Dies ist eine alte Version des Dokuments!
Exercise E1 Cylindrical Coil
(written test, approx. 6 % of a 120-minute written test, SS2021)
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)
\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*}
b) the magnetic flux density (2 points)
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*}
c) the magnetic flux (2 points)
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*}