Q53P - Fundamentals Of Physics

Question

A Rowland ring is formed of ferromagnetic material. It is circular in cross section, with an inner radius of $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{cm}$  and an outer radius of  $\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{cm}$, and is wound with $\mathbf{400}$  turns of wire. (a) What current must be set up in the windings to attain a toroidal field of magnitude  ${\mathbf{B}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{ }\mathbf{mT}$ ?  (b) A secondary coil wound around the toroid has  $\mathbf{50}\mathbf{ }\mathbf{Turns}$ and resistance $\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\Omega }$ . If, for this value of ${\mathbf{B}}_{\mathbf{0}}$ , we have ${\mathbf{B}}_{\mathbf{M}}\mathbf{=}\mathbf{800}\mathbf{ }{\mathbf{B}}_{\mathbf{0}}$  , how much charge moves through the secondary coil when the current in the toroid windings is turned on?

Step-by-Step Solution

Answer

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

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

Step 1: Listing the given quantities

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

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

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  $\mathrm{R}=8.0\mathrm{\Omega }$

Step 2: Understanding the concepts of toroid

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

Toroidal field is given as-

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

Current is given as-

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

Number of turns per unit length-

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

Step 3: (a) Calculations of the current in the windings to attain a toroidal field of magnitude B 0 = 0 . 20 mT

Using the equation,

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

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

 ${\mathrm{i}}_{\mathrm{p}}=\frac{{\mathrm{B}}_0}{{\mathrm{\mu }}_0\mathrm{n}}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Here, $\mathrm{n}$  is given as-

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

Length of the circular path is $=2{\mathrm{\pi r}}_{\mathrm{avg}}$  .

 Here we have to take the average radius.

 $\begin{array}{c}{\mathrm{r}}_{\mathrm{avg}}=\frac{({\mathrm{R}}_{\mathrm{i}}+{\mathrm{R}}_0)}{2} \\ =\frac{(5.0\text{\hspace{0.17em}cm}+6.0\text{\hspace{0.17em}cm})}{2}\\ =\frac{11}{2}\text{\hspace{0.17em}cm}\\ =5.5 \mathrm{cm}\end{array}$

We can convert it in m:

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

We get the length as

  $\begin{array}{c}\mathrm{l}=2\mathrm{\pi }(5.5\times {10}^{-2}\text{\hspace{0.17em}m})\\ =0.34\mathrm{m}\end{array}$

Substituting it in equation of n, we get

$\begin{array}{c}\mathrm{n}=\frac{400}{0.34\text{\hspace{0.17em}m}}\\ =1.16\times {10}^3 \mathrm{turns}/\mathrm{m}\end{array}$

Using the given values in equation (1),

 $\begin{array}{c}{\mathrm{i}}_{\mathrm{p}}=\frac{0.20\times {10}^{-3}\text{\hspace{0.17em}T}}{(4\mathrm{\pi }\times {10}^{-7}\text{\hspace{0.17em}H}/\text{m})(1.16\times {10}^3\text{\hspace{0.17em}Turns}/\text{m})}\\ =\frac{0.20\times {10}^{-3}}{1.46\times {10}^{-3}}\text{\hspace{0.17em}A}\\ =0.14 \mathrm{A}\end{array}$

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

Step 4: (b) Calculations of the charge that moves through the secondary coil when the current in the toroid windings is turned on

Emf induced in secondary coil is

 $\mathrm{\epsilon }={\mathrm{N}}_2\frac{\mathrm{d\varphi }}{\mathrm{dt}}$

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

 ${\mathrm{i}}_{\mathrm{s}}\mathrm{R}={\mathrm{N}}_2\frac{\mathrm{d\varphi }}{\mathrm{dt}}$

${\mathrm{i}}_{\mathrm{s}}=\frac{\left({\mathrm{N}}_2\frac{\mathrm{d\varphi }}{\mathrm{dt}}\right)}{\mathrm{R}}$

 

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.

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

Plugging the value of current, we get

 $\begin{array}{c}\mathrm{q}=\int \frac{\left({\mathrm{N}}_2\frac{\mathrm{d\varphi }}{\mathrm{dt}}\right)}{\mathrm{R}}\mathrm{dt} \\ =\frac{{\mathrm{N}}_2\mathrm{\varphi }}{\mathrm{R}}\end{array}$

We know magnetic flux through the ring is

$\begin{array}{c}\mathrm{\varphi }=\int \overrightarrow{\mathrm{B}}.\overrightarrow{\mathrm{ds}}\\ ={\mathrm{B\pi r}}^2\end{array}$

Using this in the above equation of charge, we get

 $\mathrm{q}=\frac{{\mathrm{N}}_2{\mathrm{B\pi r}}^2}{\mathrm{R}}$

Here the radius of the ring is

 $\begin{array}{c}\mathrm{r}=\frac{0.06\text{\hspace{0.17em}m}-0.05\text{\hspace{0.17em}m}}{2}\\ =\frac{0.01\text{\hspace{0.17em}m}}{2}\\ =0.5\times {10}^{-2} \mathrm{m}\end{array}$

Substituting the values, we get

$\begin{array}{c}\mathrm{q}=\frac{50\left(801{\mathrm{B}}_0\right){\mathrm{\pi r}}^2}{\mathrm{R}}\\ =\frac{50\left(801(0.20\times {10}^{-3}\text{\hspace{0.17em}T}\right)3.14{(0.5\times {10}^{-2}\text{\hspace{0.17em}m})}^2}{8.0\mathrm{\Omega }}\\ =\frac{6.2878\times {10}^{-4}}{8.0}\text{\hspace{0.17em}C}\\ =7.9\times {10}^{-5} \mathrm{C}\end{array}$

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