{{tag>magnetostatic flux_density magnetic_field_strength coil flux exam_ee2_SS2021}}{{include_n>1190}}
 
#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Cylindrical Coil\\
<fs medium>(written test, approx. 6 % of a 120-minute written test, SS2021)</fs> #@TaskText_HTML@#

A cylindrical coil with the following information is given: \\
  * Length $𝑙 = 30 {~\rm cm}$, 
  * Winding diameter $𝑑 = 390 {~\rm mm}$, 
  * Number of windings $𝑤 = 240$ ,
  * Current through the conductor $𝐼 = 500 {~\rm mA}$, 
  * Material inside: Air 
  * $\mu_0 = 4\pi\cdot 10^{-7} {{\rm Vs}\over{\rm Am}}$
The proportion of the magnetic voltage outside the coil can be neglected. 
Determine the following for the inside of the coil:  

a) the magnetic field strength (2 points) 

#@HiddenBegin_HTML~0J7AcCfimmeMYtq9_11,Path~@#

\begin{align*}
 H &= {{N \cdot I}\over{l}} = {{w \cdot I}\over{l}}  
\end{align*}

Putting in the numbers: 
\begin{align*}
 H &= {{240 \cdot 0.5 {~\rm A}}\over{0.3 {~\rm m}}} 
\end{align*}
#@HiddenEnd_HTML~0J7AcCfimmeMYtq9_11,Path ~@#

#@HiddenBegin_HTML~0J7AcCfimmeMYtq9_12,Result~@#
$H = 400 ~\rm A/m$
#@HiddenEnd_HTML~0J7AcCfimmeMYtq9_12,Result~@#

b) the magnetic flux density (2 points) 

#@HiddenBegin_HTML~0J7AcCfimmeMYtq9_21,Path~@#

The magnetic field strength is $B = \mu_0 \mu_{\rm r} \cdot H$: 

\begin{align*}
 B =  \mu_0 \mu_{\rm r} H 
\end{align*}

Putting in the numbers: 
\begin{align*}
 B &= 4\pi\cdot 10^{-7} {{\rm Vs}\over{\rm Am}} \cdot  400 ~\rm {{A}\over{m}} \\
   &= 0.0005026... {{\rm Vs}\over{\rm m^2}}
\end{align*}
#@HiddenEnd_HTML~0J7AcCfimmeMYtq9_21,Path ~@#

#@HiddenBegin_HTML~0J7AcCfimmeMYtq9_22,Result~@#
$B = 0.50 ~\rm mT$
#@HiddenEnd_HTML~0J7AcCfimmeMYtq9_22,Result~@#

c) the magnetic flux (2 points) 

#@HiddenBegin_HTML~0J7AcCfimmeMYtq9_31,Path~@#

The magnetic flux is given as:

\begin{align*}
 \Phi &= B \cdot A
\end{align*}

Since the coil is cylindrical, the cross-sectional area is given as

\begin{align*}
 A = \pi r^2 = \pi \left( {{d}\over{2}} \right)^2
\end{align*}

Therefore:
\begin{align*}
 \Phi &= B \cdot \pi \left( {{d}\over{2}} \right)^2
\end{align*}

Putting in the numbers: 
\begin{align*}
 \Phi &= 0.0005026... {{\rm Vs}\over{\rm m^2}} \cdot \pi \left( {{0.39{\rm m}}\over{2}} \right)^2 \\
      &= 0.00006004... {\rm Vs}
\end{align*}
#@HiddenEnd_HTML~0J7AcCfimmeMYtq9_31,Path ~@#

#@HiddenBegin_HTML~0J7AcCfimmeMYtq9_32,Result~@#
$\Phi = 60 ~\rm \mu Wb$
#@HiddenEnd_HTML~0J7AcCfimmeMYtq9_32,Result~@#

#@TaskEnd_HTML@#