_{\mathrm{s}}=\frac{\left({\mathrm{N}}_2\frac{\mathrm{»åÏ•}}{\mathrm{dt}}\right)}{\mathrm{R}}$

We know the magnetic field inside the coil is greater, so we can write

$\begin{array}{c}\mathrm{B}={\mathrm{B}}_0+800{\mathrm{B}}_0\\ =801{\mathrm{B}}_0\end{array}$

Now we use the current and charge relation:

$\mathrm{i}=\frac{\mathrm{dq}}{\mathrm{dt}}$

We can rearrange it for dq as

$\mathrm{dq}=\mathrm{idt}$

Using the current in secondary, we can integrate it.

$∫\mathrm{dq}=∫{\mathrm{i}}_{\mathrm{s}}\mathrm{dt}$

Plugging the value of current, we get

$\begin{array}{c}\mathrm{q}=∫\frac{\left({\mathrm{N}}_2\frac{\mathrm{»åÏ•}}{\mathrm{dt}}\right)}{\mathrm{R}}\mathrm{dt}\\ =\frac{{\mathrm{N}}_2\mathrm{Ï•}}{\mathrm{R}}\end{array}$

We know magnetic flux through the ring is

$\begin{array}{c}\mathrm{Ï•}=∫\stackrel{→}{\mathrm{B}}.\stackrel{→}{\mathrm{ds}}\\ ={\mathrm{BÏ€°ù}}^2\end{array}$

Using this in the above equation of charge, we get

$\mathrm{q}=\frac{{\mathrm{N}}_2{\mathrm{BÏ€°ù}}^2}{\mathrm{R}}$

Here the radius of the ring is

$\begin{array}{c}\mathrm{r}=\frac{0.06\text{″¾}−0.05\text{″¾}}{2}\\ =\frac{0.01\text{″¾}}{2}\\ =0.5×{10}^{-2}\mathrm{m}\end{array}$

Substituting the values, we get

$\begin{array}{c}\mathrm{q}=\frac{50\left(801{\mathrm{B}}_0\right){\mathrm{Ï€°ù}}^2}{\mathrm{R}}\\ =\frac{50\left(801(0.20×{10}^{−3}\text{â€Í¿}\right)3.14{(0.5×{10}^{−2}\text{″¾})}^2}{8.0\text{ }\mathrm{Î\copyright }}\\ =\frac{6.2878×{10}^{−4}}{8.0}\text{ C}\\ =7.9×{10}^{-5}\mathrm{C}\end{array}$

The charge that moves through the secondary coil when the current in the toroid windings is turned onis $\mathrm{q}=7.9×{10}^{-5}\mathrm{C}$.